从int变量提取字节时高位字节符号扩展异常问题排查

尝试从一个int变量中提取4个char值，但所有尝试均得到相同结果：高低字的低字节读取正确，但高字节被扩展为int时填充了全1。以下是使用GCC编译的测试代码：

#include <stdio.h>

#define PRINT   printf("b1 %u\n", bb1);\
            printf("b2 %u\n", bb2);\
            printf("b3 %u\n", bb3);\
            printf("b4 %u\n", bb4)

typedef union{
    int integer;
    struct{
        char b1;
        char b2;
        char b3;
        char b4;
    };
}bytes;

typedef struct{
    char b1:8;
    char b2:8;
    char b3:8;
    char b4:8;
}bitfield;


int main()
{
    int a;
    printf("int size = %u\n", a=sizeof(int));
    printf("char size = %u\n", a=sizeof(char));
    bytes in_union;
    bitfield in_struct;
    // values stored in bytes 
    // hword.hi 240 hword.lo 15
    // lword.hi 129 lword.lo 126
    int value = 4027548030;
    
    // #1 solution with union
    in_union.integer = value;
    
    char bb1 = in_union.b1;
    char bb2 = in_union.b2;
    char bb3 = in_union.b3;
    char bb4 = in_union.b4;
    PRINT;
    
    // #2 with pointer to union
    char* uptr;
    uptr = (char*)&in_union;
    bb1 = *uptr;
    bb2 = *(uptr+1);
    bb3 = *(uptr+2);
    bb4 = *(uptr+3);
    PRINT;
    
    // #3 with void pointer to union
    void* vptr;
    vptr = &in_union;
    bb1 = *(char*)vptr;
    bb2 = *(char*)(vptr+1);
    bb3 = *(char*)(vptr+2);
    bb4 = *(char*)(vptr+3);
    PRINT;
    
    // #4 with pointer to value variable
    char* ptr = (char*)&value;
    bb1 = *ptr;
    bb2 = *(ptr+1);
    bb3 = *(ptr+2);
    bb4 = *(ptr+3);
    PRINT;
    
    // #5 with bitfield
    in_struct.b1 = (char)value;
    in_struct.b2 = (char)(value>>8);
    in_struct.b3 = (char)(value>>16);
    in_struct.b4 = (char)(value>>24);
    bb1 = in_struct.b1;
    bb2 = in_struct.b2;
    bb3 = in_struct.b3;
    bb4 = in_struct.b4;
    PRINT;
    
return 0;
}

实际输出：

b1 126
b2 4294967169
b3 15
b4 4294967280

预期输出：

b1 126
b2 129
b3 15
b4 240

核心原因 #

问题出在**char类型的符号扩展**：

• 多数编译器（包括GCC）中，char默认是有符号类型（signed char）。
• 当提取的字节值大于127（比如129、240）时，这些值在有符号char中会被解析为负数（129对应-127，240对应-16）。
• 把这些负数的char传给printf的%u（无符号整数）格式符时，会先进行整数提升：负数以符号扩展方式填充高位为全1，最终转换为超大的无符号整数，这就是你看到的异常结果。

解决方法 #

有两种直接有效的修复方式：

方法1：使用unsigned char替代char #

将代码中所有char类型替换为unsigned char，包括结构体、联合体成员及临时变量：

typedef union{
    int integer;
    struct{
        unsigned char b1;
        unsigned char b2;
        unsigned char b3;
        unsigned char b4;
    };
}bytes;

typedef struct{
    unsigned char b1:8;
    unsigned char b2:8;
    unsigned char b3:8;
    unsigned char b4:8;
}bitfield;

// 临时变量同步修改
unsigned char bb1, bb2, bb3, bb4;

unsigned char会把所有字节值当作无符号数处理，不会产生符号扩展，传递给%u时能正确输出原始字节值。

方法2：输出时强制转换为unsigned char #

如果不想修改类型定义，可在printf中把char强制转换为unsigned char，再转为unsigned int匹配格式：
修改PRINT宏：

#define PRINT   printf("b1 %u\n", (unsigned int)(unsigned char)bb1);\
                printf("b2 %u\n", (unsigned int)(unsigned char)bb2);\
                printf("b3 %u\n", (unsigned int)(unsigned char)bb3);\
                printf("b4 %u\n", (unsigned int)(unsigned char)bb4)

先转换为unsigned char消除符号扩展，再转为unsigned int适配%u的格式要求。

注：你的预期输出中bb2应为129而非127，属于注释对应值的笔误。

内容的提问来源于stack exchange，提问作者Tomasz Nowak