C/C ++以十六进制打印字节，获取奇怪的十六进制值

问题描述

我正在使用以下命令以十六进制的形式打印出数字:

I am using the following to print out numbers from an array in hex:

char buff[1000];

// Populate array....

int i;
for(i = 0; i < 1000; ++i)
{
    printf("[%d] %02x\n", i,buff[i]);
}

但是我有时会打印出奇怪的值:

but I sometimes print weird values:

byte[280] 00
byte[281] 00
byte[282] 0a
byte[283] fffffff4         // Why is this a different length?
byte[284] 4e
byte[285] 66
byte[286] 0a

为什么会打印'fffffff4'?

Why does this print 'fffffff4'?

推荐答案

使用％02hhx 作为格式字符串.

从 CppReference 中，％02x 接受 unsigned int .当您将参数传递给 printf()时，这是 variadic函数， buff [i] 会自动转换为 int .然后格式说明符％02x 使 printf()将该值解释为 int ，因此潜在的负值如(char)-1被解释并打印为(int)-1 ，这就是您观察到的原因.

From CppReference, %02x accepts unsigned int. When you pass the arguments to printf(), which is a variadic function, buff[i] is automatically converted to int. Then the format specifier %02x makes printf() interprets the value as int, so potential negative values like (char)-1 get interpreted and printed as (int)-1, which is the cause of what you observed.

还可以推断出您的平台已签名 char 类型和32位 int 类型.

It can also be inferred that your platform has signed char type, and a 32-bit int type.

长度修饰符 hh 将告诉 printf()解释为 char 类型提供的任何内容，因此％hhx 是 unsigned char 的正确格式说明符.

The length modifier hh will tell printf() to interpret whatever supplied as char type, so %hhx is the correct format specifier for unsigned char.

或者，您可以在打印之前将数据转换为 unsigned char .喜欢

Alternatively, you can cast the data to unsigned char before printing. Like

printf("[%d] %02x\n", i, (unsigned char)buff[i]);

这还可以防止出现负值，因为 int 可以(几乎)始终包含 unsigned char 值.

This can also prevent negative values from showing up as too long, as int can (almost) always contain unsigned char value.

请参见以下示例:

#include <stdio.h>

int main(){
    signed char a = +1, b = -1;
    printf("%02x %02x %02hhx %02hhx\n", a, b, a, b);
    return 0;
}

以上程序的输出为:

01 ffffffff 01 ff

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