What is the difference between int, System.Int16, System.Int32 and System.Int64 other than their sizes?
Each type of integer has a different range of storage capacity
Type Capacity
Int16 -- (-32,768 to +32,767)
Int32 -- (-2,147,483,648 to +2,147,483,647)
Int64 -- (-9,223,372,036,854,775,808 to +9,223,372,036,854,775,807)
As stated by James Sutherland in his answer:
int and Int32 are indeed synonymous; int will be a little more
familiar looking, Int32 makes the 32-bitness more explicit to those
reading your code. I would be inclined to use int where I just need
'an integer', Int32 where the size is important (cryptographic code,
structures) so future maintainers will know it's safe to enlarge an
int if appropriate, but should take care changing Int32 variables
in the same way.
The resulting code will be identical: the difference is purely one of
readability or code appearance.
The only real difference here is the size. All of the int types here are signed integer values which have varying sizes
Int16: 2 bytes
Int32 and int: 4 bytes
Int64 : 8 bytes
There is one small difference between Int64 and the rest. On a 32 bit platform assignments to an Int64 storage location are not guaranteed to be atomic. It is guaranteed for all of the other types.
int
It is a primitive data type defined in C#.
It is mapped to Int32 of FCL type.
It is a value type and represent System.Int32 struct.
It is signed and takes 32 bits.
It has minimum -2147483648 and maximum +2147483647 value.
Int16
It is a FCL type.
In C#, short is mapped to Int16.
It is a value type and represent System.Int16 struct.
It is signed and takes 16 bits.
It has minimum -32768 and maximum +32767 value.
Int32
It is a FCL type.
In C#, int is mapped to Int32.
It is a value type and represent System.Int32 struct.
It is signed and takes 32 bits.
It has minimum -2147483648 and maximum +2147483647 value.
Int64
It is a FCL type.
In C#, long is mapped to Int64.
It is a value type and represent System.Int64 struct.
It is signed and takes 64 bits.
It has minimum –9,223,372,036,854,775,808 and maximum 9,223,372,036,854,775,807 value.
According to Jeffrey Richter(one of the contributors of .NET framework development)'s book 'CLR via C#':
int is a primitive type allowed by the C# compiler, whereas Int32 is the Framework Class Library type (available across languages that abide by CLS). In fact, int translates to Int32 during compilation.
Also,
In C#, long maps to System.Int64, but in a different programming
language, long could map to Int16 or Int32. In fact, C++/CLI does
treat long as Int32.
In fact, most (.NET) languages won't even treat long as a keyword and won't
compile code that uses it.
I have seen this author, and many standard literature on .NET preferring FCL types(i.e., Int32) to the language-specific primitive types(i.e., int), mainly on such interoperability concerns.
They tell what size can be stored in a integer variable. To remember the size you can think in terms of :-) 2 beers (2 bytes), 4 beers (4 bytes) or 8 beers (8 bytes).
Int16 :- 2 beers/bytes = 16 bit = 2^16 = 65536 = 65536/2 = -32768 to 32767
Int32 :- 4 beers/bytes = 32 bit = 2^32 = 4294967296 = 4294967296/2 = -2147483648 to 2147483647
Int64 :- 8 beers/bytes = 64 bit = 2^64 = 18446744073709551616 = 18446744073709551616/2 = -9223372036854775808 to 9223372036854775807
In short you can not store more than 32767 value in int16 , more than
2147483647 value in int32 and more than 9223372036854775807 value in
int64.
To understand above calculation you can check out this video int16 vs int32 vs int64
A very important note on the 16, 32 and 64 types:
if you run this query...
Array.IndexOf(new Int16[]{1,2,3}, 1)
you are suppose to get zero(0) because you are asking... is 1 within the array of 1, 2 or 3.
if you get -1 as answer, it means 1 is not within the array of 1, 2 or 3.
Well check out what I found:
All the following should give you 0 and not -1
(I've tested this in all framework versions 2.0, 3.0, 3.5, 4.0)
C#:
Array.IndexOf(new Int16[]{1,2,3}, 1) = -1 (not correct)
Array.IndexOf(new Int32[]{1,2,3}, 1) = 0 (correct)
Array.IndexOf(new Int64[]{1,2,3}, 1) = 0 (correct)
VB.NET:
Array.IndexOf(new Int16(){1,2,3}, 1) = -1 (not correct)
Array.IndexOf(new Int32(){1,2,3}, 1) = 0 (correct)
Array.IndexOf(new Int64(){1,2,3}, 1) = -1 (not correct)
So my point is, for Array.IndexOf comparisons, only trust Int32!
