Why can a T* be passed in register, but a unique_ptr cannot?

manpreet Best Answer 2 years ago

_x000D_ In C++, non-trivial destructor -> must have addressable storage (An illustration of a point in @MaximEgorushkin's answer using @harold's example in a comment.) If you compile: struct Foo { int bar; }; Foo test(Foo byval) { return byval; } you get: test(Foo): mov eax, edi ret i.e. the Foo object is passed to test in a register (edi) and also returned in a register (eax). but this is actually an exception to the rule. Generally, the C++ standard requires an object's storage to be addressable. Only if there's a trivial destructor [CITATION NEEDED] are we allowed the "as-if shortcut" of passing data through the registers. When the destructor is not trivial (like the std::unique_ptr example of OP's) - addressable storage means we have to put the object on the stack. This is true even if the destructor does not at all use the object's address! Thus if you compile: struct Foo2 { int bar; ~Foo2() { } }; Foo2 test(Foo2 byval) { return byval; } you get: test(Foo2): mov edx, DWORD PTR [rsi] mov rax, rdi mov DWORD PTR [rdi], edx ret with useless loading and storing.