C# short/long/int literal format?

Question

In C/C#/etc. you can tell the compiler that a literal number is not what it appears to be (ie., float instead of double, unsigned long instead of int):

var d = 1.0;  // double
var f = 1.0f; // float
var u = 1UL;  // unsigned long

etc.

Could someone point me to a list of these? I'm specifically looking for a suffix for short or Int16.

see here: http://stackoverflow.com/questions/166752/c-compiler-number-literals — fixagon, Apr 28 '11 at 15:07
possible duplicate of [Defining different types of numbers in C#](http://stackoverflow.com/questions/262555/defining-different-types-of-numbers-in-c) — Daniel Renshaw, Apr 28 '11 at 15:08

score 435 · Accepted Answer · edited May 17 '18 at 09:55

435

var d  = 1.0d;  // double
var d0 = 1.0;   // double
var d1 = 1e+3;  // double
var d2 = 1e-3;  // double
var f  = 1.0f;  // float
var m  = 1.0m;  // decimal
var i  = 1;     // int
var ui = 1U;    // uint
var ul = 1UL;   // ulong
var l  = 1L;    // long

I think that's all... there are no literal specifiers for short/ushort/byte/sbyte

edited May 17 '18 at 09:55

AustinWBryan

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answered Apr 28 '11 at 15:07

Thomas Levesque

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Does this mean you have to cast everywhere you use short/ushort/byte/sbyte? Eg.: somebyte = somebool ? (byte) 1 : (byte) 0; – mola Feb 07 '13 at 13:19
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@mola, yes, unless the desired type is unambiguous (e.g. `byte b = 42;`) – Thomas Levesque Feb 07 '13 at 16:01
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@mola `somebyte = (byte)(somebool ? 1 : 0);` – or hor May 30 '16 at 10:40
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Just to add that upper case and lower case of these literal suffixes are equivalent e.g. `1l` and `1L` both will be treated as long integer but certainly `1L` is more readable than `1l`. – RBT Mar 21 '17 at 03:14
I think you missed 1e+10 / 1e-10 – Ivandro Jao Nov 28 '17 at 12:30
these lack https://github.com/dotnet/csharplang/issues/1058 – Dzmitry Lahoda Jun 16 '20 at 07:30
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@DzmitryLahoda what do you mean? The suffixes discussed in that issue are not implemented yet. – Thomas Levesque Jun 16 '20 at 07:49

score 54 · Answer 2 · edited Feb 26 '22 at 12:38

From Integer literals:

The type of an integer literal is determined as follows:

If the literal has no suffix, it has the first of these types in which its value can be represented: int, uint, long, ulong.

If the literal is suffixed by U or u, it has the first of these types in which its value can be represented: uint, ulong.

If the literal is suffixed by L or l, it has the first of these types in which its value can be represented: long, ulong.

If the literal is suffixed by UL, Ul, uL, ul, LU, Lu, lU, or lu, it is of type ulong.

And from Real literals:

If no real type suffix is specified, the type of the real literal is double. Otherwise, the real type suffix determines the type of the real literal, as follows:

A real literal suffixed by F or f is of type float. For example, the literals 1f, 1.5f, 1e10f, and 123.456F are all of type float.

A real literal suffixed by D or d is of type double. For example, the literals 1d, 1.5d, 1e10d, and 123.456D are all of type double.

A real literal suffixed by M or m is of type decimal. For example, the literals 1m, 1.5m, 1e10m, and 123.456M are all of type decimal. This literal is converted to a decimal value by taking the exact value, and, if necessary, rounding to the nearest representable value using banker's rounding (Section 4.1.7). Any scale apparent in the literal is preserved unless the value is rounded or the value is zero (in which latter case the sign and scale will be 0). Hence, the literal 2.900m will be parsed to form the decimal with sign 0, coefficient 2900, and scale 3.

score 10 · Answer 3 · answered Mar 14 '13 at 16:00

10

If your variable isn't already a short, you have to cast it explicitly :

Object s = (Int16) 1;

answered Mar 14 '13 at 16:00

Chris

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Side note: I think this causes a boxing conversion. – 3Dave Jan 07 '14 at 16:45
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It does cause boxing because Int16 is a value type and Object is a reference type. – Scott Marcus Sep 24 '14 at 16:22

score 6 · Answer 4 · answered Apr 28 '11 at 15:07

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There isn't one for short. Just use short s = 1;.

answered Apr 28 '11 at 15:07

Random832

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Interestingly this compiles: `short z1 = (0 == 1 ? 0 : 1);` But this does not: `short y = 1; short z2 = (0 == 1 ? 0 : y);` (sorry, no line breaks allowed in comments) – yoyo Mar 08 '16 at 22:24
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@yoyo: That's because the `0` in the if branch is converted to an `int` **before** the assignment to `z2`. If you use two `short`s in your branches, the result will again be a `short`. With the if/else, the compiler can't know that your `int` can be represented as `short`. – Oliver Apr 03 '17 at 08:58

score 1 · Answer 5 · edited Nov 18 '22 at 18:43

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var myValue = unchecked((short)0x7F00);

The literal is int and thus must be cast to target type. If a value overflow occurs unchecked is required.

edited Nov 18 '22 at 18:43

Greg Burghardt

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answered Jan 26 '22 at 05:51

Tom Schardt

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This is the answer for the original question – Gerzson Mar 13 '23 at 08:45

C# short/long/int literal format?

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