Program Implementation - AP Computer Science A
Card 0 of 693
Class Person {
int height;
float weight;
``
public:
int getHeight();
void setHeight(int);
float getWeight();
void setWeight(float);
};
What is the access level of height and weight?
Class Person {
int height;
float weight;
``
public:
int getHeight();
void setHeight(int);
float getWeight();
void setWeight(float);
};
What is the access level of height and weight?
Until an access specifier is given, all class members are implicitly private. All members defined after an access specifier is used will will have that access level until another access specifier is explicitly invoked. Since no access specifier was used, weight and height are automatically private.
Note that virtual is not an access keyword.
Until an access specifier is given, all class members are implicitly private. All members defined after an access specifier is used will will have that access level until another access specifier is explicitly invoked. Since no access specifier was used, weight and height are automatically private.
Note that virtual is not an access keyword.
Compare your answer with the correct one above
Consider the following code:
public static class Clock {
private int seconds;
``
public Clock(int s) {
seconds = s;
}
``
public void setTime(int s) {
seconds = s;
}
``
public void setSeconds(int s) {
int hoursMinutes = seconds - seconds % 60;
seconds = hoursMinutes + s;
}
``
public void setMinutes(int min) {
int hours = seconds / 3600;
int currentSeconds = seconds % 60;
seconds = hours + min * 60 + currentSeconds;
}
}
Which of the following represents a method that returns the minute value of the clock?
Consider the following code:
public static class Clock {
private int seconds;
``
public Clock(int s) {
seconds = s;
}
``
public void setTime(int s) {
seconds = s;
}
``
public void setSeconds(int s) {
int hoursMinutes = seconds - seconds % 60;
seconds = hoursMinutes + s;
}
``
public void setMinutes(int min) {
int hours = seconds / 3600;
int currentSeconds = seconds % 60;
seconds = hours + min * 60 + currentSeconds;
}
}
Which of the following represents a method that returns the minute value of the clock?
This clock class is defined as having only seconds in its fields, so you have to convert this value for any of the accessors and mutators. Therefore, you need to perform careful (though simple) mathematics for this conversion. To calculate the number of minutes in a given number of seconds, you first need to remove the hours from total count. First, compute the hours using integer division (which will drop the fractional portion of the decimal). This is:
int hours = seconds / 3600;
Next, you need to figure out how many seconds are in that value of hours. This is helpful precisely because integer division drops the decimal portion. Thus, you will subtract off:
hours * 3600
from the total seconds.
This gives you the seconds that apply to the minutes and seconds of the time. Therefore, you will finally need to divide by 60 to isolate the minutes.
This clock class is defined as having only seconds in its fields, so you have to convert this value for any of the accessors and mutators. Therefore, you need to perform careful (though simple) mathematics for this conversion. To calculate the number of minutes in a given number of seconds, you first need to remove the hours from total count. First, compute the hours using integer division (which will drop the fractional portion of the decimal). This is:
int hours = seconds / 3600;
Next, you need to figure out how many seconds are in that value of hours. This is helpful precisely because integer division drops the decimal portion. Thus, you will subtract off:
hours * 3600
from the total seconds.
This gives you the seconds that apply to the minutes and seconds of the time. Therefore, you will finally need to divide by 60 to isolate the minutes.
Compare your answer with the correct one above
Consider the following code :
public class Clock {
private int seconds;
``
public Clock(int s) {
seconds = s;
}
``
public void setTime(int s) {
seconds = s;
}
``
public void setSeconds(int s) {
int hoursMinutes = seconds - seconds % 60;
seconds = hoursMinutes + s;
}
``
public void setMinutes(int min) {
int hours = seconds / 3600;
int currentSeconds = seconds % 60;
seconds = hours + min * 60 + currentSeconds;
}
}
Which of the following defines a toString method that will output the time in 24-hour format in the following form:
1:51:03
(Notice that you need to pad the minutes and seconds. You can call 12 midnight "0".)
