c - 将 char 字符串分配给 uint32_t(仅限 C，不适用于 C++)

Question

我想知道如何将 uint32_t 变量的值赋给“输入”变量。

请参阅下面的代码。

char *HEXlastName = "0x00D1";
uint32_t *in, input[];

char *utf8;
char word[50] = { "" };
    for(in = input; *in; ++in) {
        utf8 = to_utf8(*in);
        strcat(word, utf8);
        (void)userlog("%s:word--> %s",prc, word);
    }

char *to_utf8(const uint32_t cp)
{
    static char ret[5];
    const int bytes = codepoint_len(cp);

    int shift = utf[0]->bits_stored * (bytes - 1);
    ret[0] = (cp >> shift & utf[bytes]->mask) | utf[bytes]->lead;
    shift -= utf[0]->bits_stored;
    int i;
    for(i=1; i < bytes; ++i) {
        ret[i] = (cp >> shift & utf[0]->mask) | utf[0]->lead;
        shift -= utf[0]->bits_stored;
    }
    ret[bytes] = '\0';
    return ret;
}

我想将“HEXlastName”的值分配给var.: input(uint32_t)。我该怎么做？

char *HEXlastName = "0x00D1";
input = HEXlastName ;   //--> This sentente is wrong. 
                       //How I could move the content of HEXlastName to input

也许这是一个愚蠢的问题，但我是 C 语言新手。

问候

Answer 1

感谢pmg先生的回复，给了我答案。

我有一个名为“input”(uint32_t)的变量，和另一个名为“HEXlocalName”的变量，其中包含带有unicodes (0x0000010x000002)的数组字符串。

此例程将 var“hexlocalname”的内容移动到 var“input”(unit32_t)

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

int main(void) {
    char v[UCHAR_MAX + 1] = {0};
    v['0'] = 0;
    v['1'] = 1;
    v['2'] = 2;
    v['3'] = 3;
    v['4'] = 4;
    v['5'] = 5;
    v['6'] = 6;
    v['7'] = 7;
    v['8'] = 8;
    v['9'] = 9;
    v['a'] = v['A'] = 10;
    v['b'] = v['B'] = 11;
    v['c'] = v['C'] = 12;
    v['d'] = v['D'] = 13;
    v['e'] = v['E'] = 14;
    v['f'] = v['F'] = 15;
    char *HEXlastName = "0x00D10x00C10x00450x00C10x00D1";
    char *p = HEXlastName;
    uint32_t input[33];
    size_t n = 0;
    while (*p) {
        input[n++] = (v[(unsigned)p[2]]<<12) + (v[(unsigned)p[3]]<<8) + (v[(unsigned)p[4]]<<4) + (v[(unsigned)p[5]]);
        p += 6;
    }
    for (int k = 0; k < n; k++) {
        printf("input[%d] is 0x%04X\n", k, input[k]);
    }
    return 0;
}