有没有任何C实现有一个无用的单位`int`位域？

发布时间：2020-12-16 09:58:08 所属栏目：百科来源：网络整理

导读：6.7.2.1p9 of n1570说： A member of a structure or union may have any complete object type other than a variably modified type.123) In addition,a member may be declared to consist of a specified number of bits (including a sign bit,if any).

6.7.2.1p9 of n1570说：

A member of a structure or union may have any complete object type other than a
variably modified type.123) In addition,a member may be declared to consist of a
specified number of bits (including a sign bit,if any). Such a member is called a
bit-field ;124) its width is preceded by a colon.

我是否正确理解这表明结构中的单个成员{int bit：1;可能是一个标志位？

如果是这种情况,则遵循这样的位字段可能存储在某些实现上的唯一值是0和-0,其中-0可能与0存储或陷阱表示无法区分.

是否有任何实际的实现只能将一个值分配给这样的位域？

解决方法

gcc 4.9.2怎么样？

/* gcc -std=c11 -pedantic-errors -Wextra -Werror=all test.c */
#include <stdio.h>
#include <string.h>

struct foo {
    int bit:1;
};

int main() {
    struct foo f;
    f.bit = 0;
    f.bit = 1;

    printf("%in",f.bit);
    return 0;
}

编译它会发出：

$gcc -std=c11 -pedantic-errors -Wextra -Werror=all test.c
test.c: In function ‘main’: test.c:12:10: warning: overflow in
implicit constant conversion [-Woverflow]
  f.bit = 1;
          ^

运行它发出：

$./a.out 
-1

（编辑：李大同）

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