防爆pressions＆QUOT; J = ++（I | I）且j = + +（I和ⅰ）;应该是一个左值错误？ [英] Expressions "j = ++(i | i); and j = ++(i & i); should be a lvalue error?

问题描述

我期待在我的下面code：

I was expecting that in my following code:

#include<stdio.h> 
int main(){
    int i = 10; 
    int j = 10;

    j = ++(i | i);
    printf("%d %d\n", j, i);

    j = ++(i & i);
    printf("%d %d\n", j, i);

    return 1;
}

EX pressions J = ++（I | I）; 和 J = ++（I和我）; 将产生左值误差如下：

expressions j = ++(i | i); and j = ++(i & i); will produce lvalue errors as below:

x.c: In function ‘main’:
x.c:6: error: lvalue required as increment operand
x.c:9: error: lvalue required as increment operand   

但我惊讶的是上面code编译成功，如下：

But I surprised that above code compiled successfully, as below:

~$ gcc x.c -Wall
~$ ./a.out 
11 11
12 12   

上述code工作正确。

虽然其他运营商产生错误（据我所知）。 //$c$cpad.org/：（检查其他的 J = ++（I ^ I）中，即使位运算符XOR原因g08e8GkH相对=nofollow>运营商产生在编译时左值误差）。

While other operators produce error (as I understand). Even bitwise operator XOR causes of an error j = ++(i ^ i); (check other operators produce an lvalue error at compilation time).

这是什么原因？这是未指定的或不确定的？或按位或和经营者有什么不同？

What is the reason? Is this is unspecified or undefined ? or bitwise OR AND operators are different?

编译器版本：

gcc version 4.4.5 (Ubuntu/Linaro 4.4.4-14ubuntu5)

<子>但我相信编译器版本不应该理性非均匀行为。如果 ^ 未编译那么 | 和＆安培; 也没有。否则应该为所有工作

_{But I believe compiler version shouldn't reason for non-uniform behavior. If ^ not compiled then | and & also not. otherwise should work for all}

它不是这个编译器C99模式下的错误： GCC x.c -Wall -std = C99 。

Its not an error with this compiler in c99 mode: gcc x.c -Wall -std=c99.

推荐答案

您是正确的，它不应该编译，并在大多数编译器，它不编译。结果
<罢工>的（请准确指定的编译器/版本不是给你一个编译器错误）的

You are right that it should not compile, and on most compilers, it does not compile.
(Please specify exactly which compiler/version is NOT giving you a compiler error)

我只能假设，编译器知道身份的（I | I）== I 和（I和我）==我并使用这些身份优化掉除权pression，正要离开变量 I 。

I can only hypothesize that the compiler knows the identities that (i | i) == i and (i & i) == i and is using those identities to optimize away the expression, just leaving behind the variable i.

这只是一个猜测，但它使一个很大的意义对我来说。

This is just a guess, but it makes a lot of sense to me.

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