需要输出说明 [英] Require explanation for the output

问题描述

代码:

#include<stdio.h>
int main()
{
    int j = 7, i = 4;
    j = j || ++i && printf("you can");
    printf("%d %d",i,j);
    return 0;
}

输出: 4 1

[代码链接] [1]

[Code Link][1]

1. 前缀运算符的优先级高于逻辑运算符. 2.逻辑&&的优先级高于逻辑||.
在逻辑AND(&&)中，如果第一个操作数的计算结果比第二个错误，则不计算；在逻辑OR(||))中，如果第一个操作数的计算结果为true，则第二个值将不计算.
2. 完整表达式的计算结果为true，因此j为1.

1. The precedence of prefix operator is higher than logical operators. 2.Logical && has higher precedence than logical ||.
2. In Logical AND(&&) if first operand evaluates to false than second will not be evaluated and In Logical OR(||) if first operand evaluates to true, then second will not be evaluated.
3. The complete expression is evaluating to true, therefore j is 1 .

疑问:

1. 为什么这里没有遵循第一个规则?不是吗?

1. Why the first rule is not followed here? Shouldn't it be correct?

j=(j||((++i) &&printf("you can")));

因此，在printf语句中，i的值变为5.

Therefore value of i becomes 5, in the printf statement.

为什么这里违反了通用优先规则?当两个运算符的优先级相同时，关联性将起作用.编译器是否应该首先查看要评估||还是&&?

Why are the general precedence rules are violated here? Associativity comes into action when precedence of two operators is same. Shouldn't the compiler first see whether to evaluate || or &&?

如果首先评估||，而不是据我所知，那么结果是正确的.但是，如果不首先评估它，则应该打印can51.

If || is evaluated first, which shouldn't be as per my knowledge, then result is correct. However, if it is not evaluated first, then you can51 should be printed.

推荐答案

在此表达式中:

j = j || ++i && printf("you can");

||之后有一个序列点，并且从左到右进行评估.由于j为非零值，因此不对表达式的其余部分求值.因此，j || (....)变为1.由于未评估++i，所以我仍然为4.因此，输出为4, 1.

There's a sequence point after the || and it is evaluated from left to right. Since j is non-zero, the rest of the expression is not evaluated. Hence, j || (....) becomes true which is 1. Since is ++i is not evaluated i remains 4. Hence, the output is 4, 1.

根据C标准:

附件C

-以下运算符的第一个操作数的结尾:逻辑AND && (6.5.13)；逻辑或|| (6.5.14)；有条件的? (6.5.15)；逗号， (6.5.17).

— The end of the first operand of the following operators: logical AND && (6.5.13); logical OR || (6.5.14); conditional ? (6.5.15); comma , (6.5.17).

如果j为零，则将评估++i && printf("you can")，并且i将变为5，并且还将打印you can.您对++的优先级大于||是正确的，但是由于存在序列点，因此首先会忽略j||.

If you j was zero then ++i && printf("you can") would have been evaluated and i would become 5 and you can will also be printed. You are correct about the precedence of ++ being greater than ||, but since there's a sequence point, j|| is evalauted first.

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