C语言无法操控1bit的位域 | 操控1bit时, 其他也会改变

最新推荐文章于 2025-04-24 11:23:14 发布

原创最新推荐文章于 2025-04-24 11:23:14 发布 · 615 阅读

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#c语言

文章探讨了在C语言中使用union和位字段时遇到的问题，即修改位字段后，其他位字段的值被意外改变。作者尝试了添加volatile声明符来防止编译器优化，但未解决问题。最终，通过添加staticvolatile解决了问题，这表明static关键字在此场景下阻止了变量在函数调用间的默认行为，确保了变量状态的持久性。

问题

Test 1

#include <stdio.h>

union bits
{
    struct __bits
    {
        unsigned char i0 : 1;
        unsigned char i1 : 1;
        unsigned char i2 : 1;
        unsigned char i3 : 1;
        unsigned char i4 : 1;
        unsigned char i5 : 1;
        unsigned char i6 : 1;
        unsigned char i7 : 1;
    };
    unsigned char var;
};

int main()
{

    union bits a;
    unsigned char t;

    printf("size = %d\n",sizeof(union  bits));

    scanf("%d", &t);
    a.i0 = t;
    printf("%d\n", a.var);

    scanf("%d", &t);
    a.i1 = t;
    printf("%d\n", a.var);

    scanf("%d", &t);
    a.i2 = t;
    printf("%d\n", a.var);

    scanf("%d", &t);
    a.i3 = t;
    printf("%d\n", a.var);

    scanf("%d", &t);
    a.i4 = t;
    printf("%d\n", a.var);

    scanf("%d", &t);
    a.i5 = t;
    printf("%d\n", a.var);

    scanf("%d", &t);
    a.i6 = t;
    printf("%d\n", a.var);

    scanf("%d", &t);
    a.i7 = t;
    printf("%d\n", a.var);
	return 0;
}

期望结果应是自由操纵每一位,而不改变其他位
实际结果是只能改变一位
result

Test 2

#include <stdio.h>

union bits
{
    struct __bits
    {
        unsigned char i0 : 1;
        unsigned char i1 : 1;
        unsigned char i2 : 1;
        unsigned char i3 : 1;
        unsigned char i4 : 1;
        unsigned char i5 : 1;
        unsigned char i6 : 1;
        unsigned char i7 : 1;
    };
    unsigned char var;
};

int main()
{

    union bits a;
    unsigned char t;

    printf("size = %d\n",sizeof(union  bits));


    a.i0 = 1;
    a.i1 = 1;
    printf("%d\n", a.var);

    a.i0 = 1;
    a.i1 = 1;
    scanf("%d", &t);
    printf("%d\n", a.var);
    
	return 0;
}

当我进行如上测试时,发现执行 scanf 会改变a的值

解决方案

思路1 - 添加 volatile 声明符

#include <stdio.h>

union bits
{
    struct __bits
    {
        unsigned char i0 : 1;
        unsigned char i1 : 1;
        unsigned char i2 : 1;
        unsigned char i3 : 1;
        unsigned char i4 : 1;
        unsigned char i5 : 1;
        unsigned char i6 : 1;
        unsigned char i7 : 1;
    };
    unsigned char var;
};

int main()
{

    volatile union bits a;
    unsigned char t;

    printf("size = %d\n",sizeof(union  bits));

    a.i0 = 1;
    a.i1 = 1;
    printf("%d\n", a.var);

    a.i0 = 1;
    a.i1 = 1;
    scanf("%d", &t);
    printf("%d\n", a.var);

	return 0;
}

结果还是老样子

思路2 - 添加 static volatile 声明符

我无意间想到是不是可以加个static试一下

#include <stdio.h>

union bits
{
    struct __bits
    {
        unsigned char i0 : 1;
        unsigned char i1 : 1;
        unsigned char i2 : 1;
        unsigned char i3 : 1;
        unsigned char i4 : 1;
        unsigned char i5 : 1;
        unsigned char i6 : 1;
        unsigned char i7 : 1;
    };
    unsigned char var;
};

int main()
{

    static volatile union bits a;
    unsigned char t;

    printf("size = %d\n",sizeof(union  bits));

    a.i0 = 1;
    a.i1 = 1;
    printf("%d\n", a.var);

    a.i0 = 1;
    a.i1 = 1;
    scanf("%d", &t);
    printf("%d\n", a.var);

	return 0;
}