C语言 在C中将十六进制转换为字符串？
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Converting hex to string in C?
提问by Dale Reed
Hello I am using digi dynamic c. I am trying to convert this in to string
您好，我正在使用 digi dynamic c。我正在尝试将其转换为字符串
char readingreg[4];
readingreg[0] = 4a;
readingreg[1] = aa;
readingreg[2] = aa;
readingreg[3] = a0;
Currently when I do printfstatements it has to be like this:
目前，当我做printf陈述时，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 printfstatement like this
但我想要这个字符串，所以我可以使用这样的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，为此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，如果有人能告诉我如何一次处理每个元素而不是整个数组，那也没关系，因为只有 4 个元素。
回答by David Ranieri
0xaaoverflows when plain charis signed, use unsigned char:
0xaa当普通char签名时溢出，请使用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;
}
回答by Fiddling Bits
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:
如果您的机器是小端，则必须交换字节顺序：
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.
如果考虑可移植性，则无论字节顺序如何，都应使用方法二。
I get the following output:
我得到以下输出：
printf("0x%s\n", str);
0x4aaaaaa0
0x4aaaaa0
