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What will be the output of the following code?1234567891011121314151617#include <stdio.h>int main() { int i = 0, product = 10;start: if (i < 4) { product *= (i + 1); i++; if (i != 2) goto start; } printf("%d", product); return 0;}

Question

What will be the output of the following code?1234567891011121314151617#include <stdio.h>int main() { int i = 0, product = 10;start: if (i < 4) { product *= (i + 1); i++; if (i != 2) goto start; } printf("%d", product); return 0;}

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Solution

The output of the code will be 20.

Here's the step-by-step explanation:

  1. The program starts with the initialization of two integer variables, i and product, with values 0 and 10 respectively.

  2. Then it enters a labeled section of code called start.

  3. Inside start, there's an if condition that checks if i is less than 4. Since i is initially 0, the condition is true and the program enters the if block.

  4. Inside the if block, product is multiplied by i + 1 (which is 1 at the start), so product becomes 10.

  5. Then i is incremented by 1, so i becomes 1.

  6. The next if condition checks if i is not equal to 2. Since i is 1, the condition is true and the program jumps back to start.

  7. The process repeats. This time product is multiplied by i + 1 (which is 2 now), so product becomes 20.

  8. i is incremented again and becomes 2.

  9. The next if condition checks if i is not equal to 2. This time i is 2, so the condition is false and the program doesn't jump back to start.

  10. The program then prints the value of product, which is 20, and returns 0 to end the program.

This problem has been solved

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