Consider the following code :
public class Clock {
private int seconds;
``
public Clock(int s) {
seconds = s;
}
``
public void setTime(int s) {
seconds = s;
}
``
public void setSeconds(int s) {
int hoursMinutes = seconds - seconds % 60;
seconds = hoursMinutes + s;
}
``
public void setMinutes(int min) {
int hours = seconds / 3600;
int currentSeconds = seconds % 60;
seconds = hours + min * 60 + currentSeconds;
}
}
Which of the following defines a toString method that will output the time in 24-hour format in the following form:
1:51:03
(Notice that you need to pad the minutes and seconds. You can call 12 midnight "0".)
There are several points to be noted in this code. First, you need to calculate each of the constituent parts from the seconds stored in the class. The "display value" of seconds is relatively easy. This is just the remainder of a division by 60. Consider:
61 seconds => This is really 1 minute and 1 second.
155 seconds => This is really 2 minutes and 35 seconds.
So, you know that the display seconds are:
seconds % 60
You must compute the hours using integer division (which will drop the fractional portion of the decimal). This is:
int hours = seconds / 3600;
Next, you need to figure out how many seconds are in that value of hours. This is helpful precisely because integer division drops the decimal portion. Thus, you will subtract off:
hours * 3600
from the total seconds.
This gives you the seconds that apply to the minutes and seconds of the time. Therefore, you will finally need to divide by 60 to isolate the minutes.
Then, you must pad the values. This is not done using +=, which would add the "0" character to the end of the string. Instead, it is done by the form:
secString = "0" + secString;
There are several points to be noted in this code. First, you need to calculate each of the constituent parts from the seconds stored in the class. The "display value" of seconds is relatively easy. This is just the remainder of a division by 60. Consider:
61 seconds => This is really 1 minute and 1 second.
155 seconds => This is really 2 minutes and 35 seconds.
So, you know that the display seconds are:
seconds % 60
You must compute the hours using integer division (which will drop the fractional portion of the decimal). This is:
int hours = seconds / 3600;
Next, you need to figure out how many seconds are in that value of hours. This is helpful precisely because integer division drops the decimal portion. Thus, you will subtract off:
hours * 3600
from the total seconds.
This gives you the seconds that apply to the minutes and seconds of the time. Therefore, you will finally need to divide by 60 to isolate the minutes.
Then, you must pad the values. This is not done using +=, which would add the "0" character to the end of the string. Instead, it is done by the form:
secString = "0" + secString;
Compare your answer with the correct one above
Consider the following C++ pseudocode:
Class Car {
const int wheels = 4;
float milesPerGallon;
string make;
string model;
}
Car sportscar = new Car;
What is the difference between the class car, and the object sportscar?
Consider the following C++ pseudocode:
Class Car {
const int wheels = 4;
float milesPerGallon;
string make;
string model;
}
Car sportscar = new Car;
What is the difference between the class car, and the object sportscar?
When programming within the Object-Oriented paradigm, think of the class as a blueprint, and the object as a house built from that blueprint.
The class Car is a abstract specification that does not refer to any one particular instance. It is merely a protocol that all objects wishing to be cars should follow. The sportscar object is a realization of the car blueprint. It is a specific instance. In programming jargon, "sportscar is instantiated from Car."
When programming within the Object-Oriented paradigm, think of the class as a blueprint, and the object as a house built from that blueprint.
The class Car is a abstract specification that does not refer to any one particular instance. It is merely a protocol that all objects wishing to be cars should follow. The sportscar object is a realization of the car blueprint. It is a specific instance. In programming jargon, "sportscar is instantiated from Car."
Compare your answer with the correct one above
JAVA EXPLICIT TYPE CASTING
Consider the following JAVA code:
**public static void main(String[] args)**
**{**
**double number1 = 99.05;**
**int number2 = (int) number1;**
**System.out.println("number1 = " + number1);**
****System.out.println("number2 = " + number2);**
**}**
What would be the console output?
