int foo(lazy int x) pure{ return x()+x(); } void main(){ auto a=foo((writeln("impure"),1)); } This compiles and prints impure impure
note that int foo(int delegate() x) pure{ return x()+x(); } void main(){ auto a=foo({return writeln("impure"),1;}); } fails with Error: pure function 'foo' cannot call impure delegate 'x'
The pureness of lazy parameter belongs to the *caller side*, not callee side. It is a design. One use case is std.exception.enforce. It receives a condition as a lazy parameter, but whole enforce function can become pure with the design. Delegate parameter is similar to lazy parameter, but it is different in this point.