Metaprogramming with constexpr or struct

Question

We just started learning template meta programming in C++11. As an exercise we wrote a program that outputs the binary representation of an int value. We came up with two possible implementation. The first one uses recursion with enum values whereas the second method uses a constexpr function.

Our expectation was that both implementation results in executables of the same size. However, the first implementation leads to 9064 bytes whereas the second has 9096 bytes. We don't mind the tiny difference in bytes but do not understand what causes the difference.

We compiled the program with GCC 4.8.2 without optimization flag, however, the same results are found the -O2 flag.

#include <iostream>
using namespace std;

template <int val>
struct Bin
{
    enum { value = 10 * Bin<(val >> 1)>::value + (val & 1) };
};

template <>
struct Bin<0>
{
    enum { value = 0 };
};

constexpr int bin(int val)
{
  return val == 0 ? 0 : (10 * bin(val >> 1) + (val & 1));
}


int main()
{
  // Option 1
  cout << Bin<5>::value  << '\n'
       << Bin<27>::value << '\n';

  // Option 2
  cout << bin(5) << '\n'
       << bin(27) << '\n';
}

This program looks simplistic enough to dissassemble and analyze. — Borsunho
– Borsunho, Commented Aug 14, 2015 at 10:51
constexpr functions are not guaranteed to be evaluated at compile time unless used in a context that requires a constant expression. — T.C.
– T.C., Commented Aug 14, 2015 at 10:51

edmz · Accepted Answer · 2015-08-14 11:15:51Z

3

constexpr functions may be evaluated at compile-time. They are not required to.

For the code you provided, the compiler isn't indeed doing that and bin is getting called at runtime; this means the function cannot be thrown away from the assembly. By explicitly requiring the values to be constexpr with

constexpr auto i = bin(5), j = bin(27);

the calls to bin are done at compile-time as shown here. With

  cout << bin(5) << '\n'
      << bin(27) << '\n';

the relevant emitted code is

movl $5, %edi # Parameter
callq   bin(int) # Here's the call to bin
movl    std::cout, %edi
movl    %eax, %esi
callq   std::basic_ostream<char, std::char_traits<char> >::operator<<(int)
[...]
movl    $27, %edi # parameter
callq   bin(int) # call to bin
movq    %rbx, %rdi
movl    %eax, %esi
callq   std::basic_ostream<char, std::char_traits<char> >::operator<<(int)

When the call is elided, the size is the same for both versions.

edited Aug 14, 2015 at 11:15

answered Aug 14, 2015 at 10:52

edmz

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5 Comments

Michiel uit het Broek Over a year ago

Thanks for your response. Is there a way to explicitly requiring the values to be constexpr without storing them into a variable?

edmz Over a year ago

@MichielUitHetBroek The available ways to force constexpr functions to be evaluated at compile-time require additional coding (often optimized out) and the easiest is using a compile-time value. See this and related.

KeyC0de Over a year ago

@edmz How do you see that the calls are done at compile time? What I see is callqs instructions, which are basically calling functions. How do you know this occurs at compile time by looking at the generated assembly code? Thanks

edmz Over a year ago

@Nik-Lz: Have a look at the link I provided (3 yo and still working, cool): you'll see the compiler is issuing movl $101, %esi instead of a callq to bin(5).

KeyC0de Over a year ago

@edmz Yeah. All callqs are calls to std::cout which of course have to be done at runtime. Everything else is handled through movs and comparisons. No function calls. So yes, this indicates compile time computations.

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