JAVA EXPLICIT TYPE CASTING
Consider the following JAVA code:
**public static void main(String[] args)**
**{**
**double number1 = 99.05;**
**int number2 = (int) number1;**
**System.out.println("number1 = " + number1);**
****System.out.println("number2 = " + number2);**
**}**
What would be the console output?
Type casting deals with assigning the value of a variable to another variable that is of a different type. In this case we have two variables one that is a double, and another that is an integer. number1 is a double that has a value of 99.05. However, when number2 is assigned the value of number1, there is some explicit type casting going on. When an integer is assigned the value of a double, it drops off the decimal places. This means that number2 has a value of 99.
Type casting deals with assigning the value of a variable to another variable that is of a different type. In this case we have two variables one that is a double, and another that is an integer. number1 is a double that has a value of 99.05. However, when number2 is assigned the value of number1, there is some explicit type casting going on. When an integer is assigned the value of a double, it drops off the decimal places. This means that number2 has a value of 99.
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Given a vector of ints called "intVec", write a "ranged for" loop, also sometimes known as a "for each" loop, to double the values fof all elements in a vector. (In C++)
Given a vector of ints called "intVec", write a "ranged for" loop, also sometimes known as a "for each" loop, to double the values fof all elements in a vector. (In C++)
A "for each" loop was introduced in C++ 11 which allows the user to loops through a container of items without writing a traditional loop. Let's take a look at all the choices.
for( int i : intVec(){
i * = 2;
}
Although this looks like the correct answer, it's not. The int i in the first part of the for loop is passed in "by value", which means that a copy of intVec will be made and the integers inside the copy will be doubled. The integers inside the original intVec vector will be unaffected. To fix this problem, int i needs to be passed into the for loop "by reference", which is done by adding a "&" symbol after "int".
The correct loop is:
for( int& i : intVec(){
i * = 2;
}
Let's take a look at the other choices:
for( int& i : intVec(){
i = 2;
}
This one will not work because it is setting all items in intVec to 2, not doubling them.
All the other for loops are not "for each" loops so they are incorrect even if they accomplish the same output.
A "for each" loop was introduced in C++ 11 which allows the user to loops through a container of items without writing a traditional loop. Let's take a look at all the choices.
for( int i : intVec(){
i * = 2;
}
Although this looks like the correct answer, it's not. The int i in the first part of the for loop is passed in "by value", which means that a copy of intVec will be made and the integers inside the copy will be doubled. The integers inside the original intVec vector will be unaffected. To fix this problem, int i needs to be passed into the for loop "by reference", which is done by adding a "&" symbol after "int".
The correct loop is:
for( int& i : intVec(){
i * = 2;
}
Let's take a look at the other choices:
for( int& i : intVec(){
i = 2;
}
This one will not work because it is setting all items in intVec to 2, not doubling them.
All the other for loops are not "for each" loops so they are incorrect even if they accomplish the same output.
Compare your answer with the correct one above
Consider the following code:
int\[\] vals = {6,1,41,5,1};
int\[\]\[\] newVals = new int\[vals.length\]\[\];
for(int i = 0; i < vals.length; i++) {
newVals\[i\] = new int\[vals\[i\]\];
for(int j = 0; j < vals\[i\];j++) {
newVals\[i\]\[j\] = vals\[i\] * (j+1);
}
}
What does the code above do?
Consider the following code:
int\[\] vals = {6,1,41,5,1};
int\[\]\[\] newVals = new int\[vals.length\]\[\];
for(int i = 0; i < vals.length; i++) {
newVals\[i\] = new int\[vals\[i\]\];
for(int j = 0; j < vals\[i\];j++) {
newVals\[i\]\[j\] = vals\[i\] * (j+1);
}
}
What does the code above do?