EDIT: This isn't quite true for C#, a tag I missed when I answered this question - if there is a more C# specific answer, please vote for that instead!
They all represent integer numbers of varying sizes.
However, there's a very very tiny difference.
int16, int32 and int64 all have a fixed size.
The size of an int depends on the architecture you are compiling for - the C spec only defines an int as larger or equal to a short though in practice it's the width of the processor you're targeting, which is probably 32bit but you should know that it might not be.
Nothing. The sole difference between the types is their size (and, hence, the range of values they can represent).
int and int32 are one and the same (32-bit integer)
int16 is short int (2 bytes or 16-bits)
int64 is the long datatype (8 bytes or 64-bits)
They both are indeed synonymous, However i found the small difference between them,
1)You cannot use Int32 while creatingenum
enum Test : Int32
{ XXX = 1 // gives you compilation error
}
enum Test : int
{ XXX = 1 // Works fine
}
2) Int32 comes under System declaration. if you remove using.System you will get compilation error but not in case for int
The answers by the above people are about right. int, int16, int32... differs based on their data holding capacity. But here is why the compilers have to deal with these - it is to solve the potential Year 2038 problem. Check out the link to learn more about it.
https://en.wikipedia.org/wiki/Year_2038_problem
Int=Int32 --> Original long type
Int16 --> Original int
Int64 --> New data type become available after 64 bit systems
"int" is only available for backward compatibility. We should be really using new int types to make our programs more precise.
---------------
One more thing I noticed along the way is there is no class named Int similar to Int16, Int32 and Int64. All the helpful functions like TryParse for integer come from Int32.TryParse.
Related
I have a 32 bit int and I want to address only the lower half of this variable. I know I can convert to bit array and to int16, but is there any more straight forward way to do that?
It you want only the lower half, you can just cast it: (Int16)my32BitInt
In general, if you're extending/truncating bit patterns like this, then you do need to be careful about signed types - unsigned types may cause fewer surprises.
As mentioned in the comments - if you've enclosed your code in a 'checked' context, or changed your compiler options so that the default is 'checked', then you can't truncate a number like this without an exception being thrown if there are any non-zero bits being discarded - in that situation you'd need to do:
(UInt16)(my32BitInt & 0xffff)
(The option of using signed types is gone in this case, because you'd have to use & 0x7fff which then preserves only 15 bits)
just use this function
Convert.ToInt16()
or just
(Int16)valueasint
You can use implicit conversation to Int16 like;
(Int16)2;
but be careful when you do that. Because Int16 can't hold all possible Int32 values.
For example this won't work;
(Int16)2147483683;
because Int16 can hold 32787 as maximum value. You can use unchecked (C# Reference) keyword such this cases.
If you force an unchecked operation, a cast should work:
int r = 0xF000001;
short trimmed = unchecked((short) r);
This will truncate the value of r to fit in a short.
If the value of r should always fit in a short, you can just do a normal cast and let an exception be thrown.
If you need a 16 bit value and you happen to know something specific like that the number will never be less than zero, you could use a UINT16 value. That conversion looks like:
int x = 0;
UInt16 value = (UInt16)x;
This has the full (positive) range of an integer.
Well, first, make sure you actually want to have the value signed. uint and ushort are there for a reason. Then:
ushort ret = (ushort)(val & ((1 << 16) - 1));
In Conversions (on Chapter 2) in C# topic in C# 5.0 in a Nutshell, the author says:
...Conversions can be either implicit or explicit: implicit conversions happen automatically, and explicit conversions require a cast. In the following example, we implicitly convert an int to long type (which has twice the bitwise capacity of an int)...
This is the example:
int x = 12345; // int is a 32-bit integer
long y = x; // Implicit conversion to 64-bit integer
short z = (short)x; // Explicit conversion to 16-bit integer
Is there a relationship between bitwise capacity and bit capacity? or, what is author´s point respect to bitwise capacity?
I think, he wants to differntiate between "bitwise capacity" and "numeric capacity".
