C中“或”和“与”运算符的优先级

Question

In the following code, I am getting 10 | 1 | 1 as a result. But according to precedence rules shouldn't 'and' operator must be evaluated first?(and yield c=9) Like : d = a || (--b)&&(--c) since 'and' has higher precedence. ( or shortcutting breaks precedence rules ?)Thanks in advance.

#include <stdio.h>
#include <stdlib.h>

int main(){
    int a,b,c,d;
    a =1;
    b = c = 10;
    d = a|| --b&&--c;
    printf("%d\n",c);
    printf("%d\n",a);
    printf("%d\n",d);
    return 0;
}

Answer 1

Precedence and order of evaluation are two different things. From Logical OR documentation (emphasis mine):

There is a sequence point after the evaluation of lhs. If the result of lhs compares unequal to zero, then rhs is not evaluated at all (so-called short-circuit evaluation).

In case of exp1 || exp2 exp1 || exp2 , exp1 is always evaluated first as there is a sequence point after it and if exp1 is non-zero then exp2 is not evaluated.

Answer 2

Precedence only determines which operands are grouped with which operators - it does not control the order in which expressions are evaluated.

In your example, it means the expression is parsed as

a || (––b && ––c)

Both || and && force left-to-right evaluation ¹ . Both introduce a sequence point (IOW, the left hand operand will be evaluated and all side effects will be applied before the right hand operand is evaluated).

Both operators short-circuit - if the left operand of || evaluates to non-zero, then the result of the expression is 1 (true) regardless of the value of the right operand, so the right operand isn't evaluated at all. If the left operand of && is 0, then the result of the expression is 0 (false) regardless of the value of the right operand, so the right operand isn't evaluated at all.

In your expression, a is evaluated first. It has a non-zero value (1), so ––b && ––c is not evaluated.

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1. ^{Along with the ?: and comma operators. All other operators (arithmetic, equality, subscript, etc.) do not force a particular order of evaluation.}