Let's look at the main loop in this program:
for(int i = 0; i < vals.length; i++) {
newVals\[i\] = new int\[vals\[i\]\];
for(int j = 0; j < vals\[i\];j++) {
newVals\[i\]\[j\] = vals\[i\] * (j+1);
}
}
The first loop clearly runs through the vals array for the number of items in that array. After this, it creates in the newVals 2D array a second dimension that is as long as the value in the input array vals. Thus, for 41, it creates a 3rd row in newVals that is 41 columns wide.
Then, we go to the second loop. This one iterates from 0 to the value stored in the current location in vals. It places in the given 2D array location the multiple of the value. Think of j + 1. This will go from 1 to 41 for the case of 41 (and likewise for all the values). Thus, you create a 2D array with all the multiples of the initial vals array.
Let's look at the main loop in this program:
for(int i = 0; i < vals.length; i++) {
newVals\[i\] = new int\[vals\[i\]\];
for(int j = 0; j < vals\[i\];j++) {
newVals\[i\]\[j\] = vals\[i\] * (j+1);
}
}
The first loop clearly runs through the vals array for the number of items in that array. After this, it creates in the newVals 2D array a second dimension that is as long as the value in the input array vals. Thus, for 41, it creates a 3rd row in newVals that is 41 columns wide.
Then, we go to the second loop. This one iterates from 0 to the value stored in the current location in vals. It places in the given 2D array location the multiple of the value. Think of j + 1. This will go from 1 to 41 for the case of 41 (and likewise for all the values). Thus, you create a 2D array with all the multiples of the initial vals array.
Compare your answer with the correct one above
Which of the following blocks of code will output an evenly-spaced 2D array (i.e. an evenly-spaced matrix)? For example, a simple evenly-spaced matrix is:
1 2 3
4 5 6
7 8 9
Which of the following blocks of code will output an evenly-spaced 2D array (i.e. an evenly-spaced matrix)? For example, a simple evenly-spaced matrix is:
1 2 3
4 5 6
7 8 9
There are several critical points to look at in the code given as possible answers above.
First, you need to make sure that the main output is not a new line. Thus, the following is incorrect:
System.out.println(matrix\[i\]\[j\]);
This would create a new line for every single value!
Next, you need to make sure that a new line is output after every row of the matrix. This is the final output statement. Thus, it cannot be:
System.out.print(" ");
This would only output a character with no new line.
Finally, for everything except for the first element of each row, you will need to output sufficient padding before the given element element (so as to make the matrix to be evenly-spaced). We are not going to be too particular here about the problem of very big numbers—the values are given to us as is. Note that it is insufficient to use a mere space for this. You should use a tab character (defined as '\t'). A space will not guarantee enough spacing (no pun intended)!
The correct answer has the following logic:
if(j != 0) {
System.out.print("\t\t");
}
This means "when the column is not the first column" (i.e. j != 0), "then output tabs to space sufficiently."
There are several critical points to look at in the code given as possible answers above.
First, you need to make sure that the main output is not a new line. Thus, the following is incorrect:
System.out.println(matrix\[i\]\[j\]);
This would create a new line for every single value!
Next, you need to make sure that a new line is output after every row of the matrix. This is the final output statement. Thus, it cannot be:
System.out.print(" ");
This would only output a character with no new line.
Finally, for everything except for the first element of each row, you will need to output sufficient padding before the given element element (so as to make the matrix to be evenly-spaced). We are not going to be too particular here about the problem of very big numbers—the values are given to us as is. Note that it is insufficient to use a mere space for this. You should use a tab character (defined as '\t'). A space will not guarantee enough spacing (no pun intended)!
The correct answer has the following logic:
if(j != 0) {
System.out.print("\t\t");
}
This means "when the column is not the first column" (i.e. j != 0), "then output tabs to space sufficiently."
