Console Output - Computer Science
Card 0 of 12
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."
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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.
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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.
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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.
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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):
* *
* * * *
*
* * *
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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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