Difference Between 0 And 0u - Stack Overflow

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Learn more about Labs Difference between 0 and 0u Ask Question Asked 9 years, 7 months ago Modified 9 years, 7 months ago Viewed 13k times 5

Text from a textbook:

The uint type is 32-bit unsigned integer type. Its default value is the number 0u or 0U (the two are equivalent). The 'u' letter indicates that the number is of type uint (otherwise it is understood as int).

Then is there a difference between 0 and 0u? If not, then why is the default value '0u'. What is the advantage of suffixing an integer with 'u'?

Why would I ever have to use, say '5u', instead of just using '5'?

• c#
• integer
• default-value
• unsigned

Share Improve this question Follow asked Apr 12, 2015 at 3:22 discussedtreediscussedtree 24322 gold badges44 silver badges1313 bronze badges 5

• To show that it is unsigned. Some values are better represented as an absolute, like distances. – Carcigenicate Commented Apr 12, 2015 at 3:24
• One possible use of '5u' is if you like to use the keyword 'var' a lot. If you write " var j = 5; ", then it would be stored as an int and if you write " var j = 5u; ", then it will be stored as a uint. – discussedtree Commented Apr 12, 2015 at 3:32
• @Carcigenicate " uint distance = 730480; ". Didn't need to use the suffix 'u' here. – discussedtree Commented Apr 12, 2015 at 3:35
• 1. Hungarian notation (putting a type identifier in the name) is usually frowned up. 2. The idea is, if you know that the number will never require a sign, it's good practice to use a type that shows that rule. That extra bit you gain doubles the number's capacity too for the same amount of space, which can be useful. – Carcigenicate Commented Apr 12, 2015 at 3:39
• 1. Where did I put a type identifier in the name? 'uint' was meant to be the type identifier, it is not included in the variable name. The variable name is 'distance' 2. The type identifier 'uint' means 'unsigned integer' and automatically gives me an extra bit. I wouldn't have to suffix 'u' for that. – discussedtree Commented Apr 12, 2015 at 3:54

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2 Answers 2

Sorted by: Reset to default Highest score (default) Trending (recent votes count more) Date modified (newest first) Date created (oldest first) 5

By default, the compiler infers a numeric literal to be either double or integral type.

• If the literal contains a decimal point or the exponential symbol, it is a double.
• Otherwise, the literal's type is the first type in this list that can fit the literal's value: int,uint,long,ulong.

The suffixes u (or U) and l (or L) are rarely necessary, because the uint, long and ulong types can nearly always be either inferred or implicitly converted from int:

long i=5; //implicit lossless conversion from int literal to long

Similarly d suffix is also redundant, in that all literals with a decimal point are inferred to be double.

double x= 4.0d; //(Here, d is redundant)

The f (float) and m (decimal) suffixes are the most useful and should always be applied when specifying float or decimal literals. Without the f suffix following lines would not compile, because 3.5 would be inferred to be type double, which has no implicit conversion to float.

float f= 3.5f;

Similarly, the following line would not compile without the m (decimal) suffix.

decimal d=-1.25m;

In conclusion, suffixes instructs compilers to consider integral literal (like 123) to be certain type of number- for example long. Similarly, double literal to be considered other types - for example float, decimal and so on.

Share Improve this answer Follow answered Apr 12, 2015 at 4:50 ANewGuyInTownANewGuyInTown 6,38755 gold badges3535 silver badges4646 bronze badges 2

• In long i=5, is 5 an int literal or is it inferred as a long literal? If the former, does that mean conversion is performed when it is assigned to i? – bgfvdu3w Commented Jan 31, 2018 at 23:40
• 1 @Mark, yes there is implicit (lossless) conversion from int to long. – ANewGuyInTown Commented May 16, 2019 at 2:02

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Their type is different. While an unsuffixed integer literal can be interpreted as either uint or int depending on the context, the u suffix forces it to an unsigned type. For instance, you can't assign 5u to an int variable without a cast because it has a uint value.

You would use it when the type of the expression matters. For instance, var i = 5 makes i an int, but var i = 5u makes i a uint.

You can also use it to select a uint overload when passing a literal as a parameter. For instance:

void Foo(int x); void Foo(uint x); Foo(5); Foo(5u);

In general, though, you don't need it often.

Share Improve this answer Follow answered Apr 12, 2015 at 3:33 zneakzneak 138k4646 gold badges265265 silver badges335335 bronze badges 2

• Why is the default value of uint '0u', and not '0'? – discussedtree Commented Apr 12, 2015 at 3:40
• 1 Because 0 and 0u don't have the same type, and 0u is the uint one. – zneak Commented Apr 12, 2015 at 3:51

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