Compare your answer with the correct one above
Consider the following code:
for(int i = 1; i <= 10; i++) {
for(int j = 0; j < (20 - i * 2) / 2; j++) {
System.out.print(" ");
}
for(int j = 0; j < i * 2; j++) {
System.out.print("*");
}
for(int j = 0; j < (20 - i * 2) / 2; j++) {
System.out.print(" ");
}
System.out.println();
}
Describe the output of the code above.
Consider the following code:
for(int i = 1; i <= 10; i++) {
for(int j = 0; j < (20 - i * 2) / 2; j++) {
System.out.print(" ");
}
for(int j = 0; j < i * 2; j++) {
System.out.print("*");
}
for(int j = 0; j < (20 - i * 2) / 2; j++) {
System.out.print(" ");
}
System.out.println();
}
Describe the output of the code above.
This code is best explained by analyzing it directly. Comments will be provided in bold below:
// The loop runs from 1 to 10
for(int i = 1; i <= 10; i++) {
// First, you run through 0 to some number
// That number is computed by taking the current value of i, doubling it,
// subtracting that from 20, then dividing by 2.
// So, consider the following values for i:
// 0: 20 - 0 = 20; Divided by 2: 10
// 1: 20 - 2 = 18; Divided by 2: 9
// 2: 20 - 2 = 16; Divided by 2: 8
// etc...
// Thus, you will output single spaces that number of times (10, 9, 8...)
for(int j = 0; j < (20 - i * 2) / 2; j++) {
System.out.print(" ");
}
// Next, you will output an asterix i * 2 number of times. Thus, your
// smallest value is 2 and your largest 20.
for(int j = 0; j < i * 2; j++) {
System.out.print("*");
}
// This is just a repeat of the same loop from above
for(int j = 0; j < (20 - i * 2) / 2; j++) {
System.out.print(" ");
}
// This bumps us down to the next line
System.out.println();
}
Thus, you have output that basically has even padding on the left and right, with an increasing number of asterix characters (2, 4, 6, etc). This is an upward-facing arrow, having a point of width 2 and a base of width 20.
This code is best explained by analyzing it directly. Comments will be provided in bold below:
// The loop runs from 1 to 10
for(int i = 1; i <= 10; i++) {
// First, you run through 0 to some number
// That number is computed by taking the current value of i, doubling it,
// subtracting that from 20, then dividing by 2.
// So, consider the following values for i:
// 0: 20 - 0 = 20; Divided by 2: 10
// 1: 20 - 2 = 18; Divided by 2: 9
// 2: 20 - 2 = 16; Divided by 2: 8
// etc...
// Thus, you will output single spaces that number of times (10, 9, 8...)
for(int j = 0; j < (20 - i * 2) / 2; j++) {
System.out.print(" ");
}
// Next, you will output an asterix i * 2 number of times. Thus, your
// smallest value is 2 and your largest 20.
for(int j = 0; j < i * 2; j++) {
System.out.print("*");
}
// This is just a repeat of the same loop from above
for(int j = 0; j < (20 - i * 2) / 2; j++) {
System.out.print(" ");
}
// This bumps us down to the next line
System.out.println();
}
Thus, you have output that basically has even padding on the left and right, with an increasing number of asterix characters (2, 4, 6, etc). This is an upward-facing arrow, having a point of width 2 and a base of width 20.
Compare your answer with the correct one above
Consider the following code:
What is the output of the method call mystery("Green eggs and ham")
Consider the following code:
What is the output of the method call mystery("Green eggs and ham")
The method String.split() splits a String into an array of Strings separated according to the expression within the method argument. The expression String.split("") splits the String at every character. The "for" loop concatenates the elements of the String array together, separated by a comma.
The method String.split() splits a String into an array of Strings separated according to the expression within the method argument. The expression String.split("") splits the String at every character. The "for" loop concatenates the elements of the String array together, separated by a comma.
