const challenge

[Christopher Wright]

Janice Caron wrote:
> Not really. We could have two functions, cat and icat, whereby cat
> always returns a mutable, and icat always returns an invariant. The
> concatenation operator, by contrast, always returns the same type as
> its inputs.

The issue with the concatenation operator is that its operands need not 
be the same type. You can use invariant(char)[] and char[], and 
invariant(char)[] takes precedence.

I'm not sure how often this issue comes up.

If you have a buffer, you can append to it with ~=. You might want to 
prepend something, though, and in that case, the current behavior will 
be problematic.

If you want to prepend something to a string, but you want to build the 
stuff you prepend...this seems like a marginal use case, more so than 
the buffer use case. It's easy enough to do:
char[] build;
const(char)[] prepend = build;
str = prepend ~ build;

The workaround for a mutable buffer is:
char[] tmpBuffer = new char[buffer.length + prepend.length];
tmpBuffer[0..prepend.length] = prepend[];
tmpBuffer[prepend.length..$] = buffer[];
buffer = tmpBuffer;

That is extremely ugly.