What will the following code display? #include using namespace std; void doSomething (int); int main () { int x{2}; cout << x << '/n'; doSomething (x); cout << x << '/n'; return 0; } void doSomething (int num) { num = 0; cout << num << '/n';

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Computers & Technology

11 June, 08:52

What will the following code display? #include using namespace std; void doSomething (int); int main () { int x{2}; cout << x << '/n'; doSomething (x); cout << x << '/n'; return 0; } void doSomething (int num) { num = 0; cout << num << '/n';

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Half Pint · Answer 1

The code will display the following on the screen:-

2

0

2

Explanation:

The output is like this because x{2} assigns the value 2 to the integer x. First the value of x is printed on the screen that is 2 and that newline is used to go to the next line. After that function is called doSomething with x passed as the argument. doSomething function assigns the value zero to the variable passed and prints it on the screen. So the next line is 0. Since x is passed by value so no changes will appear in the original argument. Hence the last line printed is 2.