求一份C语言，const char* 转uint8_t数组方法

liven001

已知数据：
const char* keyStr = "7E360165C72B"；

想转换成如下数据
uint8_t keya[6] = { 0x7E, 0x36, 0x01, 0x65, 0xC7, 0x2B };

ZB_陈

百度搜索 C hex 转数组

苏紫方璇

循环取2个字符用strtol转换

无闻无问

循环sscanf也可以实现

wildfire_810

记得做好容错

pzx521521

楼上正解, 这种问题自己搜一下更好

[C] 纯文本查看 复制代码

/**
 * 字符串转字节数组
 * [url=home.php?mod=space&uid=952169]@Param[/url] source
 *  字符串
 * @param length
 *  字符串长度
 * @param target
 *  字节数组
 * @param n
 *  转换后长度
 */
void hex_2_uint8(const char* source, size_t length, uint8_t* target, size_t n) {
    n = 0;
    for (int i = 0; i < length; i += 2) {
        if (source[i] >= 'a' && source[i] <= 'f') {
            target[n] = static_cast<unsigned char>(source[i] - 'a' + 10);
        }
        else if (source[i] >= 'A' && source[i] <= 'F') {
            target[n] = static_cast<unsigned char>(source[i] - 'A' + 10);
        }
        else {
            target[n] = static_cast<unsigned char>(source[i] - '0');
        }
        if (source[i + 1] >= 'a' && source[i + 1] <= 'f') {
            target[n] = static_cast<unsigned char>((target[n] << 4) | (source[i + 1] - 'a' + 10));
        }
        else if (source[i + 1] >= 'A' && source[i + 1] <= 'F') {
            target[n] = static_cast<unsigned char>((target[n] << 4) | (source[i + 1] - 'A' + 10));
        }
        else {
            target[n] = (target[n] << 4) | (source[i + 1] - '0');
        }
        ++n;
    }
}
int main() {
    const char* keyStr = "7E360165C72B";
    uint8_t keya[6] = { 0x7E, 0x36, 0x01, 0x65, 0xC7, 0x2B };
    uint8_t test[6];
    hex_2_uint8(keyStr, 12, test, 6);
}

zsjalive

baidu搜索，很多的。

HEIMEI

int string_to_hex(char *string, uint8_t *hex)
{
        char *p;
        int i;
        uint8_t tmp;

        for (i = 1, p = string; *p != '\0'; p++, i++)
        {
                if (*p > '9')
                {
                        tmp = *p - 55;
                }
                else
                {
                        tmp = *p - '0';
                }
               
                printf("tmp = %#x\n", tmp);

                if (!(i % 2))
                {
                        *hex |= tmp;
                        hex++;       
                }
                else
                {
                        *hex = tmp << 4;
                }
        }

        return 0;
}

似水無痕

[C] 纯文本查看 复制代码

#include <stdio.h>
#include <stdint.h>

#define ArraySize 6

int stringToUint8(const char* str, uint8_t* array, int arraySize);

int main(void)
{
	const char* keyStr = "7E360165C72B";
	uint8_t keya[ArraySize] = {0};
	int arrayValidLen=stringToUint8(keyStr, keya, ArraySize);
	for (int i = 0; i < arrayValidLen; i++)
		printf("%#x\t", keya[i]);
	return 0;
}

int stringToUint8(const char* str, uint8_t* array, int arraySize)
{
	int arrayValidLen = 0;
	for (int arrayElement = 0; arrayValidLen < arraySize && str[arrayValidLen * 2] != '\0'; arrayValidLen++)
	{
		sscanf(&str[arrayValidLen * 2], "%2x", &arrayElement);
		array[arrayValidLen] = arrayElement;
	}
	return arrayValidLen;
}

c03xp

我也发一个,也是以前网上找的，先把字符串整体转换成数字，然后通过位运算得到各个分量，可能数字太大会溢出，可以把函数改良成64位的

[C] 纯文本查看 复制代码

static UCHAR a2xtable[] = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 0, 0, 0, 0, 0,
0,10,11,12,13,14,15, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0,10,11,12,13,14,15, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, };

UINT util_atox(const char* s) {
    UINT res = 0;
    while (*s) {
        res = (res << 4) | a2xtable[*s++];  //利用二进制和十六进制之间的换算方法
    }
    return res;
}