字符型数据0x80强转成unsigned int为什么会扩展符号位？

Question

#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

Answer 1

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?]