问题
Currently, I can round a double to an output stream using:
output.setf(std::ios::fixed,std::ios::floatfield);
output.precision(3);
But I'm given a double and I need to make the conversion before I insert it to a vector. So for instance, if the number -0.00078 appears then it equals to 0.000 and I won't need to save it. On the other hand, 1.0009 will become 1.001 (same as the precision function handles it).
How can I convert doubles like that in C++?
回答1:
A common trick is to do it with maths:
value = round( value * 1000.0 ) / 1000.0;
Where round will handle negative and positive values correctly... Something like this (untested):
inline double round( double val )
{
if( val < 0 ) return ceil(val - 0.5);
return floor(val + 0.5);
}
You'll still want to set the decimal places to 3 during output, due to floating point precision problems.
回答2:
Other answers here have given you a technique. But it's important to mention that not all values can be exactly represented in floating-point. 1.001 is a good example; the nearest possible value is 1.00099999999999988987.
So if your aim is to get strictly 3 decimal places, then the answer is: that's not possible.
回答3:
I know this is a very old post but I was looking for a solution to the same problem. However, I did not want to create a special function for it so I came up with the following:
#include <sstream>
#include <iomanip>
...
...
...
double val = 3.14159;
stringstream tmp;
tmp << setprecision(3) << fixed << val;
double new_val = stod(tmp.str()); // new_val = 3.143
tmp.str(string()); // clear tmp for future use
Not sure if this is the best way to do it but it worked for me!
回答4:
You can multiply it by 1000 and then round (or truncate) it; this will give you a value 1000 times the 3-decimal place value. Note that, if you divide it by 1000 to get the 'rounded' value, you may end up w/ more than 3 decimal places (due to round off error).
来源：https://stackoverflow.com/questions/14369673/round-double-to-3-points-decimal