CODEWITHSUNDEEP
c++compiler-optimizationregister-allocation
grok
grok

Reputation: 210

Performance difference between POD and non-POD classes

I'm struggling to understand why my compilers (g++ 8.1.0 and clang++ 6.0.0) treat POD (plain-old-data) and non-POD code differently.

Test code:

#include <iostream>

struct slong {
  int i;
  ~slong() { i = 0; }
};

int get1(slong x) { return 1+x.i; }

int main() {
  std::cerr << "is_pod(slong) = " << std::is_pod<slong>::value << std::endl;
}

defines a class slong with a destructor (hence not POD) and the compiler, with -Ofast, will produce for get1

        movl    (%rdi), %eax
        incl    %eax

but when I comment out the destructor (so slong becomes POD) I get

        leal    1(%rdi), %eax

Of course the performance issue is minor; still I'd like to understand. In other (more complicated) cases I also noticed more significant code differences.

Upvotes: 3

Views: 534

Answers (1)

anatolyg
anatolyg

Reputation: 28278

Note that movl accesses memory, while leal doesn't.

When passing a struct to a function by value, ABI can stuff it into a register (rdi) if it's POD.

If the struct is not POD, ABI must pass it on stack (presumably because the code may need its address to call the destructor, access the vtable and do other complicated stuff). So accessing its member requires indirection.

Upvotes: 4

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