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What happens if we compile the below code?Note: This question helps in clearing AMCAT test.1234567891011#include <stdio.h> void function(int *a) { int y = 13; y == *a ? printf("%d", y + 10) : printf("%d", *a);}int main() { int x = 13; function(&x);}

Question

What happens if we compile the below code?Note: This question helps in clearing AMCAT test.1234567891011#include <stdio.h> void function(int *a) { int y = 13; y == *a ? printf("%d", y + 10) : printf("%d", *a);}int main() { int x = 13; function(&x);}

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Solution

When you compile and run the given C code, here's what happens:

  1. The main() function is called first because it is the entry point of the program.

  2. Inside the main() function, an integer x is declared and initialized with the value 13.

  3. The function function(&x) is called with the address of x as the argument. This means the function is expecting a pointer to an integer as its parameter.

  4. Inside the function(), another integer y is declared and initialized with the value 13.

  5. The ternary operator ? : is used to check if y is equal to the value pointed by a (which is 13 in this case). If the condition is true, it prints y + 10 which is 23. If the condition is false, it prints the value pointed by a which is 13.

  6. Since y and the value pointed by a are both 13, the condition is true. Therefore, 23 is printed to the console.

So, the output of the program will be 23.

This problem has been solved

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