What will be the output of the following code? int main() {int x = 5;int * ptr = &x;*ptr + 1;cout << (*ptr)++;}
Question
What will be the output of the following code? int main() {int x = 5;int * ptr = &x;*ptr + 1;cout << (*ptr)++;}
Solution
The output of the code will be 5.
Here's the step by step explanation:
-
int x = 5;This line declares an integer variablexand assigns it the value 5. -
int * ptr = &x;This line declares a pointerptrand assigns it the address ofx. Nowptris pointing tox. -
*ptr + 1;This line increments the value pointed byptrby 1, but it doesn't change the value ofxbecause the result is not stored anywhere. -
cout << (*ptr)++;This line prints the value pointed byptrwhich is the value ofx(5), and then increments the value ofxby 1. But the increment happens after the value is printed, so it doesn't affect the output.
So, the output of the code will be 5.
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