有关函数pthread_join（）＆c语言返回值的问题

Question

我正在使用pthread_join（）以C语言获取返回值，其第二个参数不是null，而是2级指针。因为我想获得一个矩阵。

但是，每次我在返回线程后调试以检查矩阵，并且其内部的数据完全错误时，问题是线程函数块中的结果是正确的，这可能意味着错误发生在期间。返回的过程。

 #include<stdio.h>
  #include<stdlib.h>
  #include<pthread.h>
  #include<time.h>
  int cti(char* c)
  {
  
    const char* ptr = c;
    int priceNum = 0;
    while (*ptr)
    {
        priceNum *= 10;
        priceNum += *ptr - '0';
        ptr++;
    }
    return priceNum;
  }
  struct Mat
  {
    int a[2][10];
    int b[10][10];
    int c;
  };
  void* Mul(void* para)
  {
    struct Mat* p = (struct Mat*)para;
    static int t[10][10] = { 0 };
    int i, j, k;
    for (i = 0; i < 2; i++)
        for (j = 0; j < 10; j++)
            for (k = 0; k < 10; k++)
                t[p->c + i][j] += (p->a[i][k]) * (p->b[k][j]);
    return (void*)t;
  }
  int main(int argc, char* argv[])
  {
    clock_t start, stop;
    start = clock();
    FILE* fpa, * fpb, * fpc;
    int count = 0, n = 0, i, j, k;
    char ch, c[3] = { 0 };
    fpa = fopen(argv[1], "r");
    fpb = fopen(argv[2], "r");
    fpc = fopen("t10.txt", "w");
    while (!feof(fpa))
    {
        ch = fgetc(fpa);
        if (ch == '#')
        {
            fseek(fpa, 15, SEEK_CUR);
            count++;
        }
        if (count == 2)break;
        if (ch == '\n')n++;
    }
    rewind(fpa);
    int** Ma = (int**)malloc(sizeof(int*) * n);
    for (i = 0; i < n; i++)
        Ma[i] = (int*)malloc(sizeof(int*) * n);
    count = 0;
    for (i = 0; i < n; i++)
        for (j = 0; j < n;)
        {
            ch = fgetc(fpa);
            if (ch == '#')
            {
                fseek(fpa, 15, SEEK_CUR);
                ch = fgetc(fpa);
            }
  
            if (ch != '\n')
            {
                c[count] = ch;
                count++;
            }
            else
            {
                c[count] = '\0';
                Ma[i][j] = cti(c);
                j++;
                count = 0;
            }
        }
  
    for (i = 0; i < n; i++)
        for (j = 0; j < n; j++)
        {
            printf("%d\t", Ma[i][j]);
            if (j == n - 1)printf("\n");
        }
    int** Mb = (int**)malloc(sizeof(int*) * n);
    for (i = 0; i < n; i++)
        Mb[i] = (int*)malloc(sizeof(int*) * n);
    for (i = 0; i < n; i++)
        for (j = 0; j < n;)
        {
            ch = fgetc(fpb);
            if (ch == '#')
            {
                fseek(fpb, 15, SEEK_CUR);
                ch = fgetc(fpb);
            }
  
            if (ch != '\n')
            {
                c[count] = ch;
                count++;
            }
            else
            {
                c[count] = '\0';
                Mb[i][j] = cti(c);
                j++;
                count = 0;
            }
        }
    for (i = 0; i < n; i++)
        for (j = 0; j < n; j++)
        {
            printf("%d\t", Mb[i][j]);
            if (j == n - 1)printf("\n");
        }
    int** Mc = (int**)malloc(sizeof(int*) * n);
    for (i = 0; i < n; i++)
        Mc[i] = (int*)malloc(sizeof(int*) * n);
    pthread_t p[5];
    struct Mat m[5];
    for (k = 0; k < 5; k++)
    {
        m[k].c = k * 2;
        for (j = 0; j < n; j++)
        {
            m[k].a[0][j] = Ma[2 * k][j];
            m[k].a[1][j] = Ma[2 * k + 1][j];
        }
    }
    for (k = 0; k < 5; k++)
        for (i = 0; i < n; i++)
            for (j = 0; j < n; j++)
                m[k].b[i][j] = Mb[i][j];
    for (k = 0; k < 5; k++)
    {
        pthread_create(&p[k], NULL, &Mul, &m[k]);
    }
    int r[n][n];
    for (k = 0; k < 5; k++)
    {
        pthread_join(p[k], (void*)&r);
        for (i = 0; i < n; i++)
            for (j = 0; j < n; j++)
                Mc[i][j] = r[i][j];
    }
    for (i = 0; i < n; i++)
    {
        for (j = 0; j < n; j++)
        {
            //printf("\t%d",mat[i][j]);
            if (j == 0)fprintf(fpc, "#This is Row %d:\n", i);
            fprintf(fpc, "%d\n", Mc[i][j]);
            //if(j==N-1)fprintf(fpa,"\n");
        }
    }
    fclose(fpa);
    fclose(fpb);
    fclose(fpc);
    stop = clock();
    double duration = ((double)(stop - start)) / CLOCKS_PER_SEC;
    printf("Time used = %.2lf\n", duration);
    return 0;
  }

来自24-34是线程函数 来自146-152是问题