Compare your answer with the correct one above
public static void main(String[] args) {
int[][] x = {{4,5,6,7},{91,15,14,13}};
int[][] y = {{-13,4,41,14},{14,5,13,3}};
int[][] z = doWork(x,y);
for(int i = 0; i < z.length; i++) {
for(int j = 0; j < z[0].length;j++) {
System.out.print(z[i][j] + " ");
}
System.out.println();
}
}
``
public static int[][] doWork(int[][] a, int[][] b) {
int[][] ret = new int[a.length][a[0].length];
for(int i = 0; i < a.length; i++) {
for(int j = 0; j < a[i].length; j++) {
ret[i][j] = a[i][j] + b[i][j];
}
}
return ret;
}
What is the console output for the code above?
public static void main(String[] args) {
int[][] x = {{4,5,6,7},{91,15,14,13}};
int[][] y = {{-13,4,41,14},{14,5,13,3}};
int[][] z = doWork(x,y);
for(int i = 0; i < z.length; i++) {
for(int j = 0; j < z[0].length;j++) {
System.out.print(z[i][j] + " ");
}
System.out.println();
}
}
``
public static int[][] doWork(int[][] a, int[][] b) {
int[][] ret = new int[a.length][a[0].length];
for(int i = 0; i < a.length; i++) {
for(int j = 0; j < a[i].length; j++) {
ret[i][j] = a[i][j] + b[i][j];
}
}
return ret;
}
What is the console output for the code above?
The doWork method implements a relatively standard type of 2D array iteration, one that goes through each element. The outer loop goes through the first dimension of the array. Then, the inner loop provides index for the second dimension. From this, you get the total 2D index [i][j]. The line ret[i][j] = a[i][j] + b[i][j];actually performs the operation at the given index. Here, it performs an addition, combining the values at [i][j] found in the two arrays. This gives you the sum at each index of ret. The main method outputs each of these values, following the same standard algorithm for 2D array traversal.
The doWork method implements a relatively standard type of 2D array iteration, one that goes through each element. The outer loop goes through the first dimension of the array. Then, the inner loop provides index for the second dimension. From this, you get the total 2D index [i][j]. The line ret[i][j] = a[i][j] + b[i][j];actually performs the operation at the given index. Here, it performs an addition, combining the values at [i][j] found in the two arrays. This gives you the sum at each index of ret. The main method outputs each of these values, following the same standard algorithm for 2D array traversal.
Compare your answer with the correct one above
LOGIC WITH 2-D ARRAYS
Given:
int\[\]\[\] myArray = { {1, 2},
{3, 4} };
What would the following statement print out to the console?
System.out.print(myArray\[1\]\[1\] + 10);
LOGIC WITH 2-D ARRAYS
Given:
int\[\]\[\] myArray = { {1, 2},
{3, 4} };
What would the following statement print out to the console?
System.out.print(myArray\[1\]\[1\] + 10);
Before anything is printed out into the console, the following is first evaluated:
myArray\[1\]\[1\] + 10
myArray\[1\]\[1\] is referring to the item that is in row=1 and column=1 of myArray. Taking note that arrays start with row 0 and column 0, we see that the item in row 1 column 1 is the number 4. Now we have the following: 4+10. This evaluates to 14. Therefore, the Java print statement will print out the number 14 on the console.
Before anything is printed out into the console, the following is first evaluated:
myArray\[1\]\[1\] + 10
myArray\[1\]\[1\] is referring to the item that is in row=1 and column=1 of myArray. Taking note that arrays start with row 0 and column 0, we see that the item in row 1 column 1 is the number 4. Now we have the following: 4+10. This evaluates to 14. Therefore, the Java print statement will print out the number 14 on the console.
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int a = 10, b = 5;
for(int i = 0; i < a; i+=2) {
for(int j = 0; j < i % b; j++) {
System.out.print("* ");
}
System.out.println();
}
What will be the console output for the code above?
int a = 10, b = 5;
for(int i = 0; i < a; i+=2) {
for(int j = 0; j < i % b; j++) {
System.out.print("* ");
}
System.out.println();
}
What will be the console output for the code above?
