c语言-10进制字符串转16进制字符串

代码1-动态分配:

#include <stdio.h>
#include <string.h>
#include <malloc.h>
char *DecToHex(char *pHex,char *pDec,int Declen)
{
	int       i,Hexlen = Declen *2;
	char        hex[] = "0123456789";
	pHex = (char *)malloc(sizeof(char)*Hexlen);
	for (i = 0; i < Declen; i++)
	{
		*pHex++ = hex[*pDec>>4];
		*pHex++ = hex[*pDec++ & 0x0F];
	}
	*pHex = '\0';
	return pHex -Hexlen;
}
int main()
{
	char Dec[] = "01234";
	char *pHex = NULL;
	printf("-------------10进制字符串----------\n");
	printf("%s\n",Dec);
	printf("-------------16进制字符串----------\n");
	printf("%s\n", DecToHex(pHex,Dec,strlen(Dec)));
	printf("\n\n\n");
}

代码2-非动态分配:

#include <stdio.h>
#include <string.h>
#include <malloc.h>
char *DecToHex(char *pHex,char *pDec,int Declen)
{
	int       i,Hexlen = Declen *2;
	char        hex[] = "0123456789";
	for (i = 0; i < Declen; i++)
	{
		*pHex++ = hex[*pDec>>4];
		*pHex++ = hex[*pDec++ & 0x0F];
	}
	*pHex = '\0';
	return pHex -Hexlen;
}
int main()
{
	char Dec[] = "01234";
	char Hex[11];
	printf("-------------10进制字符串----------\n");
	printf("%s\n",Dec);
	printf("-------------16进制字符串----------\n");
	printf("%s\n", DecToHex(Hex,Dec,strlen(Dec)));
	printf("\n\n\n");
}

当然,如果你想完成Ascii码到Hex的转换,只需要将hex[]="0123456789"改成,hex[]="0123456789ABCDEF"。这也是为什么上面没做保护性判断,因为这个函数主要就是完成Ascii到Hex的转换,只是在这里是一种更特殊的情况。

运行结果:

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转载自blog.csdn.net/u013598629/article/details/82966366