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What is the output for the following code?12345678910111213141516171819#include <stdio.h> int main() { int a[5] = {1, 2, 3, 4, 6}; switch (a[1]) { case 1: int r1 = a[1] + a[2]; printf("%d", r1); break; case 2: int r2 = a[3] * a[4]; printf("%d", r2); break; default: int d = a[0] * a[4]; printf("%d", d); } return 0;}

Question

What is the output for the following code?12345678910111213141516171819#include <stdio.h> int main() { int a[5] = {1, 2, 3, 4, 6}; switch (a[1]) { case 1: int r1 = a[1] + a[2]; printf("%d", r1); break; case 2: int r2 = a[3] * a[4]; printf("%d", r2); break; default: int d = a[0] * a[4]; printf("%d", d); } return 0;}

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Solution

The output of the given code will be 24.

Here's the step-by-step explanation:

  1. The array a is initialized with the values {1, 2, 3, 4, 6}.
  2. The switch statement checks the value of a[1], which is 2.
  3. Since the value is 2, it matches with the case 2: and the code inside this case is executed.
  4. Inside case 2:, a new variable r2 is declared and initialized with the product of a[3] and a[4], which is 4 * 6 = 24.
  5. The value of r2 is then printed, so the output is 24.
  6. After executing the code inside case 2:, the program breaks out of the switch statement.
  7. The program then ends with return 0;, indicating successful execution.

This problem has been solved

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