On Friday, 9 February 2018 at 15:50:24 UTC, Ralph Doncaster wrote:
On Friday, 9 February 2018 at 15:24:27 UTC, Mike Parker wrote:
On Friday, 9 February 2018 at 15:05:33 UTC, Ralph Doncaster
wrote:
This seems odd to me. Is there a way I can make a function
that takes an array of any type but only of a specific size
in bytes?
void.d(8): Error: function void.foo (void[12] arr) is not
callable using argument types (uint[3])
Failed: ["/usr/bin/dmd", "-v", "-o-", "void.d", "-I."]
void foo(void [12] arr)
{
}
void main()
{
uint[3] arr;
foo(arr);
}
void has no size, so what does it mean to have 12 of them?
according to the docs and my testing, the size of a void array
element is 1,
Correct.
so the following code prints 12:
import std.stdio;
void foo(void [] arr)
{
writeln("length: " arr.length);
}
void main()
{
uint[3] arr;
foo(arr);
}
I thought about using templates, but I was looking for a simple
way of making a function that takes an array of 12 bytes,
whether it is uint[3], ubyte[12], or ushort[6].
uint[3] and void[12] have the same size but are different types
so the compiler will reject it.
you can reinterpret cast them(i.e. *cast(void[12])(&arr) ), but
this is a rather blunt tool. Is it safe in the case that the
types size is at least 12 but not safe in general.
Is there a way I can make a function that takes an array of any
type but only of a specific size in bytes?
With a template that constrains the types size:
void foo(T)(T t) if (T.sizeof == 12)
{
//...
}
alternately reject incorrect values at runtime
void foo(ubyte[] t)
in
{
assert(t.length == 12);
}
do
{
//...
}
