Question

I was recently came across a piece of code

// Program to overcome division by zero

int a=0;
int b=100;

int c= a==0 || b/a ;

printf("Hello");

//Output : Hello

My theory:
According to the precedence, operator / has higher precedence than ||. So b/a must get executed first and we should get a run time error.

I assume what is happening though is :

short-circuit operator || , evaluates the LHS a==0, which is true and hence does not execute b/a.

Is my theory wrong?. I am pretty sure this is something very simple that i just can’t figure out right now

Answer 1

Precedence doesn’t imply evaluation order, only grouping (parentheses).

There is a sequence point (old parlance) after the evluation of the first operand of the ||, so the first operand of || must be evaluated before the second, regardless of what these operands are. Since in this case the overall result of the expression a == 0 || b/a was determined by the first operand, the second isn’t evaluated at all.