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    Question

    Consider the following C code snippet:      

      #include     int recursiveSum(int n) {         static int sum = 0; // Static variable         if (n > 0) {             sum += n;             recursiveSum(n - 1);         }         return sum;     }     int main() {         printf("First call: %d\n", recursiveSum(3));         printf("Second call: %d\n", recursiveSum(2));         return 0;     }     What will be the output of the main function?
    A First call: 6 Second call: 9 Correct Answer Incorrect Answer
    B First call: 6 Second call: 3 Correct Answer Incorrect Answer
    C First call: 6 Second call: 6 Correct Answer Incorrect Answer
    D First call: 3 Second call: 2 Correct Answer Incorrect Answer
    E First call: 0 Second call: 0 Correct Answer Incorrect Answer

    Solution

    The key here is the static int sum = 0; declaration. A static variable is initialized only once when the program starts and retains its value across multiple function calls.       First call: recursiveSum(3) sum is initialized to 0 (only once). recursiveSum(3): n=3. sum becomes 0 + 3 = 3. Calls recursiveSum(2). recursiveSum(2): n=2. sum becomes 3 + 2 = 5. Calls recursiveSum(1). recursiveSum(1): n=1. sum becomes 5 + 1 = 6. Calls recursiveSum(0). recursiveSum(0): n=0. The if (n > 0) condition is false. Returns the current value of sum, which is 6. All preceding calls return this value. So, printf("First call: %d\n", recursiveSum(3)); prints "First call: 6".     Second call: recursiveSum(2) sum does not reset to 0. It retains its value of 6 from the previous execution. recursiveSum(2): n=2. sum becomes 6 + 2 = 8. Calls recursiveSum(1). recursiveSum(1): n=1. sum becomes 8 + 1 = 9. Calls recursiveSum(0). recursiveSum(0): n=0. The if (n > 0) condition is false. Returns the current value of sum, which is 9. All preceding calls return this value. So, printf("Second call: %d\n", recursiveSum(2)); prints "Second call: 9".     Therefore, the output is:     First call: 6     Second call: 9

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