Re: Why do "const inout" and "const inout shared" exist?

[ag0aep6g via Digitalmars-d]

On 07/02/2017 10:55 AM, Walter Bright wrote:
If it is declared as:
     inout(char)[] foo(bool condition, inout(char)[] chars);

your specific case will work as expected. Perhaps you meant:

No, it doesn't. The function doesn't compile with that signature.

----
inout(char)[] foo(bool condition, inout(char)[] chars)
{
    if (!condition)
        return "condition failed!"; /* Error: cannot implicitly convert 
expression "condition failed!" of type string to inout(char)[] */
    return chars;
}
----