C ++将字符串转换为十六进制，反之亦然 [英] C++ convert string to hexadecimal and vice versa

问题描述

在C ++中将字符串转换为十六进制和反之亦然的最好方法是什么？

What is the best way to convert a string to hex and vice versa in C++?

示例：

• 字符串Hello World到十六进制格式： 48656C6C6F20576F726C64


• 从十六进制 48656C6C6F20576F726C64 到字符串：Hello World

• A string like "Hello World" to hex format: 48656C6C6F20576F726C64
• And from hex 48656C6C6F20576F726C64 to string: "Hello World"

推荐答案

一个字符串，如Hello World，十六进制格式：48656C6C6F20576F726C64。 p>

A string like "Hello World" to hex format: 48656C6C6F20576F726C64.

啊，你在这里：

#include <string>

std::string string_to_hex(const std::string& input)
{
    static const char* const lut = "0123456789ABCDEF";
    size_t len = input.length();

    std::string output;
    output.reserve(2 * len);
    for (size_t i = 0; i < len; ++i)
    {
        const unsigned char c = input[i];
        output.push_back(lut[c >> 4]);
        output.push_back(lut[c & 15]);
    }
    return output;
}

#include <algorithm>
#include <stdexcept>

std::string hex_to_string(const std::string& input)
{
    static const char* const lut = "0123456789ABCDEF";
    size_t len = input.length();
    if (len & 1) throw std::invalid_argument("odd length");

    std::string output;
    output.reserve(len / 2);
    for (size_t i = 0; i < len; i += 2)
    {
        char a = input[i];
        const char* p = std::lower_bound(lut, lut + 16, a);
        if (*p != a) throw std::invalid_argument("not a hex digit");

        char b = input[i + 1];
        const char* q = std::lower_bound(lut, lut + 16, b);
        if (*q != b) throw std::invalid_argument("not a hex digit");

        output.push_back(((p - lut) << 4) | (q - lut));
    }
    return output;
}

（假设一个char有8位，但你可以从这里带走。）

(This assumes that a char has 8 bits, so it's not very portable, but you can take it from here.)

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