将浮点数赋给unsigned char时发生了什么？ [英] what happened when assign a float to unsigned char ?

问题描述

#include <stdio.h>
#include <iostream>
using namespace std;

int main() {
    float f = -1;
    unsigned char c = f;
    return 0;
}

我检查& f和& c的内容如下：

I check the contents of &f and &c as follows:

(gdb) x/xw &f
0x7fffffffe2b8: 0xbf800000
(gdb) x/xw &c
0x7fffffffe2bf: 0x000000ff

为什么& c看起来像这样？

why does &c look like this ?

推荐答案

转换 -1 浮点到无符号字符，它被强制转换为 0xffffffff 的整数，所以将它存储在一个字符（字节）中给出 0xff ，如您所见。

Converting -1 in floating point to an unsigned character it is coerced to an integer which would be 0xffffffff, so storing that in a character (byte) gives 0xff, displayed as you see.

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