Is there a C++ cross-platform library that provides me with a portable maximum integer number?
I want to declare:
const int MAX_NUM = /* call some library here */;
I use MSVC 2008 unmanaged.
+6
A:
See limits.h (C) or climits (C++). In this case you would want the INT_MAX constant.
Lukáš Lalinský
2009-11-13 21:24:37
+28
A:
In the C++ standard library header <limits>, you will find:
std::numeric_limits<int>::max()
Which will tell you the maximum value that can be stored in a variable of type int. numeric_limits is a class template, and you can pass it any of the numeric types to get the maximum value that they can hold.
The numeric_limits class template has a lot of other information about numeric types as well.
James McNellis
2009-11-13 21:25:52
Actually it's called a "function template", not the other way around. But otheriwse that's very good advice. +1
sbi
2009-11-13 21:47:44
@sbi: Thank you for the correction. The C++ standard does indeed call them "function templates" (14.8.1), but also uses the phrase "nontemplate function," which makes me think that the phrase "template function" might also be acceptable; even highly reputable C++ references (http://www.parashift.com/c++-faq-lite/templates.html#faq-35.13) use the phrase "template function." In any case, I've updated the answer with the unambiguously correct terminology. :-)
James McNellis
2009-11-13 21:57:41
Comeau has a faq entry on that: http://www.comeaucomputing.com/techtalk/templates/#terms
Georg Fritzsche
2009-11-13 22:03:56
@gf: Thanks; I haven't seen that FAQ before (I've not used Comeau), but there is a lot of interesting stuff there. I have had to correct my answer again, though, since there aren't any template functions or function templates in it. `numeric_limits` is a class template.
James McNellis
2009-11-13 22:28:55
In a similar vein, talking about word-order, this "portably provides a maximum integer", it doesn't "provide a portable maximum integer". There does not exist an integer which is a portable maximum integer, for precisely the reason `numeric_limits` exists, which is that it's different on different implementations ;-)
Steve Jessop
2009-11-14 03:54:44
A:
I know the answer has been given but I just want to know from my old days, I used to do
int max = (unsigned int)-1
Will it give the same as
std::numeric_limits<int>::max()
?
Mike Gleason jr Couturier
2009-11-13 21:51:17
For future reference, StackOverflow works best as a 1 Question -> Many Answer type of thing. If you have a question (and the one you bring up seems reasonable), feel free to post a brand new question. Here, it looks like you're trying to offer advice that others perseive as incorrect.
Outlaw Programmer
2009-11-13 22:25:39
To expand on what Andrey says: no, it won't give the same. `(int)(unsigned int)-1` is -1 (well, actually it's implementation-defined, but it's usually -1), whereas `std::numeric_limits<int>::max()` is a large positive `int` value. However, `(unsigned int)-1` is equal to `std::numeric_limits<unsigned int>::max()`, always.
Steve Jessop
2009-11-14 04:00:31