Why does indexing a string inside of a recursive call yield a different result?

[ag0aep6g]

On 10.05.20 12:02, Adnan wrote:
> ulong editDistance(const string a, const string b) {
>      if (a.length == 0)
>          return b.length;
>      if (b.length == 0)
>          return a.length;
> 
>      const auto delt = a[$ - 1] == b[$ - 1] ? 0 : 1;
> 
>      import std.algorithm : min;
> 
>      return min(
>          editDistance(a[0 .. $ - 1], b[0 .. $ - 1]) + delt,
>          editDistance(a, b[0 .. $ - 1]) + 1,
>          editDistance(a[0 .. $ - 1], b) + 1
>      );
> }
> 
> This yields the expected results but if I replace delt with its 
> definition it always returns 1 on non-empty strings:
> 
> ulong editDistance(const string a, const string b) {
>      if (a.length == 0)
>          return b.length;
>      if (b.length == 0)
>          return a.length;
> 
>      //const auto delt = a[$ - 1] == b[$ - 1] ? 0 : 1;
> 
>      import std.algorithm : min;
> 
>      return min(
>          editDistance(a[0 .. $ - 1], b[0 .. $ - 1]) + a[$ - 1] == b[$ - 
> 1] ? 0 : 1, //delt,
>          editDistance(a, b[0 .. $ - 1]) + 1,
>          editDistance(a[0 .. $ - 1], b) + 1
>      );
> }
> 
> Why does this result change?

You're going from this (simplified):

     delt = a == b ? 0 : 1
     result = x + delt

to this:

     result = x + a == b ? 0 : 1

But that new one isn't equivalent to the old one. The new one actually 
means:

     result = (x + a == b) ? 0 : 1

You need parentheses around the ternary expression:

     result = x + (a == b ? 0 : 1)