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字符型数据0x80强转成unsigned int为什么会扩展符号位?

#include <stdio.h>

int main(int argc, char *argv[])
{
    printf("%x
", (unsigned int)(unsigned char)0x80);
    printf("%x
", (unsigned int)(signed char)0x80);
    
    return 0;
}

输出:

80
ffffff80

第一个输出80可以理解,毕竟是从0x80转换为无符号int
第二个输出ffffff80我无法理解,从1字节数据扩展为4字节数据,并且是扩展到无符号类型,为什么符号位也跟着扩展了?

P.S. C/C++中输出结果一致


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1 Answer

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by (71.8m points)

负数转 unsigned 就是这样规定的啊。

原话是这样的(C++17)

conv.integral:

If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2n where n is the number of bits used to represent the unsigned type). [?Note: In a two's complement representation, this conversion is conceptual and there is no change in the bit pattern (if there is no truncation). ?—?end note?]

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