On Wednesday, 26 April 2023 at 18:24:08 UTC, DLearner wrote:
Consider:
```
struct S1 {
int A;
int B;
int foo() {
return(A+B);
}
}
struct S2 {
int A;
int B;
}
int fnAddS2(S2 X) {
return (X.A + X.B);
}
void main() {
import std.stdio : writeln;
S1 Var1 = S1(1, 2);
writeln("Total Var1 = ", Var1.foo());
S2 Var2 = S2(1, 2);
writeln("Total Var2 = ", fnAddS2(Var2));
return;
}
```
Of the two ways shown of producing the total from the same
underlying structure, which is the better style?
To make both version really equivalent you should rather write
int fnAddS2(ref S2 X) {
return (X.A + X.B);
}
the `this` for the member functions is a reference as obviously
you want to eventually mutate the members and not their copy.
What happened in your `fnAddS2` is that the whole stuff was
blitted and mutation of the members would have no effect on the
argument used in the call.
Further, do we care about the situation where there are many
variables of type 'S', which presumably means the function code
generated from S1 gets duplicated many times, but not so with
S2?
Your presumption is wrong. The code generated for a function is
not regenerated per instance. One version is enough to handle all
the instances as the instance is itself a parameter of the
function. What is happening for `foo()`, i.e the member function,
is that there is an hidden parameter. Depending on the language
the way the hidden argument is read might be slightly different
but you really should consider that
struct S1 {
int A;
int B;
int foo() {
return(A+B);
}
}
is like
struct S1 {
int A;
int B;
static int foo(ref S1 that) {
return(that.A+that.B);
}
}
or
struct S1 {
int A;
int B;
}
int foo(ref S1 that) {
return(that.A+that.B);
}
One problem of the hidden parameter is that for example
int foo(const ref S1 that) {
return(that.A+that.B);
}
cannot be expressed (at first glance)... how to qualify `const`
something that is hidden ? It's actually possible using a member
function attribute:
struct S1 {
int A;
int B;
int foo() const /*const is for the hidden parameter*/ {
return(A+B);
}
}