Why lower precedence operator executes first?

Question

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Problem with operator precedence

we know that precedence of prefix is greater than "LOGICAL AND" (&&) and precedence of "LOGICAL AND" is greater than "LOGICAL OR" (||).

Below program seems to violate it:

int main()

{
    int i=-3,j=2,k=0,m;
    m=++i||++j&&++k;
    printf("%d %d %d %d",i,j,k,m);
    return 0;
}

If precedence of ++ is more than && and || then all prefix should execute first. After this i=-2,j=3,k=1 and then && will execute first. why output shows : -2 2 0 1 ?

The behavior of the program is also same on ubuntu v12.04.

http://en.wikipedia.org/wiki/Short-circuit_evaluation – Mysticial Aug 28 '12 at 02:48 — Mysticial, Aug 28 '12 at 02:48

score 5 · Accepted Answer · answered Aug 28 '12 at 02:49

5

The && and || operators are "short-circuiting". That is, if the value on the left is FALSE for && or TRUE for || then the expression on the right is not executed (since it's not needed to determine the value of the overall expression).

answered Aug 28 '12 at 02:49

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what about the precedence of ++ . – karthik Aug 28 '12 at 02:56
precedence affects what the operators apply to. It has no effect on the order in which they are executed. – Chris Dodd Aug 28 '12 at 03:08
@aakasha -- You're confusing precedence and associativity. – Hot Licks Aug 28 '12 at 11:59

nofomopls · Answer 2 · 2012-09-12T01:19:25.653

2

It's correct because Short-circuiting definition.

m = ++i||++j&&++k

First, left part ++i is always TRUE so now i is -2 and it doesn't execute the right part of expression, the value of j,k don't change.

edited Sep 12 '12 at 01:19

answered Aug 28 '12 at 04:03

nofomopls

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Why lower precedence operator executes first?

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