Using replaceInPlace, string and char[]

[Ali Çehreli via Digitalmars-d-learn]

This looks like a bug to me. The template constraints of the two 
overloads are pretty complicated. This case should match only one of them.

On 08/15/2015 12:43 AM, TSalm wrote:

 > Don't understand why this doesn't work: it compiles fine and runs
 > perfectly if I change "char[]" by "string"

You mean, this:

import std.array;
import std.stdio;

void main()
{
     string a = "mon texte 1";    // <-- now string
     writeln(a.ptr);              // added
     char[] b = "abc".dup;
     size_t x   = 4;
     size_t y   = 9;
     replaceInPlace( a, x , y, b );
     writeln( a );
     writeln(a.ptr);              // added
}

The output:

4BC480
mon abc 1
7FC2AB867210    <-- different

 > ... don't understand why
 > since the documentation says :
 >        String literals are immutable (read only).
 >
 > How this function can change a type that is immutable ?

It cannot change the characters of the original string. replaceInPlace 
takes its first parameter by reference. What changes is 'a' itself. As 
evidenced by the output of the program, 'a' is now a slice to a new set 
of immutable characters.

If there were other slices to "mon texte 1", they wouldn't see a change.

Ali