Meaning of U suffix

Question

What does the postfix (or suffix) U mean for the following values?

0U
100U

Note that where you see U, it's common to use the lowercase version "u", because when it's capitalized it can be easily be confused with hex constants. I.e. 0xFFFFFFFFu is more obvious than 0xFFFFFFFFU. — Billy ONeal, Dec 07 '10 at 19:41
the title says all, so why i need to write the question body? — lovespring, Dec 07 '10 at 20:25
@Sugar: I guess I use F a lot more than 0 and 1 in hex constants :) — Billy ONeal, Feb 26 '18 at 22:49

score 76 · Accepted Answer · edited Mar 18 '19 at 12:15

76

It stands for unsigned.

When you declare a constant, you can also specify its type. Another common example is L, which stands for long. (and you have to put it twice to specify a 64-bit constant).

Example: 1ULL.

It helps in avoiding explicit casts.

edited Mar 18 '19 at 12:15

sixter

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answered Dec 07 '10 at 19:22

ruslik

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There are also cases where it's necessary. For instance, integral constants are interpreted as integers by the compiler, so a constant like `0xffffffffffffffff` will lose its high 32 bits without the `ll` suffix. – zneak Dec 14 '11 at 18:48
2

@zneak actually, it won't. Integral literals have the type that can hold the value (if such a type exists). See [The type of the literal](https://en.cppreference.com/w/cpp/language/integer_literal#The_type_of_the_literal). Thus, the type of your example literal is `long unsigned int` on GCC 8.4.0 x86_64. – Ruslan Sep 04 '20 at 12:44

score 46 · Answer 2 · answered Dec 07 '10 at 19:27

46

Integer constants in C and C++ can optionally have several suffixes:

123u      the value 123 is an unsigned int
123l       (that's a lowercase L) 123 is a signed long
123L      ditto
123uL    unsigned long
123LL    a signed long long, a 64 bit or 128 bit value (depending on the environment)
123uLL unsigned long long

answered Dec 07 '10 at 19:27

wallyk

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[cppreference](http://en.cppreference.com/w/cpp/language/integer_literal) says for "12345...u" that "the type is unsigned long long even without a long long suffix". Quite confusing, since your answer says 123u is unsigned int. Who is right, or where is the misunderstanding? – Ela782 Mar 23 '18 at 22:47
@Ela782: I looked at cpp reference, but couldn't find anything which says "u" implies long long. Did I misunderstand? – wallyk Mar 23 '18 at 23:59
If you click my link in the comment above, near the bottom of that page, as a comment in the code example, it says: `<< 12345678901234567890u << '\n'; // the type is unsigned long long even without a long long suffix`. Did you find that? It's certainly possible that I misunderstand that sentence. – Ela782 Mar 24 '18 at 10:42
@Ela782: Oh, I see. That comment presumes it is compiled for a machine which *has* `long long` and since the constant exceeds the range for `long` it is already a `long long`. – wallyk Mar 26 '18 at 17:59
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Ooh I get it now, I see! Thank you! You mean "since it exceeds the range for (unsigned) `int`" though, because `u` would make it an (unsigned) `int` - but because the number is so large, it'll make an (unsigned) `long long`. Correct? – Ela782 Mar 27 '18 at 19:45
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@Ela782: Yep. I am not sure off the top of my head, but without the `u` it could well be a signed long long. (checked) Nope, it overflows a 64-bit signed int. The `u` is needed. If the architecture it were compiled for had 128-bit `int`s one could choose either signed or unsigned. – wallyk Mar 28 '18 at 01:38
Ok cool, thank you very much for this explanation! Makes it very clear now :-) – Ela782 Mar 29 '18 at 00:08

Meaning of U suffix

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