使用g ++ -fopenmp编译的代码不会运行并行

Question

如果我用

编译下面的代码

g++ -std=c++11 -o omp_for_loop omp_for_loop.cpp -fopenmp -lpthread

或

g++ -std=c++11 -o omp_for_loop omp_for_loop.cpp -fopenmp -lpthread -L/usr/lib64/gcc/x86_64-suse-linux/4.8 -lgomp

它只使用一个核心。我哪里出错了？

#include <string>
#include <iostream>
#include <sstream>
#include <fstream>
#include <vector>
#include <array>
#include <time.h>
#include <stdio.h>
#include <math.h>
#include <omp.h>

int main(){
    omp_set_num_threads(4);
    int nthreads, tid;
    #pragma omp parallel shared(nthreads) private(tid)
    {
        tid = omp_get_thread_num();
        if (tid == 0)
        {
            nthreads = omp_get_num_threads();
            printf("Number of threads = %d\n", nthreads);
        }
        printf("Thread %d starting...\n",tid);

    }

    int N=4096;
    std::vector<double> values (N, 0.0);

    std::ifstream F("time_spec_values.txt");
    {
        for (int q=0; q<N; q++)
        {
            F >> values[q];
        }
    }

/*for(int w=0; w<512; w++)
  {
  std::cout << values[w] << std::endl;
  }*/
//std::cout << " ====================================================== " << std::endl;

    double PI2=2*M_PI;
    int square=N*N; //
    std::vector<int> index (square, 0.0);
    std::vector<double> outvalues (square, 0.0);
    std::vector<double> RESULT (square, 0.0);

    int i;
    #pragma omp parallel for private(i)
    for (i=0; i<square; i++)
    {
        index[i]=i;
    }

    int k,l, idx=0;

    #pragma omp parallel for private(k,l,idx) shared(outvalues)  schedule(dynamic,1024)

    for (k=0; k<N; k++)
    {
        for (l=0; l<N; l++)
        {
            idx=(k*N)+l;
            double VV=PI2*(values[k]-values[l]);
            outvalues[idx]=VV;
        }
    }


    int m;
    #pragma omp parallel for private(m) shared(RESULT,outvalues) schedule(dynamic,1024)

    for (m=0; m<square; m++)
    {
        RESULT[m]=outvalues[m];
    }


    std::cout << "index size = " << index.size()<< "outvalues size = " << outvalues.size() << std::endl;

    for(int n=0; n<square; n++)
        std::cout << RESULT[n] << std::endl;
}