Setting minimum number of decimal places for std::ostream precision

问题

Is there a way to set the "minimum" number of decimal places that a std::ostream will output?

For example, say I have two unknown double variables that I want to print (values added here for the sake of illustration):

double a = 0;
double b = 0.123456789;

I can set my maximum decimal precision so that I output b exactly

std::cout << std::setprecision(9) << b << std::endl;
>>> 0.123456789

Is there a way to set a "minimum" precision (a minimum number of decimal places), while retaining the "maximum" precision, so that

std::cout << a << std::endl << b << std::endl;

yields

0.0
0.123456789

not

0
0.123456789

?

Thanks! Phil

---

the short answer to this is "No". The stream has only one precision setting, with no facility to differentiate between maximum and minimum precision. Thanks all for your generous advice!

回答1:

I do not think there is a way to achieve what you are asking without turning the number into a string (with high precision), and stripping off the trailing zeros.

This is appropriate, because just because there are trailing zeros doesn't mean there isn't precision there, and the run-time can't tell that.

For example, if I measure the weight of an object with a cheap scale, it may be 1.0 kgs.

If I weigh it with a high-accuracy scale, it may be 1.00000 kgs. Just because there are trailing zeros, doesn't mean that accuracy should be discarded.

回答2:

cout << setprecision(1) << fixed << d << endl;

Used fixed after setprecision.

Edit: This is what you want. It would change precision based on d.

cout << setprecision(d?9:1) << fixed << d << endl;

来源：https://stackoverflow.com/questions/2489627/setting-minimum-number-of-decimal-places-for-stdostream-precision