在 C 中将大位字符串转换为十六进制

Question

我需要像这样转换大位字符串：

unsigned char* key = "0111010111010101010101010100101011010";

像这样的十六进制字符串：

unsigned char* string = EBAAAA95A;

问题是我的密钥通常长于 50 或 60 位，当我尝试用

strtoll(key, NULL, 2)

转换它们时，返回的数字远远大于

long long int

甚至可以容纳的数字。有没有一种有效的方法可以直接将其转换为十六进制？

非常感谢您的帮助！

Answer 1

我认为你需要推出自己的实现。这是一个从@mediocrevegetable 获得灵感的实现

#include <stdlib.h>
#include <string.h>

static inline int bitstr_to_int(const char *str, int len)
{
        int h = 0;
        for (int i = 0; i < len; i++) {
                switch (str[i]) {
                        case '0': h *= 2; continue;
                        case '1': h *= 2; h++; continue;
                        default: return -1;
                }
        }
        return h;
}

char * bitstr_to_hexstr(const char *str, int len)
{
        static const char hex_digit[] = {
                '0', '1', '2', '3',
                '4', '5', '6', '7',
                '8', '9', 'A', 'B',
                'C', 'D', 'E', 'F'
        };

        int i = 0; //traces the input chars in bitstr
        int j = 0; //traces the output chars in hexstr
        int h = 0;
        int r = len % 4;
        int q = len / 4;

        char *hexstr = malloc(q + !!r + 1);
        if (hexstr == NULL) return NULL;

        if (r != 0) {
                h = bitstr_to_int(str, r);
                if (h == -1) goto err;
                hexstr[j++] = hex_digit[h];
        }

        for (i = r; i < len; i += 4) {
                h = bitstr_to_int(str+i, 4);
                if (h == -1) goto err;
                hexstr[j++] = hex_digit[h];
        }

        hexstr[j] = '\0';
        return hexstr;
err:
        free(hexstr);
        return NULL;
}

int main(void)
{
    const char *key = "0111010111010101010101010100101011010";

    char *res = bitstr_to_hexstr(key, strlen(key));
    printf("%s\n", res);
}

Answer 2

如果您需要处理大数字，一个好的库（也许是首选库）是 GMP。这是一个简单的示例，演示了您询问的转换：

#include <stdio.h>
#include <gmp.h>

int main()
{
    char *instr = "0111010111010101010101010100101011010";
    char outstr[100];
    mpz_t x;
    mpz_init(x);
    mpz_set_str(x, instr, 2);                     /* convert from base 2 */
    printf("%s\n", mpz_get_str(outstr, 16, x));   /* convert to base 16 */
}

Answer 3

如果字符串超过 64 位，您可以一次转换一个十六进制数字，确保只有第一个数字使用少于 4 位。

这是一个带有测试用例的实现：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

/* convert len bits to a number */
static unsigned bin2hex_val(const char *str, size_t len) {
    unsigned v = 0;
    while (len --> 0) {
        v = v * 2 + (*str++ - '0');
    }
    return v;
}

char *bin2hex_str(const char *str) {
    static const char hex_digits[16] = "0123456789ABCDEF";
    size_t len = strspn(str, "01"); /* number of binary digits */
    size_t hexlen = (len + 3) / 4;     /* number of hex digits */
    char *p = malloc(hexlen + 1);   /* allocate the hex string */
    size_t i = 0;

    if (p) {
        /* special case the first hex digit if fewer than 4 bits */
        if (len % 4) {
            p[i++] = hex_digits[bin2hex_val(str, len % 4)];
            str += len % 4;
            len -= len % 4;
        }
        /* convert remaining groups of 4 bits as hex digits */
        while (len > 0) {
            p[i++] = hex_digits[bin2hex_val(str, 4)];
            str += 4;
            len -= 4;
        }
        p[i] = '\0';
    }
    return p;
}

void test(const char *str) {
    char *p = bin2hex_str(str);
    printf("%s -> %s\n", str, p);
    free(p);
}

int main() {
    test("");
    test("0");
    test("1");
    test("01");
    test("011");
    test("0111");
    test("01110");
    test("011101");
    test("0111010");
    test("01110101");
    test("01110101110101010101010101001010");
    test("0111010111010101010101010100101011010011101011101010101010101010");
    test("01110101110101010101010101001010110100111010111010101010101010100101011010");
    test("ABC");
    test("0.0");
    test("123");
    return 0;
}

输出：

 ->
0 -> 0
1 -> 1
01 -> 1
011 -> 3
0111 -> 7
01110 -> 0E
011101 -> 1D
0111010 -> 3A
01110101 -> 75
01110101110101010101010101001010 -> 75D5554A
0111010111010101010101010100101011010011101011101010101010101010 -> 75D5554AD3AEAAAA
01110101110101010101010101001010110100111010111010101010101010100101011010 -> 1D755552B4EBAAAA95A
ABC ->
0.0 -> 0
123 -> 1