This is more of a question of curiosity but does anyone know how negative precision values are handled in C++? For example:
double pi = 3.14159265;
cout.precision(-10);
cout.setf(ios::fixed, ios::floatfield);
cout << pi << endl;
I ve tried this out and using GCC and it seems that the precision value is ignored but I was curious if there is some official line on what happens in this situation.
Strangely (and wrongly, IMHO) the C++ Standard specifies a signed type (streamsize) as the parameter for precision, so it won t be converted to a large number. However, the standard is silent on what a negative number might mean, if anything.
I can t find any specification of this behaviour neither in C++03 nor in draft of C++0x (N3092). However, the C89 standard says
7.19.6.1 The fprintf function
A negative precision argument is taken as if the precision were omitted.
I would expect that C++ std::ostream behaviour is consistent with the C I/O printf. fprintf and related functions.
A quick test with Visual C++ 10.0 and GCC 4.4.1 suggests it to be this way and negative precision means precision omitted. And, precision omitted means a default value which is 6 places as specified in 27.5.4.1 basic_ios constructors in Table 125
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