In the example, the data types differ in bitwise capacity: int has 32, long 64 and short 16. In this case, conversions to data types with higher capacity happen implicit, conversions to data types with lower bitwise capacity happen explicit.
On the other hand, there's something like "numeric capacity" where int and uint do share the same number of bits (they have the same "bitwise capacity"), but are still not fully compatible in terms of values you can store (uint has no support for negative values).
I think they mean “capacity, with respect to bits”. If they had left out the “bitwise” part, then it could easily be interpreted as “this type holds twice as many values as the other type”, which is wrong: it holds much more than twice the number of values. It holds twice the number of bits, which increases the number of values exponentially.
It is the same thing. It just means that you have twice as many bits to represent your value, which means you can store much larger numbers. Numeric capacity is therefore tied to bitwise capacity since the more bits the higher numeric capacity.
With a 64 bit data type you can represent your value using 64 bit binary numbers
I am doing some classification and I am not sure:
INT is a primitive datatype with keyword "int"
But I can use Int16,Int32 or Int64 - I know C# has its own names for them. But are those data types as well or it is still INT? And mainly, can we say "short" is a datatype or INT16 is a datatype?
Thanks :)
In C#, the following things are always true:
short == Int16
ushort == UInt16
int == Int32
uint == UInt32
long == Int64
ulong == UInt64
Both versions are data types. All of the above are integers of various lengths and signed-ness.
The main difference between the two versions (as far as I know) is what colour they are highlighted as in Visual Studio.
short is a data type representing 16-bit integers (1 order below int, which is 32-bit).
Int16 is in fact also a data type and is synonymous with short. That is,
Int16.Parse(someNumber);
also returns a short, same as:
short.Parse(someNumber)
Same goes with Int32 for int and Int64 for long.
In C#, int is just a shorter way of saying System.Int32.
In .NET, even the primitive data types are actually objects (derived from System.Object).
So an int in C# = an Integer in VB.Net = System.Int32.
there's a chart of all the .NET data types here: http://msdn.microsoft.com/en-us/library/47zceaw7%28VS.71%29.aspx
This is part of the .NET Common Type System that allows seamless interoperability between .NET languages.
An int (Int32) has a memory footprint of 4 bytes. But what is the memory footprint of:
int? i = null;
and :
int? i = 3;
Is this in general or type dependent?
I'm not 100% sure, but I believe it should be 8 Bytes, 4 bytes for the int32, and (since every thing has to be 4-Byte aligned on a 32 bit machine) another 4 bytes for a boolean indicating whether the integer value has been specified or not.
Note, thanks to #sensorSmith, I am now aware that newer releases of .Net allow nullable values to be stored in smaller footprints (when the hardware memory design allows smaller chunks of memory to be independently allocated). On a 64 Bit machine it would still be 8 bytes (64 bits) since that is the smallest chunk of memory that can be addressed...
A nullable for example only requires a single bit for the boolean, and another single bit for the IsNull flag and so the total storage requirements is less than a byte it theoretically could be stored in a single byte, however, as usual, if the smallest chunk of memory that can be allocated is 8 bytes (like on a 64 bit machine), then it will still take 8 bytes of memory.
The size of Nullable<T> is definitely type dependent. The structure has two members
boolean: For the hasValue
value: for the underlying value
The size of the structure will typically map out to 4 plus the size of the type parameter T.
int? a = 3;
00000038 lea ecx,[ebp-48h]
0000003b mov edx,3
00000040 call 78BFD740
00000045 nop
a = null;
00000046 lea edi,[ebp-48h]
00000049 pxor xmm0,xmm0
0000004d movq mmword ptr [edi],xmm0
It seems that first dword is for the value, and the second one is for null flag. So, 8 bytes total.
Curious, BinaryWritter doesn't like to write nullable types. I was wandering if it could pack it tighter then 8 bytes...
The .NET (and most other languages/frameworks) default behavior is to align struct fields to a multiple of their size and structs themselves to a multiple of the size of their largest field. Reference: StructLayout
Nullable<T> has a bool flag and the T value. Since bool takes just 1 byte, the size of the largest field is the size of T; and Nullable doubles the space needed compared to a T alone. Reference:Nullable Source
Clarification: If T is itself a non-primitive struct rather than a primitive type, Nullable increases the space needed by the size of the largest primitive field within T or, recursively, within any of T's non-primitive fields. So, the size of a Nullable<Nullable<bool>> is 3, not 4.
