In the below code:
#include <stdio.h>
int main(void)
{
int i=-3,j=2,k=0,m;
m=++i||++j&&++k;
printf("%d %d %d %d",i,j,k,m);
}
Output:
-2 2 0 1
Why is k = 0? because I think k also gets executed because of && operator?
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Raktim Biswas
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Marco Roberts
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2Why are you writing code that is difficult to read? – Ed Heal Jul 30 '16 at 11:32
1 Answers
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C uses short circuit-logic - since ++i isn't zero, it's true, and since it's the left-hand side of an || operator, we know no matter what's on the right-hand side, it will result to true. Hence, C (and a bunch of similar languages) don't even bother evaluating the right hand side, and just quickly return true. Since ++k is never evaluated, k remains unchanged, and is still 0 after the m=++i||++j&&++k; statement.
Mureinik
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he might need to know that [`&&` has higher precedence than `||`](http://en.cppreference.com/w/cpp/language/operator_precedence) therefore the expression is parsed as `(++i) || (++j && ++k)` – phuclv Jul 30 '16 at 11:33