正确的方法是使用二进制位域打印数字

Question

这是我使用二进制位字段打印 int 数字找到的代码，但我也仅使用 unsigned、int 读取该位字段的类型，因此，它是否合法此代码使用 char 类型？

struct bits
{
    unsigned char ch1 : 1;//01
    unsigned char ch2 : 1;
    unsigned char ch3 : 1;
    unsigned char ch4 : 1;
    unsigned char ch5 : 1;
    unsigned char ch6 : 1;
    unsigned char ch7 : 1;
    unsigned char ch8 : 1;
};

void main()
{
    int data = -1;
    int length = 4;

    struct bits *p = &data;

    while (length--)
    {
        printf("%d%d%d%d %d%d%d%d ",
            (p + length)->ch8,
            (p + length)->ch7,
            (p + length)->ch6,
            (p + length)->ch5,
            (p + length)->ch4,
            (p + length)->ch3,
            (p + length)->ch2,
            (p + length)->ch1
            );
    }
    system("pause");
}

Answer 1

GCC 是allowing作为位域的任何整数类型，与 C99 和 C11 标准相反：

Allowable bit-field types other than _Bool, signed int, and unsigned int (C99 and C11 6.7.2.1).
Other integer types, such as long int, and enumerated types are permitted even in strictly conforming mode.

Answer 2

A bit-field shall have a type that is a qualified or unqualified version of _Bool, signed int, unsigned int, or some other implementation-defined type.

某些编译器允许使用 unsigned char 作为 "implementation-defined type"。

(p + length)->ch8 将通过通常的 整数提升 作为可变参数函数的参数。因此 ch8 将被提升为匹配 "%d".

的 int

代码可以使用 unsigned

struct bits {
    unsigned ch1 : 1;//01
    unsigned ch2 : 1;
    unsigned ch3 : 1;
    unsigned ch4 : 1;
    unsigned ch5 : 1;
    unsigned ch6 : 1;
    unsigned ch7 : 1;
    unsigned ch8 : 1;
};

正确的方法是使用二进制位域打印数字

the right way use bit field print number in binary

c

bit-fields