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Hello,
What are the differences between Int128 and other Int groups such as Int16, Int32, and Int64?
The sizeof operator doesn't work on Int128.
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You have the error:
'Int128' does not have a predefined size, therefore sizeof can only be used in an unsafe context If you use unsafe and build with the unsafe option, you should get:
2
4
8
16
Thank you. I ran the program but I don't understand where the main problem comes from.
1- What does "'Int128' does not have a predefined size" mean? (differences in structure? I saw Int64 and Int128, they are the same and struct)
2- What are the differences between Int128 and Int64 or Int32?3- The compiler said "'Int128' does not have a predefined size" so If Int128 doesn't have a predefined size what happens when we use unsafe, the compiler returns to us 16? (Compiler said Int128 doesn't have a predefined size but return 16)
In my opinion, after compilation, sizeof is replaced with corresponding constant. If the program uses 16 as a size of Int128, it becomes problematic in other future releases or updates of .NET, where the structure could be changed. For example, the size of the next new imaginary implementation of Int128 is 4 (in 32-bit programs):
struct Int128
{
byte[] data; // array of 16 bytes
. . .
}
Without recompilation, the program that uses sizeof(Int128) will not work correctly.
In contrast, SByte, Int16, Int32 and Int64 are guaranteed to have an unchangeable size; they are based on corresponding 8-, 16-, 32- and 64-bit microprocessors’ registers and instructions.
Maybe, in the future, Int128 will also have a predefined fixed size.
I don't know, but as a guess, since Int128 is a struct with 2 values rather than a single defined type that's natively supported by the underlying processor architecture, I guess it's playing safe for any situation where the processor architecture wouldn't be able to pack the 2 values contiguously.
As VS and doc say, it is not a predefined type
It works inside unsafe { } or
with Marshal.SizeOf(typeof(Int128) (or Marshal.SizeOf<Int128>())
Could you please see the comment in the question? I reply and ask some questions. Thank you