十六进制字符串到无符号字符[] [英] hex string to unsigned char[]

问题描述

今天我尝试将十六进制字符串转换为无符号字符[]

today I tried to convert a hex string to an unsigned char[]

string test = "fe5f0c";
unsigned char* uchar= (unsigned char *)test.c_str();
cout << uchar << endl;

这导致了

fe5f0c

hrmpf :-(.所需的行为如下:

hrmpf :-(. The desired behaviour would be as follows:

unsigned char caTest[2]; 
caTest[0] = (unsigned char)0xfe;
caTest[1] = (unsigned char)0x5f;
caTest[2] = (unsigned char)0x0c;
cout << caTest << endl;

打印不可读的 ascii 代码.我经常做错事^^.将不胜感激任何建议.

which prints unreadable ascii code. As so often I am doing something wrong ^^. Would appreciate any suggestions.

提前致谢

推荐答案

当然，您只需在解析后隔离您感兴趣的位:

Sure, you just have to isolate the bits you are interested in after parsing:

#include <string>
#include <cstdlib>
#include <iostream>

typedef unsigned char byte;

int main()
{
    std::string test = "40414243";
    unsigned long x = strtoul(test.c_str(), 0, 16);
    byte a[] = {byte(x >> 24), byte(x >> 16), byte(x >> 8), byte(x), 0};
    std::cout << a << std::endl;
}

请注意，我将输入字符串更改为八位数字，否则数组将以值 0 开头，并且 operator<< 会将其解释为结尾，而您不会能够看到任何东西.

Note that I changed the input string to an eight digit number, since otherwise the array would start with the value 0, and operator<< would interpret that as the end and you wouldn't be able to see anything.

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