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

[ag0aep6g via Digitalmars-d-learn]

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)