问题描述
因此，我正在编写一些代码，并且在程序的一部分中得到了意外的输出，这破坏了整个系统.
So, I was writing some code and I was getting an unexpected output in one part of my program which disrupted the entire system.
我设法将问题提取并简化为基本的逻辑表达式.假设:
I managed to extract and simplify the problem to a basic logical expression. Let's say:
int i = 1, j = 1, k = 0;
printf("%d\n", ++i || ++j && k);
printf("%d, %d, %d\n", i, j, k);
return 0;
该程序的输出为:
1
2 1 0
我认为由于||运算符的短路特性，j的值未增加到2.但是我很困惑第一个"%d"的值是1.对于&&语句返回1来说，k的值是否不应该为非零?还是因为++i || ++j不是0并因此返回1而根本不执行该语句?
I am thinking that the value of j was not incremented to 2 due to the short circuit nature of the || operator. However I am confused how the value of the first "%d" is 1. Shouldn't the value of k be non-zero for the && statement to return 1? Or isn't this statement executed at all since ++i || ++j is not 0 and thus returns a 1?
- 我知道
&&是逻辑和，而 expr1 &&如果 expr1 &&的值为 expr2 ，则其值为1. expr2 均为两者非零.
- I know that
&&is a logical and, and expr1 && expr2 has the value 1 if values of expr1 && expr2 are both non-zero.
任何澄清将不胜感激，请原谅此问题的基本性质.
Any clarifications would be appreciated and please excuse the basic nature of this question.
推荐答案
&&的优先级高于||.(请参见 https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B)所以
&& has higher precedence than ||.(See https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B )so
++i || ++j && k
是
++i || (++j && k)
如果和||会短路> 6.5.14p4 .
and || shortcircuits if the first operator is truthy, as per 6.5.14p4 .
如果您使用的是gcc或clang，并使用-Wall编译代码，则编译器会轻推您将这些括号放在此处.听取建议可能是一个好主意，因为有些人对优先顺序感到困惑(我听到).
If you're on gcc or clang and compile your code with -Wall, the compiler will nudge you to put those parentheses there. It's probably a good idea to heed that advice, as some people get confused by the precedences (I hear).
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