C++实现16进制字符串转换成int整形值

开发中经常需要把16进制字符串转换成整形,写了个个代码供大家参考下:

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

//字符转换成整形
int hex2int(char c)
{
if ((c >= 'A') && (c <= 'Z'))
{
return c - 'A' + 10;
}
else if ((c >= 'a') && (c <= 'z'))
{
return c - 'a' + 10;
}
else if ((c >= '0') && (c <= '9'))
{
return c - '0';
}
}
int main()
{
//十六进制字符串转换成整形
const char* hexStr = "EFA0";
int data[32] = {0};
int count = 0;
for (int i=0; i<strlen(hexStr); i+=2)
{
int high = hex2int(hexStr[i]); //高四位
int low = hex2int(hexStr[i+1]); //低四位
data[count++] = (high<<4) + low;
}
//打印输出
for (int i=0; i<strlen(hexStr)/2; i++)
{
printf("%d ", data[i]);
}
return 1;
}

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