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

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
0.1234开发者_JAVA技巧56789

not

0
0.123456789

?

Thanks! Phil

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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!

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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;

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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.