Force all && to be executed?

Question

Consider the following variadic function

template <typename Type, typename... Types>
bool f(Type& arg, Types&... args)
{
    return f(arg) && f(args...);
}

template <typename Type>
bool f(Type& arg)
{
    // Do something
} 

If one level of recursion is false, then I suspect that the following will not be executed. Is there a trick to force the recursion on all arguments even if one of them returns false ?

Answer 1

This shouldn't be too hard:

template <typename Type, typename... Types>
bool f(Type& arg, Types&... args)
{
    bool b1 = f(arg);
    bool b2 = f(args...);
    return b1 && b2;
}

Answer 2

As the debate has evolved in a comparison of the AndyProwl and Alon solution, I've benchmarked the both solution, and the result ... depends of the number of arguments.

Compiling with:

g++-4.7 -std=c++11 -Wall -Wextra -O3 main.cpp -o main -D_FIRST

benchmarks the AndyProwl solution and compiling with :

g++-4.7 -std=c++11 -Wall -Wextra -O3 main.cpp -o main -D_SECOND

benchmarks the Alon solution.

Here is the program of the benchmark for 10 arguments.

#include <iostream>
#include <chrono>

// Function 1 : with &&
template <typename Type>
inline bool f1(const Type& arg)
{
   return arg;
}
template <typename Type, typename... Types>
inline bool f1(const Type& arg, const Types&... args)
{
    bool arg1 = f1(arg);
    bool arg2 = f1(args...);
    return arg1 && arg2;
}

// Function 2 : with &
template <typename Type>
inline bool f2(const Type& arg)
{
   return arg;
}
template <typename Type, typename... Types>
inline bool f2(const Type& arg, const Types&... args)
{
    return f2(arg) & f2(args...);
}

// Benchmark
int main(int argc, char* argv[])
{
    // Variables
    static const unsigned long long int primes[10] = {11, 13, 17, 19, 23, 29, 31, 37, 41, 43};
    static const unsigned long long int nbenchs = 50;
    static const unsigned long long int ntests = 10000000;
    unsigned long long int sum = 0;
    double result = 0;
    double mean = 0;
    std::chrono::high_resolution_clock::time_point t0 = std::chrono::high_resolution_clock::now();

    // Loop of benchmarks
    for (unsigned long long int ibench = 0; ibench < nbenchs; ++ibench) {

        // Initialization
        t0 = std::chrono::high_resolution_clock::now();
        sum = 0;

        // Loop of tests
        for (unsigned long long int itest = 1; itest <= ntests; ++itest) {
#ifdef _FIRST
            sum += f1((itest+sum)%primes[0], (itest+sum)%primes[1], (itest+sum)%primes[2], (itest+sum)%primes[3], (itest+sum)%primes[4], (itest+sum)%primes[5], (itest+sum)%primes[6], (itest+sum)%primes[7], (itest+sum)%primes[8], (itest+sum)%primes[9]);
#endif
#ifdef _SECOND
            sum += f2((itest+sum)%primes[0], (itest+sum)%primes[1], (itest+sum)%primes[2], (itest+sum)%primes[3], (itest+sum)%primes[4], (itest+sum)%primes[5], (itest+sum)%primes[6], (itest+sum)%primes[7], (itest+sum)%primes[8], (itest+sum)%primes[9]);
#endif
        }

        // Finalization
        result = std::chrono::duration_cast<std::chrono::duration<double>>(std::chrono::high_resolution_clock::now()-t0).count();
        mean += result;
        std::cout<<"time = "<<result<<" (sum = "<<sum<<")"<<std::endl;
    }

    // End
    std::cout<<"mean time = "<<mean/nbenchs<<std::endl;
    return 0;
}

With 50 benchmarks for each solution with a given number of arguments, the dispersion is very small, and the mean time over these benchmarks is a reliable indicator.

My first benchmark has been with the "right" number of arguments where the Alon solution is faster than the AndyProwl solution.

The final results are here :

So the AndyProwl solution is generally faster than the Alon one. So, now I can validate your answer. But I think that the difference is so small that it's architecture/compiler dependent.

So:

• AndyProwl+1 for your generally faster solution
• Alon+1 for your constexpr-ready solution

Answer 3

You can execute them separately and return a bool expression:

bool b0 = f(arg);
bool b1 = f(args);
return b0 && b1;