在C中将十六进制转换为字符串？

Hello I am using digi dynamic c. I am trying to convert this in to string

您好我正在使用digi动态c。我试图将其转换为字符串

char readingreg[4];
readingreg[0] = 4a;
readingreg[1] = aa;
readingreg[2] = aa;
readingreg[3] = a0;

Currently when I do printf statements it has to be like this:

目前,当我做printf语句时,它必须是这样的:

printf("This is element 0: %x\n", readingreg[0]);

But I want this in string so I can use printf statement like this

但我想在字符串中这样,所以我可以像这样使用printf语句

  printf("This is element 0: %s\n", readingreg[0]);

I am essentialy sending the readingreg array over TCP/IP Port, for which I need to have it as string. I cant seem to be able to convert it into string. Thanks for your help. Also if someone can tell me how to do each element at a time rather than whole array, that would be fine to since there will only be 4 elements.

我本质上是通过TCP / IP端口发送readingreg数组,我需要将它作为字符串。我似乎无法将其转换为字符串。谢谢你的帮助。此外,如果有人可以告诉我如何一次完成每个元素而不是整个数组,那将是很好的,因为只有4个元素。

2 个解决方案

#1

0xaa overflows when plain char is signed, use unsigned char:

签名普通char时0xaa溢出,使用unsigned char:

#include <stdio.h>

int main(void)
{
    unsigned char readingreg[4];
    readingreg[0] = 0x4a;
    readingreg[1] = 0xaa;
    readingreg[2] = 0xaa;
    readingreg[3] = 0xa0;
    char temp[4];

    sprintf(temp, "%x", readingreg[0]);
    printf("This is element 0: %s\n", temp);
    return 0;
}

#2

If your machine is big endian, you can do the following:

如果您的机器是大端,您可以执行以下操作:

char str[9];

sprintf(str, "%x", *(uint32_t *)readingreg);

If your machine is little endian you'll have to swap the byte order:

如果你的机器是小端,你将不得不交换字节顺序:

char str[9];
uint32_t host;

host = htonl(*(uint32_t *)readingreg);
sprintf(str, "%x", host);

If portability is a concern, you should use method two regardless of your endianness.

如果可移植性是一个问题,您应该使用方法二,无论您的字节顺序如何。

I get the following output:

我得到以下输出:

printf("0x%s\n", str);

0x4aaaaaa0

#1