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What is the output of the following code snippet?int x = 5;int y = 10;int *ptr = &x;*ptr = y;printf("%d", x);510015Generates a compilation error

Question

What is the output of the following code snippet?

int x = 5;
int y = 10;
int *ptr = &x;
*ptr = y;
printf("%d", x);
  • 5
  • 10
  • 15
  • Generates a compilation error
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Solution

The output of the code snippet will be 10.

Here's the step by step explanation:

  1. int x = 5; This line declares an integer variable x and assigns it the value 5.

  2. int y = 10; This line declares another integer variable y and assigns it the value 10.

  3. int *ptr = &x; This line declares a pointer variable ptr and assigns it the address of x. So, ptr is now pointing to x.

  4. *ptr = y; This line assigns the value of y to the location ptr is pointing to. Since ptr is pointing to x, this effectively changes the value of x to 10.

  5. printf("%d", x); This line prints the value of x, which is now 10. So, the output of this code snippet is 10.

This problem has been solved

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