CODEWITHSUNDEEP
c++c++11overheadpass-by-rvalue-reference
Taylor Nichols
Taylor Nichols

Reputation: 618

Do rvalue references have the same overhead as lvalue references?

Consider this example:

#include <utility>

// runtime dominated by argument passing
template <class T>
void foo(T t) {}

int main() {
    int i(0);
    foo<int>(i); // fast -- int is scalar type
    foo<int&>(i); // slow -- lvalue reference overhead
    foo<int&&>(std::move(i)); // ???
}

Is foo<int&&>(i) as fast as foo<int>(i), or does it involve pointer overhead like foo<int&>(i)?

EDIT: As suggested, running g++ -S gave me the same 51-line assembly file for foo<int>(i) and foo<int&>(i), but foo<int&&>(std::move(i)) resulted in 71 lines of assembly code (it looks like the difference came from std::move).

EDIT: Thanks to those who recommended g++ -S with different optimization levels -- using -O3 (and making foo noinline) I was able to get output which looks like xaxxon's solution.

Upvotes: 6

Views: 591

Answers (2)

Oliv
Oliv

Reputation: 18081

Efficiency of parameter passing depends on the ABI.

For example, on linux the Itanium C++ ABI specifies that references are passed as pointers to the referred object:

3.1.2 Reference Parameters

Reference parameters are handled by passing a pointer to the actual parameter.

This is independent of the reference category (rvalue/lvalue reference).

For a broader view, I have found this quote in a document from the Technical University of Denmark, calling convention, which analyzes most of the compilers:

References are treated as identical to pointers in all respects.

So rvalue and lvalue reference involve pointer overhead on all ABI.

Upvotes: 5

xaxxon
xaxxon

Reputation: 19771

In your specific situation, it's likely they are all the same. The resulting code from godbolt with gcc -O3 is https://godbolt.org/g/XQJ3Z4 for:

#include <utility>

// runtime dominated by argument passing
template <class T>
int foo(T t) { return t;}

int main() {
    int i{0};
    volatile int j;
    j = foo<int>(i); // fast -- int is scalar type
    j = foo<int&>(i); // slow -- lvalue reference overhead
    j = foo<int&&>(std::move(i)); // ???
}

is:

    mov     dword ptr [rsp - 4], 0 // foo<int>(i);
    mov     dword ptr [rsp - 4], 0 // foo<int&>(i);
    mov     dword ptr [rsp - 4], 0 // foo<int&&>(std::move(i)); 
    xor     eax, eax
    ret

The volatile int j is so that the compiler cannot optimize away all the code because it would otherwise know that the results of the calls are discarded and the whole program would optimize to nothing.

HOWEVER, if you force the function to not be inlined, then things change a bit int __attribute__ ((noinline)) foo(T t) { return t;}:

int foo<int>(int):                           # @int foo<int>(int)
        mov     eax, edi
        ret
int foo<int&>(int&):                          # @int foo<int&>(int&)
        mov     eax, dword ptr [rdi]
        ret
int foo<int&&>(int&&):                          # @int foo<int&&>(int&&)
        mov     eax, dword ptr [rdi]
        ret

above: https://godbolt.org/g/pbZ1BT

For questions like these, learn to love https://godbolt.org and https://quick-bench.com/ (quick bench requires you to learn how to properly use google test)

Upvotes: 6

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