在 C/C++ 中传递二维数组时出错-C/C++问题

Error in Passing 2d Array in C/C++(在 C/C++ 中传递二维数组时出错)

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问题描述

我收到错误:

无法将 int** 转换为 int*[5]...

Cannot convert int** to int*[5]...

在我的代码中.我知道如何将二维数组传递给函数，但我被困在这里.

in my code. I know how to pass 2 D arrays to functions but I am stuck here.

该代码用于对邻接矩阵进行深度优先遍历.

The code is intended for Depth first Traversal over a Adjacency Matrix.

    #include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>

using namespace std;

#define m 5
#define n 5

void dfs(int G[][n],int i,int visited[])
{
     visited[i]=1;
     int j;
     for(j=0;j<n;j++)
     {
         if(!visited[j]&&G[i][j]==1)
                     dfs(G,j,visited);
     }
}

int main()
{   int r,c,i;
    printf("Enter Dimensions
");
    scanf("%d %d",&r,&c);
    r = m; c = n;
    int **G = (int **)malloc(r*sizeof(int*));
    for(i=0;i<r;i++)
    {
        G[i] = (int *)malloc(c*sizeof(int));
    }
    printf("Enter the Matrix
");
    int j;

    for(i=0;i<r;i++)
        for(j=0;j<c;j++)
          scanf("%d",G[i][j]);
    int *visited = (int*)malloc(r*sizeof(int));
    memset(visited,0,sizeof(visited));

    dfs(G,0,visited);

}

推荐答案

我认为你需要的是这个，它是 ac 解决方案，如果你想要 c++ 你应该使用 std::vector 代替.

I think what you need is this, it is a c solution, if you want c++ you should probably use std::vector instead.

#include <stdlib.h>
#include <string.h>
#include <stdio.h>

using namespace std;


void dfs(int **G, int i, int *visited, int c)
{
    int j;

    visited[i] = 1;
    for(j = 0 ; j < c ; j++)
    {
        if((visited[j] == 0) && (G[i][j] == 1))
            dfs(G, j, visited, c);
    }
}

int main()
{
    int   r, c, i, j;
    int **G;
    int  *visited;

    printf("Enter Dimensions
");
    scanf("%d %d", &r, &c);

    G = malloc(r * sizeof(*G));
    if (G == NULL)
        return -1;
    visited = malloc(r * sizeof(*visited));
    if (visited == NULL)
    {
        free(G);
        return -1;
    }
    memset(visited, 0, r * sizeof(*visited));

    printf("Enter the Matrix
");
    for(i = 0 ; i < r ; i++)
    {
        G[i] = malloc(c * sizeof(**G));
        if (G[i] == NULL)
        {
            for (j = i - 1 ; j >= 0 ; --j)
                free(G[j]);
            free(G);

            return -1;
        }
        for(j = 0 ; j < c ; j++)
            scanf("%d", &(G[i][j]));
    }
    dfs(G, 0, visited, c);

    /* now free */
    free(visited);
    for(i = 0 ; i < r ; i++)
        free(G[i]);
    free(G);

    return 0;
}

这也可能是一个 C++ 解决方案

this could also be a c++ solution

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>

void dfs(int **G, int i, int *visited, int c)
{
    int j;

    visited[i] = 1;
    for(j = 0 ; j < c ; j++)
    {
        if((visited[j] == 0) && (G[i][j] == 1))
            dfs(G, j, visited, c);
    }
}

int main()
{
    int   r, c, i, j;
    int **G;
    int  *visited;

    printf("Enter Dimensions
");
    scanf("%d %d", &r, &c);

    G       = new int*[r];
    visited = new int[r];

    memset(visited, 0, r * sizeof(*visited));

    printf("Enter the Matrix
");
    for(i = 0 ; i < r ; i++)
    {
        G[i] = new int[c];
        for(j = 0 ; j < c ; j++)
            scanf("%d", &(G[i][j]));
    }
    dfs(G, 0, visited, c);

    /* now free */
    delete[] visited;
    for(i = 0 ; i < r ; i++)
        delete[] G[i];
    delete[] G;

    return 0;
}

最后是 std::vector

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>

using namespace std;
void dfs(const vector< vector<int> > &G, int i, vector<int> &visited)
{
    size_t j;

    visited[i] = 1;
    for(j = 0 ; j < visited.size() ; j++)
    {
        if((visited[j] == 0) && (G[i][j] == 1))
            dfs(G, j, visited);
    }
}

int main()
{
    int r, c, i, j;


    printf("Enter Dimensions
");
    scanf("%d %d", &r, &c);

    vector< vector<int> > G(r, vector<int>(c));
    vector<int>           visited(r, 0);

    printf("Enter the Matrix
");
    for(i = 0 ; i < r ; i++)
    {
        for(j = 0 ; j < c ; j++)
            scanf("%d", &(G[i][j]));
    }
    dfs(G, 0, visited);

    return 0;
}

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