What will be the output of the code snippet?123456789#include <stdio.h>int main() { int a = 7, b = 11; int *ptr = &a; int x = *ptr * *ptr + b; printf("%d", x); return 0;}
Question
What will be the output of the code snippet?123456789#include <stdio.h>int main() { int a = 7, b = 11; int *ptr = &a; int x = *ptr * *ptr + b; printf("%d", x); return 0;}
Solution
The output of the code snippet will be 60. Here's how:
- The code first declares two integer variables
aandb, and assigns them the values 7 and 11 respectively. - Then it declares a pointer
ptrand assigns it the address ofa. So,ptrnow points toa. - The next line declares an integer
xand assigns it the value of the expression*ptr * *ptr + b. Here,*ptris the dereference operator, which gets the value stored at the addressptris pointing to. Sinceptris pointing toa,*ptris 7. So, the expression becomes7 * 7 + 11, which equals 60. - Finally, the code prints the value of
x, which is 60.
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