The value range (minimum & maximum)

Question

What exactly the maximum and the minimum value of any definition type? Is this possible?

unsigned int maximum_uint = (maximum_value)(unsigned int);
short minimum_short = (minimum_value)(short);
float maximum_float = (maximum_value)(float);

score 2 · Answer 1 · edited Sep 12 '13 at 19:57

2

What you have written is probably not possible.

The limits of various types are provided in the C-style C++ header climits and some in the C++ header limits

See :

climits

limits

edited Sep 12 '13 at 19:57

answered Jan 30 '13 at 08:17

asheeshr

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Great, I was not knowing this. But it will really help. – Pranit P Kothari Jan 30 '13 at 08:21
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There is no `climits.h` header. There is only `limits.h` (which is deprecated), `climits` (which is the C++ version of C's `limits.h`) and `limits` (which should be preferred over the other two, anyway). – Christian Rau Jan 30 '13 at 08:29
@ChristianRau Thanks for pointing that out. Typo. I meant climits – asheeshr Jan 30 '13 at 08:30
@ChristianRau I dont understand what you mean by that. I made the change – asheeshr Jan 30 '13 at 08:35
@AshRj That change wasn't that severe, since `limits.h` exists pretty well (even if being deprecated). What doesn't exist is `climits.h`. But never mind, doing it myself. – Christian Rau Jan 30 '13 at 08:37

score 2 · Accepted Answer · answered Jan 30 '13 at 08:26

2

#include <limits>

unsigned int maximum_uint = std::numeric_limits<unsigned int>::max();
short minimum_short = std::numeric_limits<short>::min();
float maximum_float = std::numeric_limits<float>::max();

answered Jan 30 '13 at 08:26

Marius Bancila

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You forgot the `std::` qualification (or an apprpriate `using` statement). – Christian Rau Jan 30 '13 at 08:27

The value range (minimum & maximum)

2 Answers2