Why is it that when I store the DATA of a double value in a char array if I get the data from the char array to another double, it returns a Float value?
In this:
double vals = 3.141592654;
char xvals[sizeof(vals)];
memcpy(&xvals, &vals, sizeof(xvals));
double y;
memcpy(&y, &xvals, sizeof(xvals));
std::cout<<y<<"\n";
OUTPUT: 3.14159
It does not "return a float value" : std::cout will simply not print all those digits by default.
you can use std::setprecison (from <iomanip>) to change the number of digits to be printed :
#include <iostream>
#include <iomanip>
int main()
{
double vals = 3.141592654;
std::cout << vals << "\n";
std::cout << std::setprecision(10);
std::cout << vals;
return 0;
}
Output:
3.14159
3.141592654
There has been no data loss or forced conversion. The default precision for cout is 6. This will give you the answer you need
std::cout<<std::setprecision(10)<<y<<"\n";
EDIT : You need to include the header <iomanip> for std::setprecision.
double vals = 3.141592654; std::cout<<vals <<"\n";print?float? ISTM it is simply rounded for output.