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What will be the output of the following C code?
    #include <stdio.h>
    int main()
    {
        int i = 3;
        int l = i / -2;
        int k = i % -2;
        printf("%d %d\n", l, k);
        return 0;
    }
  • a)
    Compile time error
  • b)
     -1 1
  • c)
    1 -1
  • d)
    Implementation defined
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
What will be the output of the following C code? #include <stdio.h...
Understanding the Code
The provided C code performs basic arithmetic operations involving division and modulus with negative numbers. Let's break down the code step-by-step.
Code Analysis
- The variable `i` is initialized to 3.
- The division operation `i / -2` is computed next:
- In C, when dividing an integer by a negative integer, the result is floored. Thus, `3 / -2` results in `-1`.
- Next, the modulus operation `i % -2` is evaluated:
- The result of the modulus operation takes the sign of the divisor. Therefore, `3 % -2` results in `1`.
Expected Output
- Finally, the `printf` function outputs the values of `l` and `k`:
- `l` is `-1` and `k` is `1`.
Thus, the output from the `printf` statement will be:
-1 1
Conclusion
The correct answer to the question is option 'B', which outputs `-1 1`. This is due to the specific rules of integer division and modulus in C when negative integers are involved.
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Community Answer
What will be the output of the following C code? #include <stdio.h...
B
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What will be the output of the following C code? #include <stdio.h> int main() { int i = 3; int l = i / -2; int k = i % -2; printf("%d %d\n", l, k); return 0; }a)Compile time errorb)-1 1c)1 -1d)Implementation definedCorrect answer is option 'B'. Can you explain this answer?
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