c operator-precedence operators associativity

C 中的运算符优先级

发布于 2024-11-24 14:15:07 字数 171 浏览 2 评论 0原文

printf ("%d \n", 2 > !3 && 4 - 1 != 5 || 6 ) ;

有人可以向我解释一下这是如何评估的吗？我最困惑的是 3... 如何评估 2 > 前面的 ! 符号!3？

原文

printf ("%d \n", 2 > !3 && 4 - 1 != 5 || 6 ) ;

Can someone explain to me how this is evaluated ? What I am most confused about is the ! symbol in front of the 3... how to evaluate 2 > !3 ?

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傲鸠 2024-12-01 14:15:07

！逻辑上不是。这意味着它对于 0 返回 1，或者对于任何其他值返回 0。

!0 == 1

!1 == 0

!897489 == 0

所以在你的例子中 2 > !3 将被评估 2 > 0 是 1

对于你的整个表达式，你将有

2 > !3 && 4 - 1 != 5 || 6 
2 > 0  && 3 != 5 || 6
1 && 1 || 6                    && is higher precedence || but some compiler will warn (gcc)
1 || 6                         the 6 will not be evaluated (touched) because of short-circuit evaluation
1

! is the logical not. Meaning it returns 1 for 0 or 0 for any other value.

!0 == 1

!1 == 0

!897489 == 0

So in your example 2 > !3 will be evaluted 2 > 0 which is 1

For your whole expression you will have

2 > !3 && 4 - 1 != 5 || 6 
2 > 0  && 3 != 5 || 6
1 && 1 || 6                    && is higher precedence || but some compiler will warn (gcc)
1 || 6                         the 6 will not be evaluated (touched) because of short-circuit evaluation
1

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分开我的手 2024-12-01 14:15:07

下面是运算符优先级表。由此我们可以推断出在您的表达式中

! 具有最高优先级
- 接下来
> 接下来
< code>!= 接下来是
&& 接下来是
|| 接下来是

! 表示逻辑不是 !0 == 1, !anythingElse == 0 。

表达式的计算结果为

((2 > (!3)) && ((4 - 1) != 5)) || 6

通常，操作数的计算顺序是未指定的，因此在

2 > !3

编译器可以在计算 2 之前自由计算 !3

但是，对于&&和 || 时，总是首先评估左侧，以便可以使用短路（使用 && 如果左侧为 false (0)，则不评估右侧，使用 < code>|| 如果左侧为 true（除 0 以外的任何值），则不评估右侧）。在上面的代码中，6 永远不会被计算，因为 || 的左侧为 true。 6是

6 || ((2 > (!3)) && ((4 - 1) != 5))

正确的，所以右侧永远不会被评估。

Here is a table of operator precedence. From this we can deduce that in your expression

! has the highest precedence
- comes next
> comes next
!= comes next
&& comes next
|| comes next

! means logical not !0 == 1, !anythingElse == 0.

The expression is evaluated as

((2 > (!3)) && ((4 - 1) != 5)) || 6

Generally, the order of evaluation of the operands is unspecified, so in the case of

2 > !3

the compiler is free to evaluate !3 before evaluating2

However, for && and ||, the left side is always evaluated first so that short circuiting can be used (with && if the left side is false (0), the right side is not evaluated, with || if the left side is true (anything but 0), the right side is not evaluated). In the above, the 6 would never be evaluated because the left side of || is true. With