Operator precedence of |, ~ and & operator?

Question

I'm confused about these logical operators. can someone please explain the precedence and associative rules of these operators. in bit wise operations, a=011, b=010 and c=001 in d whether a should be negated first or should the evaluation be started from right to left and whose precedence is higher? the output is 4, 3,3.

#include<stdio.h>
int main()
{
 int a=3,b=2,c=1,d,e,f;
 d=~a|b&c;
 printf("d=%d\n",d);
 e=a|b&~c;
 printf("d=%d\n",e);
 f=a|b&c;
 printf("d=%d\n",f);
 return 0;
}

Funny that one of the top answers from Google will give you http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B — Ed Heal, Jun 14 '14 at 14:50
http://en.cppreference.com/w/cpp/language/operator_precedence — phuclv, Jun 14 '14 at 14:57
A minimum of research is expected of enthusiastic and professional programmers... — Deduplicator, Jun 14 '14 at 15:54

score 2 · Answer 1 · answered Jun 14 '14 at 14:51

2

C unary operators (e.g., ~) have higher precedence than binary operators.

& operator has higher precedence than | operator.

See man 7 operators for a summary of operator precedence.

answered Jun 14 '14 at 14:51

ouah

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Operator precedence of |, ~ and & operator?

1 Answers1