將十六進制字符串轉換為字節數組轉換可變長度十六進制字符串的最佳方法是什么？"01A1"到包含該數據的字節數組。即轉換為：std::string = "01A1";變成這樣char* hexArray;int hexLength;或者這個std::vector<char> hexArray;所以當我把這個寫到一個文件hexdump -C我得到的二進制數據包含01A1.
4 回答
哈士奇WWW
TA貢獻1799條經驗 獲得超6個贊
#include <cstdint>#include <initializer_list>#include <stdexcept>#include <utility>/* Quick and dirty conversion from a single character to its hex equivelent */constexpr std::uint8_t HexCharToInt(char Input){
return
((Input >= 'a') && (Input <= 'f'))
? (Input - 87)
: ((Input >= 'A') && (Input <= 'F'))
? (Input - 55)
: ((Input >= '0') && (Input <= '9'))
? (Input - 48)
: throw std::exception{};}/* Position the characters into the appropriate nibble */constexpr std::uint8_t HexChar(char High, char Low){
return (HexCharToInt(High) << 4) | (HexCharToInt(Low));}/* Adapter that performs sets of 2 characters into a single byte and combine the results into a uniform initialization list used to initialize T */template <typename T, std::size_t Length, std::size_t ... Index>constexpr T HexString(const char (&Input)[Length], const std::index_sequence<Index...>&){
return T{HexChar(Input[(Index * 2)], Input[((Index * 2) + 1)])...};}/* Entry function */template <typename T, std::size_t Length>constexpr T HexString(const char (&Input)[Length]){
return HexString<T>(Input, std::make_index_sequence<(Length / 2)>{});}constexpr auto Y = KS::Utility::HexString<std::array<std::uint8_t, 3>>("ABCDEF");
蝴蝶刀刀
TA貢獻1801條經驗 獲得超8個贊
BIGNUM *input = BN_new();int input_length = BN_hex2bn(&input, argv[2]);input_length = (input_length + 1) / 2; // BN_hex2bn() returns number of hex digitsunsigned char *input_buffer = (unsigned char*)malloc(input_length);retval = BN_bn2bin(input, input_buffer);
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