C 程序未能正确地将整数附加到向量

Question

我有一个 C 程序，我想在其中数组工作不能被进程数完全整除的进程之间划分二维数组。 然而，这是我的小程序，我计算将分配给每个进程的大二维数组的起始 x 和 y 索引（以及每个进程将获得多少行和列）并将结果附加到向量。 将元素附加到向量时出现问题（我已打印出 i_start 和 j_start 向量的程序输出）。 我不确定我做错了什么，但任何帮助都会很棒。 谢谢你。

#include <stdio.h>
#include <math.h>

int main() {
  int K = 10;
  int L = 10;
  int P = 8;

  double pco = sqrt(P);
  int pcol = (int)pco;

  while((P % pcol) != 0) {
    pcol--;
  }

  int prow = P/pcol;

  int KL = K*L;

  int coldiv = K/pcol;
  int rowdiv = L/prow;

  int colrem = K % pcol;
  int rowrem = L % prow;

  int map[L][K];

  int count = 1;
  for(int i=0; i < L; i++) {
    for(int j=0; j < K;j++) {
      map[i][j] = count;
      count++;
    }
  }

  int i_start[P];
  int j_start[P];
  int n_rows[P];
  int n_cols[P];

  int start;
  int num_cols = 0;
  int count1 = 0;

  // appending to vector here
  for(int j = 0; j < prow; j++) {
    for(int i = 0; i < pcol; i++) {
      if(i < colrem) {
        start = (i*(coldiv+i));
        num_cols = coldiv + 1;
        printf("col index start: %d\n", i*(coldiv+i));
        printf("col index finish: %d\n", i*(coldiv+i)+coldiv+1);
      }
      else {
        start = (i*coldiv+colrem);
        num_cols = coldiv;
        printf("col index start: %d\n", i*coldiv+colrem);
        printf("col index finish: %d\n", i*coldiv+colrem+coldiv);
      }
      i_start[count1] = start;
      n_cols[count1] = num_cols;
      count1++;
    }
  }

  int count2 = 0;
  for(int j = 0; j < pcol; j++) {
    for(int i = 0; i < prow; i++) {
      int start, num_rows;
      if(i < rowrem) {
        int start = i*(rowdiv+i);
        num_rows = rowdiv + 1;
        printf("row index start: %d\n", i*(rowdiv+i));
        printf("row index finish: %d\n", i*(rowdiv+i)+rowdiv+1);
        j_start[count2] = start;
      }
      else {
        int start = i*rowdiv+rowrem;
        num_rows = rowdiv;
        printf("row index start: %d\n", i*rowdiv+rowrem);
        printf("row index finish: %d\n", i*rowdiv+rowrem+rowdiv);
        j_start[count2] = start;
      }
      n_rows[count2] = num_rows;
      count2++;
    }
  }

  // error: printing vectors
  for(int i = 0; i < P; i ++) {
    printf("i: %d\n", i);
    printf("i start: %d\n",i_start[P]);
    printf("j start: %d\n",j_start[P]);
  }
}

程序输出：

col index start: 0
col index finish: 5
col index start: 5
col index finish: 10
col index start: 0
col index finish: 5
col index start: 5
col index finish: 10
col index start: 0
col index finish: 5
col index start: 5
col index finish: 10
col index start: 0
col index finish: 5
col index start: 5
col index finish: 10
row index start: 0
row index finish: 3
row index start: 3
row index finish: 6
row index start: 6
row index finish: 8
row index start: 8
row index finish: 10
row index start: 0
row index finish: 3
row index start: 3
row index finish: 6
row index start: 6
row index finish: 8
row index start: 8
row index finish: 10
i: 0
i start: 1
j start: 0
i: 1
i start: 1
j start: 0
i: 2
i start: 1
j start: 0
i: 3
i start: 1
j start: 0
i: 4
i start: 1
j start: 0
i: 5
i start: 1
j start: 0
i: 6
i start: 1
j start: 0
i: 7
i start: 1
j start: 0

Answer 1

如果我理解您的意图，那么您只是打印了错误的输出。

 for(int i = 0; i < P; i ++) { printf("i: %d\\n", i); printf("i start: %d\\n",i_start[P]); printf("j start: %d\\n",j_start[P]); }

在这里，您为i_start和j_start打印始终与P而不是i相同的索引。

更改为：

 for(int i = 0; i < P; i ++) {
     printf("i: %d\n", i);
     printf("i start: %d\n",i_start[i]);
     printf("j start: %d\n",j_start[i]);
 }

你的输出将变成：

col index start: 0
col index finish: 5
col index start: 5
col index finish: 10
col index start: 0
col index finish: 5
col index start: 5
col index finish: 10
col index start: 0
col index finish: 5
col index start: 5
col index finish: 10
col index start: 0
col index finish: 5
col index start: 5
col index finish: 10
row index start: 0
row index finish: 3
row index start: 3
row index finish: 6
row index start: 6
row index finish: 8
row index start: 8
row index finish: 10
row index start: 0
row index finish: 3
row index start: 3
row index finish: 6
row index start: 6
row index finish: 8
row index start: 8
row index finish: 10
i: 0
i start: 0
j start: 0
i: 1
i start: 5
j start: 3
i: 2
i start: 0
j start: 6
i: 3
i start: 5
j start: 8
i: 4
i start: 0
j start: 0
i: 5
i start: 5
j start: 3
i: 6
i start: 0
j start: 6
i: 7
i start: 5
j start: 8