You can check using some code similar to the one at https://www.dotnetperls.com/nullable-memory.
I got the following results:
Int32 4 bytes
Int32? 8 bytes
Int16 2 bytes
Int16? 4 bytes
Int64 8 bytes
Int64? 16 bytes
Byte 1 bytes
Byte? 2 bytes
bool 1 bytes
bool? 2 bytes
An int? is a struct containing a boolean hasValue, and an int. Therefore, it has a footprint of 5 bytes. The same applies to all instances of a nullable<T>: size = sizeof(T)+sizeof(bool)
The nullable type is a structure that contains the regular variable and a flag for the null state.
For a nullable int that would mean that it contains five bytes of data, but it's of course padded up to complete words, so it's using eight bytes.
You can generally expect that any nullable type will be four bytes larger than the regular type, except for small types like byte and boolean.
32-bit and 64-bit machines:
int == 4 bytes
int? == 8 bytes == 4 for int + 4 for the nullable type wrapper.
The nullable type wrapper requires 4 bytes of storage. And the integer
itself requires 4 bytes for each element. This is an efficient
implementation. In an array many nullable types are stored in
contiguous memory.
Based on a personal test (.NET Framework 4.6.1, x64, Release) and from – https://www.dotnetperls.com/nullable-memory
Also, if interesting: why int on x64 equals only 4 bytes?
Note: this is valid for Nullable<int> only, the size of Nullable<T> totally depends on the type.
I have a small question about structures with the LayoutKind.Explicit attribute set. I declared the struct as you can see, with a fieldTotal with 64 bits, being fieldFirst the first 32 bytes and fieldSecond the last 32 bytes. After setting both fieldfirst and fieldSecond to Int32.MaxValue, I'd expect fieldTotal to be Int64.MaxValue, which actually doesn't happen. Why is this? I know C# does not really support C++ unions, maybe it will only read the values well when interoping, but when we try to set the values ourselves it simply won't handle it really well?
[StructLayout(LayoutKind.Explicit)]
struct STRUCT {
[FieldOffset(0)]
public Int64 fieldTotal;
[FieldOffset(0)]
public Int32 fieldFirst;
[FieldOffset(32)]
public Int32 fieldSecond;
}
STRUCT str = new STRUCT();
str.fieldFirst = Int32.MaxValue;
str.fieldSecond = Int32.MaxValue;
Console.WriteLine(str.fieldTotal); // <----- I'd expect both these values
Console.WriteLine(Int64.MaxValue); // <----- to be the same.
Console.ReadKey();
The reason is that FieldOffsetAttribute takes a number of bytes as parameter -- not number of bits. This works as expected:
[StructLayout(LayoutKind.Explicit)]
struct STRUCT
{
[FieldOffset(0)]
public Int64 fieldTotal;
[FieldOffset(0)]
public Int32 fieldFirst;
[FieldOffset(4)]
public Int32 fieldSecond;
}
Looking at the hex values if Int32.MaxValue and Int64.MaxValue should provide the answer.
The key is the most significant bit. For a positive integer, the most significant bit is only set for a negative number. So the max value of Int32 is a 0 followed by a whole series of 1s. The order is unimportant, just that there will be at least a single 0 bit. The same is true of Int64.MaxValue.
Now consider how a union should work. It will essentially lay out the bits of the values next to one another. So now you have a set of bits 64 in length which contains two 0 bit values. One for each of the Int32.MaxValue instances. This cannot ever be equal to Int64.MaxValue since it can contain only a single 0 bit.
Oddly enough you will probably get the behavior you are looking for if you set fieldSecond to Int32.MinValue.
EDIT Missed that you need to make it FieldOffset(4) as well.
Ben M provided one of the more important elements - your definition is not setup correctly.
That being said, this won't work - even in C++ with a union. The values you specified won't be (and shouldn't be) the same values, since you're using signed (not unsigned) ints. With a signed int (Int32), you're going to have a 0 bit followed by 1 bits. When you do the union, you'll end up with a 0 bit, followed by a bunch of 1 bits, then another 0 bit, then a bunch of 1 bits... The second 0 bit is what's messing you up.
If you used UInt32/UInt64, this would work property, since the extra sign bit doesn't exist.