No implicit opOpAssign for structs with basic types?

[Ali Çehreli]

On 4/4/20 8:45 AM, Steven Schveighoffer wrote:

 >> Yes, sure, but in C++ I don't have to explicitly write this down. It
 >> just works. IMO that makes a lot of sense as long as all types fit.
 >> This just looks superfluously.
 >>
 >
 > steves at homebuild:~$ cat test.cpp
 > struct S
 > {
 >      float a;
 >      float b;
 > };
 > int main()
 > {
 >      S a = {1, 5};
 >      S b = {2, 5};
 >
 >      a += b;
 > }
 > steves at homebuild:~$ g++ -o test test.cpp
 > test.cpp: In function ‘int main()’:
 > test.cpp:11:4: error: no match for ‘operator+=’ (operand types are ‘S’
 > and ‘S’)
 >    a += b;
 >    ~~^~~~
 >
 > Doesn't seem to "just work" for me...

I was about to say the same. C++ does not have this feature. What it has 
as a feature and as a guideline is to define an operator+ outside of the 
type's definition. Perhaps that's what's helping in C++ in this case: 
the type looks clean but there are "interface" functions outside of it.

I've used the following trick in D for many of my types, which I've been 
copy-pasting but it can be mixed in:

struct S {
   float a;
   float b;

   int opCmp(const(typeof(this)) that) const {
     import std.typecons : tuple;
     return tuple(this.tupleof).opCmp(tuple(that.tupleof));
   }
}

unittest {
   assert(S(3) < S(4));
   assert(S(1, 2) > S(1, 1));
}

void main() {
}

Ali