All this problem requires is careful attention to the loop control variables. Notice that the first loop adds 2 to i every looping. This means that it will run for 0,2,4,6, and 8. Thus, a total of five lines will be printed.
Now, the second loop dictates the number of * characters that will be output per line. This is going to be based upon the result of the modulus i % b. Remember, modulus is a remainder calculation. Thus, you will have:
0 % 5 = 0
2 % 5 = 2
4 % 5 = 4
6 % 5 = 1
8 % 5 = 3
Hence, you will have the following (note the first line is empty, given the result of 0 % 5):
* *
* * * *
*
* * *
All this problem requires is careful attention to the loop control variables. Notice that the first loop adds 2 to i every looping. This means that it will run for 0,2,4,6, and 8. Thus, a total of five lines will be printed.
Now, the second loop dictates the number of * characters that will be output per line. This is going to be based upon the result of the modulus i % b. Remember, modulus is a remainder calculation. Thus, you will have:
0 % 5 = 0
2 % 5 = 2
4 % 5 = 4
6 % 5 = 1
8 % 5 = 3
Hence, you will have the following (note the first line is empty, given the result of 0 % 5):
* *
* * * *
*
* * *
Compare your answer with the correct one above
class Base{
public:
void foo(int n) { cout << "Base::foo(int)
"; }
};
class Derive: public Base{
public:
void foo(double n) { cout << "Derived::foo(double)
";}
};
int main(){
Derived der;
der.foo(42);
}
The above code is written in C++, what is the output of the program?
class Base{
public:
void foo(int n) { cout << "Base::foo(int)
"; }
};
class Derive: public Base{
public:
void foo(double n) { cout << "Derived::foo(double)
";}
};
int main(){
Derived der;
der.foo(42);
}
The above code is written in C++, what is the output of the program?
This is an example of inheritance. The child class (Derived) is an inherited class of the parent class (Base). When the Derived object is created and the "foo" method is called, foo in the Derived class will be called. If there is the same method in the parent class and child class, the child's method will be called.
This is an example of inheritance. The child class (Derived) is an inherited class of the parent class (Base). When the Derived object is created and the "foo" method is called, foo in the Derived class will be called. If there is the same method in the parent class and child class, the child's method will be called.
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What will the following code result in?
int main(){
int i = 7;
const int * ip = &i;
cout<< *ip << endl;
}
What will the following code result in?
int main(){
int i = 7;
const int * ip = &i;
cout<< *ip << endl;
}
Let's take a look at the code.
int i = 7;
This line assigns 7 to the variable "i".
const int * ip = &i;
This line creates a constant int pointer and assigns the address of "i" to the pointer.
cout << *ip << endl;
This line prints the dereference of the pointer, which holds the value 7.
Let's take a look at the code.
int i = 7;
This line assigns 7 to the variable "i".
const int * ip = &i;
This line creates a constant int pointer and assigns the address of "i" to the pointer.
cout << *ip << endl;
This line prints the dereference of the pointer, which holds the value 7.
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What is the output of this program?
arr = \["hello", "my", "name", "is"\]
for (int i = 0; i < arr.length - 1; i++) {
System.out.println(arr\[i\]);
}
System.out.println("Sandra");
What is the output of this program?
arr = \["hello", "my", "name", "is"\]
for (int i = 0; i < arr.length - 1; i++) {
System.out.println(arr\[i\]);
}
System.out.println("Sandra");
The words are printed on separate lines due to the System.out.println() call and then whatever is inside of the parentheses is printed on the next line. This call is different from System.out.print() which prints words on the same lines.
Therefore, the Sysytem.out.printIn gives("Sandra"):
hello
my
name
is
Sandra
The words are printed on separate lines due to the System.out.println() call and then whatever is inside of the parentheses is printed on the next line. This call is different from System.out.print() which prints words on the same lines.
Therefore, the Sysytem.out.printIn gives("Sandra"):
hello
my
name
is
Sandra
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True or False.
