C：将十六进制字符串解析为无符号长

Question

我有一个带有 IP 地址的字符串：

char *input_string = "fe80000000000000022318fffeedef59";

我需要将其转换为无符号长整数：

unsigned long l = 0xfe80000000000000022318fffeedef59;

我尝试了一些代码，但是 none 有效，这是我做的：

#include <stdio.h>
#include <stdlib.h>
void main() 
{
    char *input_string = "fe80000000000000022318fffeedef59";
    printf("input_string : %s\n\n", input_string);

    unsigned long l1 = strtol(input_string, NULL, 16);
    printf("unsigned long l1 = strtol(input_string, NULL, 16) : l1 = %lx \n\n", l1);

    unsigned long l2 = atol(input_string);
    printf("unsigned long l2 = atol(input_string) : l2 = %lx \n\n", l2);
}

输出：

input_string : fe80000000000000022318fffeedef59

unsigned long l1 = strtol(input_string, NULL, 16) : l1 = 7fffffffffffffff

unsigned long l2 = atol(input_string) : l2 = 0

Answer 1

你的电话号码太大了。您应该检查函数 strtol 和 atol 的 return 状态。

Answer 2

您输入的字符串（32 字节）对应于一个非常大的十六进制数。为了保存这样一个 32 字节的字符串，您需要一个 16 字节（128 位）的数据类型。但是 unsigned long 只有 8 个字节。这就是您得到这样输出的原因。

Answer 3

这是我最终完成的。

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

void big_hex_string_to_long_array(char *);

void main( void)
{
    char *ipv6_string = "FE800000000000000202B3FFFE1E8329";

    big_hex_string_to_long_array(ipv6_string);

}

/*
  * The ipv6 address is 128 bits long, too much to process in a single block.
  * Here I split the address in four equals parts of 32 bits which can be stored
  * in an long int (4 Bytes).
  */
void big_hex_string_to_long_array(char * hex_string)
{
    // create two buffers, one for the most significants bytes and the other for the lowest significants bytes
    char buf_1[8+1];    // +1 : the space for '[=10=]'
    char buf_2[8+1];
    char buf_3[8+1];
    char buf_4[8+1];

    // copy each half in each buffers and add ending character
    memcpy(buf_1, &(*hex_string), sizeof(char)* 8);     // copy the 8 first characters in buf_1
    buf_1[8]='[=10=]';                      // set the last character 
    memcpy(buf_2, &(*hex_string)+8, sizeof(char)* 8);   // copy the 8 next characters in buf_2
    buf_2[8]='[=10=]';                      //...                       
    memcpy(buf_3, &(*hex_string)+16, sizeof(char)* 8);
    buf_3[8]='[=10=]';
    memcpy(buf_4, &(*hex_string)+24, sizeof(char)* 8);
    buf_4[8]='[=10=]';

    printf("\nchar arrays : \nbuf1 = %s\nbuf2 = %s\nbuf3 = %s\nbuf4 = %s\n",buf_1, buf_2, buf_3, buf_4);

    // store each buffer as unsigned long
    long l1 = strtol(buf_1, NULL, 16);      // convert string to long
    long l2 = strtol(buf_2, NULL, 16);
    long l3 = strtol(buf_3, NULL, 16);
    long l4 = strtol(buf_4, NULL, 16);

    printf("\nlong int : \nl1 = %lx\nl2 = %lx\nl3 = %lx\nl4 = %lx\n",l1, l2, l3, l4);
}

输出

char arrays :

buf1 = FE800000

buf2 = 00000000

buf3 = 0202B3FF

buf4 = FE1E8329

long int :

l1 = fffffffffe800000

l2 = 0

l3 = 202b3ff

l4 = fffffffffe1e8329

Answer 4

作为 @Weather Vane 评论："You'll need a 128-bit numeric variable to hold the value from a 32-byte hex string"

可以使用sscanf()保存成2个64位整数。

#include <stdio.h>
#include <stdint.h>
int main(void) {
  uint64_t add[2];
  char *input_string = "fe80000000000000022318fffeedef59";
  if (2 != sscanf(input_string, "%16" SCNx64 "%16" SCNx64, &add[0], &add[1])) {
    return -1;
  }
  printf("%016" PRIx64 "%016" PRIx64 , add[0], add[1]);
  return 0;
}

C：将十六进制字符串解析为无符号长

C : Parse Hex string to unsigned long

c

casting

long-integer