printf char 为 c 中的十六进制 [英] printf char as hex in c

问题描述

我希望从下面的代码输出类似 \9b\d9\c0... 的东西，但我得到 \ffffff9b\ffffffd9\ffffffc0\ffffff9d\53\ffffffa9\fffffff4\49\ffffffb0\ffffffeff\ffffffd9\ffffffaa\61\fffffff7\54\ffffffb.我向 char 添加了显式转换，但它没有效果.这是怎么回事?

I expect ouput something like \9b\d9\c0... from code below, but I'm getting \ffffff9b\ffffffd9\ffffffc0\ffffff9d\53\ffffffa9\fffffff4\49\ffffffb0\ffff ffef\ffffffd9\ffffffaa\61\fffffff7\54\fffffffb. I added explicit casting to char, but it has no effect. What's going on here?

typdef struct PT {
    // ... omitted
    char GUID[16];
} PT;
PT *pt;
// ... omitted
int i;
for(i=0;i<16;i++) {
    printf("\\%02x", (char) pt->GUID[i]);
}

只转换为 (unsigned char) 对我有用.编译器在使用 %02hhx (gcc -Wall) 时向我发出警告.(unsigned int) 没有效果.

only casting to (unsigned char) worked for me. Compiler spits warnings on me when using %02hhx (gcc -Wall). (unsigned int) had no effect.

推荐答案

发生这种情况的原因是您系统上的 char 已签名.当您将它们传递给具有可变数量参数的函数时，例如 printf(在签名的固定参数部分之外)char 将转换为 int，它们在这个过程中得到了符号扩展.

The reason why this is happening is that chars on your system are signed. When you pass them to functions with variable number of arguments, such as printf (outside of fixed-argument portion of the signature) chars get converted to int, and they get sign-extended in the process.

要解决此问题，请将值强制转换为 unsigned char:

To fix this, cast the value to unsigned char:

printf("\\%02hhx", (unsigned char) pt->GUID[i]);

演示.

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