我要做的是将double转换为十六进制字符串然后再转换为double.
以下代码执行转换为双十六进制字符串.
char * double2HexString(double a)
{
char *buf = new char[17]; // double is 8-byte long,so we have 2*8 + terminating \0
char *d2c;
d2c = (char *) &a;
char *n = buf;
int i;
for(i = 0; i < 8; i++)
{
sprintf(n,"%02X",*d2c++);
n += 2;
}
*(n) = '\0';
}
这似乎工作,但是,我不知道如何将结果字符串转换回double.
请指教 ：)
解决方法
我很惊讶地看到没有人提出标准解决方案,这是ISO C99标准中的％格式说明符.
#include <iostream> #include <string> #include <stdio.h> std::string double2hexastr(double d) { char buffer[25] = { 0 }; ::snprintf(buffer,25,"%A",d); // TODO Check for errors return buffer; } double hexastr2double(const std::string& s) { double d = 0.0; ::sscanf(s.c_str(),"%lA",&d); // TODO Check for errors return d; } int main() { std::cout << "0.1 in hexadecimal: " << double2hexastr(0.1) << std::endl; std::cout << "Reading back 0X1.999999999999AP-4,it is "; std::cout << hexastr2double("0X1.999999999999AP-4") << std::endl; }
