Question

Does the output of the code given below depends on the compiler or it is guaranteed to be same for all the compilers?

int main()
{ 
    int i=5;
    i=i/3;
    printf("%d\n",i);
    return 0;
}

Answer 1

Yes, the behaviour of your example is well-defined.

However, in the case of negative values, it is less clear. Pre-C99, whether integer division was rounded towards zero or towards negative infinity was left as implementation-defined:

If either operand is negative, whether the
result of the / operator is the largest integer less than the
algebraic quotient or the smallest integer greater than the algebraic
quotient is implementation-defined

C99 mandates rounding towards zero:

When integers are divided, the result of the / operator is the algebraic quotient with any
fractional part discarded..