In limits.h, there are #defines for INT_MAX and INT_MIN (and SHRT_* and LONG_* and so on), but only UINT_MAX.
Should I define UINT_MIN myself? Is 0 (positive zero) a portable value?
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Ariel Bold
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2+1 for caring enough about portability and correctness to ask. – Thom Smith Aug 17 '10 at 04:29
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3I should give a -1 for senselessly wanting to remove the "magic number" 0. But oh well. – R.. GitHub STOP HELPING ICE Aug 17 '10 at 06:11
3 Answers
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It's an unsigned integer - by definition its smallest possible value is 0. If you want some justification besides just common sense, the standard says:
6.2.6.2 Integer types
- For unsigned integer types other than
unsigned char, the bits of the object representation shall be divided into two groups: value bits and padding bits (there need not be any of the latter). If there are N value bits, each bit shall represent a different power of 2 between 1 and 2^(N−1), so that objects of that type shall be capable of representing values from 0 to 2^(N−1) using a pure binary representation; this shall be known as the value representation. The values of any padding bits are unspecified.
Carl Norum
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@Ariel, how does it not answer your question? The minimum value for an unsigned type is defined by the standard to be 0. – Carl Norum Aug 18 '10 at 06:29
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the last `2^(N-1)` should be `(2^N)-1` instead e.g., for `N=8` `UCHAR_MAX = (2^8)-1 = 255` – jfs Nov 15 '17 at 15:49
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You could use std::numeric_limits<unsigned int>::min().
Kirill V. Lyadvinsky
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2This will work in C++, and this will not work in C. What is C/C++? – Kirill V. Lyadvinsky Aug 18 '10 at 04:57
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Our source tree contains C and C++ and so this need to work with both. This is why I tagged the question with both. – Ariel Bold Aug 18 '10 at 05:24
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If you want to be "typesafe" you could use 0U, so if you use it in an expression you will have the correct promotions to unsigned.
Jens Gustedt
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1Thinking of it, it is perhaps not so a good idea to have a file that has the same name as a standard file or to mess around with the later. I'd put it another place. If you want to have a `+` to the `0U` (why?) you should have the whole in parenthesis `(+0U)` since otherwise you might have unwanted effects when the `+` could be interpreted as addition. – Jens Gustedt Aug 18 '10 at 06:53
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You don't need parenthesis. Even in `1+UINT_MIN`, it would expand to `1+ +0U`, which is a correct expression. Macro expansion happens at token level, and will not glue the two plusses together. – MSalters Aug 18 '10 at 10:47
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@MSalters: I think you need it since it makes invalid sequences valid. Something like `5 UINT_MIN` would become valid, where it really shouldn't. – Jens Gustedt Aug 18 '10 at 11:26
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@Jens: Using that logic, I can equally say that you must _not_ have parenthesis as it makes invalid sequences valid. Something like `sin UINT_MIN` becomes valid. – MSalters Aug 18 '10 at 14:46
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@MSalters: good point. So it is a matter of taste which invalid sequences should become valid ;-) Brings us back to the question, why using the `+` in the first place? `0U` would be much simpler, then, and is really a well defined constant. – Jens Gustedt Aug 18 '10 at 15:26
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