The output of this code snippet will be "Hello, I'm hungry!"
public static void meHungry() {
String hungry = "hungry";
String iAm = "I'm";
String hello = "Hello";
String message = "";
if (hungry != null) {
message += hungry;
}
if (hello != null && iAm != null) {
message = hello + iAm + hungry;
}
System.out.println(message);
}
True or False.
The output of this code snippet will be "Hello, I'm hungry!"
public static void meHungry() {
String hungry = "hungry";
String iAm = "I'm";
String hello = "Hello";
String message = "";
if (hungry != null) {
message += hungry;
}
if (hello != null && iAm != null) {
message = hello + iAm + hungry;
}
System.out.println(message);
}
The message that is printed out is "Hello I'm hungry"
Notice there is no punctuation in the message. The code does not add punctuation to the message, but prints the words out in the same order as the phrase in the prompt. Be mindful of what's actually happening in the code.
The message that is printed out is "Hello I'm hungry"
Notice there is no punctuation in the message. The code does not add punctuation to the message, but prints the words out in the same order as the phrase in the prompt. Be mindful of what's actually happening in the code.
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In Swift (iOS), give a description of what this method does.
func getDivisors(num: Int, divisor: Int) -> Bool {
var result: Bool = False
if (num % divisor == 0) {
result = True
}
println(result)
}
In Swift (iOS), give a description of what this method does.
func getDivisors(num: Int, divisor: Int) -> Bool {
var result: Bool = False
if (num % divisor == 0) {
result = True
}
println(result)
}
The modulus function "%" determines the remainder of a division. If I have 1 % 2, the remainder is 1. If I have 2 % 2, the remainder is 0. Therefore, if the remainder is equal to 0, then the number is divisible by the function. Thus, the method returns true if the number is evenly divisble by the divisor and false otherwise.
The modulus function "%" determines the remainder of a division. If I have 1 % 2, the remainder is 1. If I have 2 % 2, the remainder is 0. Therefore, if the remainder is equal to 0, then the number is divisible by the function. Thus, the method returns true if the number is evenly divisble by the divisor and false otherwise.
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Does the code compile and if yes, what is the the output?
public class Test
{
public static void main ( String \[\] args )
{
int x = 2;
if (x = 2)
System.out.println("Good job.");
else
System.out.println("Bad job);
}
}
Does the code compile and if yes, what is the the output?
public class Test
{
public static void main ( String \[\] args )
{
int x = 2;
if (x = 2)
System.out.println("Good job.");
else
System.out.println("Bad job);
}
}
It doesn't compile. It is supposed to be x == 2 within the if statement not x = 2. The if statement checks for booleans (only true or false) and cannot convert an integer into a boolean to meet that check.
It doesn't compile. It is supposed to be x == 2 within the if statement not x = 2. The if statement checks for booleans (only true or false) and cannot convert an integer into a boolean to meet that check.
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Which of the following code samples adequately uses constants?
Which of the following code samples adequately uses constants?
Remember that a constant variable cannot be changed once it has been assigned. Normally, you assign the variable immediately on the same line as the declaration. However, if you were to create the constant and wait to assign a value, that would be find syntactically as well. Thus, the correct answer does not have a problem, though it might appear so at first if you did not know this. Now, remember that the way to declare a constant is to use the keyword "final". (The keyword "const" works in some other languages like C++.) All of the incorrect answers (other than the one using "const") alter the constant after it has been defined.
Remember that a constant variable cannot be changed once it has been assigned. Normally, you assign the variable immediately on the same line as the declaration. However, if you were to create the constant and wait to assign a value, that would be find syntactically as well. Thus, the correct answer does not have a problem, though it might appear so at first if you did not know this. Now, remember that the way to declare a constant is to use the keyword "final". (The keyword "const" works in some other languages like C++.) All of the incorrect answers (other than the one using "const") alter the constant after it has been defined.
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