Why are the loops for memset() 1M times and 10M times cost the same time?

Question

Here is my code:

#include <iostream>
#include <sys/time.h>
#include <string.h>

using namespace std;

int main()
{
    char* a = (char*)malloc(1024);
    int times = 10000000;
    struct timeval begin, end;
    gettimeofday(&begin, NULL);
    for(int i=0; i<times; i++){
        memset(a, 1, 1024);
    }
    gettimeofday(&end, NULL);
    cout << end.tv_sec - begin.tv_sec << "." << end.tv_usec - begin.tv_usec << endl;
    return 0;
}

When I set times to 1M, the output is about 0.13 second, however when I set times to 10M, the output is still about 0.13 second. What causes such circumstance? Is it caused by the optimisation of Linux or the compiler?

Answer 1

UPDATE: optimisation disabled

I think you need to use more precise chrono instead of time.h and disable compiler optimisations:

#include <iostream>
#include <string.h>
#include <chrono>

#ifdef __GNUC__
    #ifdef __clang__
        static void test() __attribute__ ((optnone)) {
    #else
        static void __attribute__((optimize("O0"))) test() {
    #endif
#elif _MSC_VER
    #pragma optimize( "", off )
        static void test() {
#else
    #warning Unknow compiler!
        static void test() {
#endif

    char* a = (char*) malloc(1024);

    auto start = std::chrono::steady_clock::now();
    for(uint64_t i = 0; i < 1000000; i++){
        memset(a, 1, 1024);
    }
    std::cout<<"Finished in "<<std::chrono::duration<double, std::milli>(std::chrono::steady_clock::now() - start).count()<<std::endl;
}

#ifdef _MSC_VER
    #pragma optimize("", on)
#endif

int main() {
    test();

    return 0;
}

10M: Finished in 259.851 1M: Finished in 26.3928