Bug 1451 – overloading methods by mixin in static if doesn't work

<code> template Foo(T) { void callme(T) { } } class Bar { mixin Foo!(int) F1; alias F1.callme callme; static if(true) { mixin Foo!(float) F2; alias F2.callme callme; } } void main() { Bar bar = new Bar; bar.callme(cast(int)0); // <- works bar.callme(cast(float)0); // <- fails (line 25) } </code> This code fails to compile with these error messages: test.d(25): function test.Bar.Foo!(int).callme ((int _param_0)) does not match parameter types (float) test.d(25): Error: cannot implicitly convert expression (0) of type float to int After commenting out the static if it compiles perfectly, like expected.

<code> struct A { static A opCall() { A ret; return ret; } mixin M; } template M() { static typeof(*this) opCall(uint i) { A ret; return ret; } } int main(char[][] args) { A a1 = A(); A a2 = A(42); return 0; } </code> Line 24: function t.A.opCall () does not match parameter types (int) Line 24: Error: expected 0 arguments, not 1 Still the same bug, simple test case above

cat > bug.d << CODE mixin template func(T) { void bar(T) {} } mixin func!(int); mixin func!(double); void baz() { bar(0); bar(1.0); } CODE dmd -c bug

The errors in the two comments are by design. Comment #1: mixin template members are hidden by members of the same name in the surrounding scope. Comment #2: mixin templates introduce distinct overload sets in order to prevent accidental function hijacking. The original bug report is valid, and the example code now compiles with DMD 2.058. If anyone disagrees, please reopen.

>Comment #2: mixin templates introduce distinct overload sets in order to prevent accidental function hijacking. Makes sense. Just for the record, I tried to unroll a TypeTuple into an overloadset, which now works fine after merging the overloads. mixin template _visit(Nodes...) if(Nodes.length) { void visit(Nodes[0] n); static if (Nodes.length > 1) { mixin _visit!(Nodes[1 .. $]) vi; alias vi.visit visit; } }