c ++将动态分配的二维向量传递给函数

Question

我正在尝试通过 C++ 中的引用将动态分配的二维向量传递给函数。

最初我试图用一个二维数组来做这个，但我被告知要尝试用一个二维向量来代替。由于转换错误，我下面的代码在 solve_point(boardExVector) 行失败。

#include <stdio.h>       /* printf */
#include <bits/stdc++.h> /* vector of strings */
using namespace std;

void solve_point(vector<char> *board){ 
    printf("solve_point\n");
    board[2][2] = 'c';
}

int main(){
    //dynamically allocate width and height
    int width = 7;
    int height = 9;
    //create 2d vector
    vector<vector<char>> boardExVector(width, vector<char>(height));
    boardExVector[1][2] = 'k';
    //pass to function by reference
    solve_point(boardExVector);
    //err: no suitable conversion function from "std::vector<std::vector<char, std::allocator<char>>, std::allocator<std::vector<char, std::allocator<char>>>>" to "std::vector<char, std::allocator<char>> *" exists
    printf("board[2][2] = %c\n", boardExVector[2][2]);
}

我刚刚回到 C++，所以指针和引用是我正在努力变得更好的东西，我已经在网上寻找解决方案并且已经尝试了一些通常涉及更改 solve_point 函数头来包含 * 或 & 但我还没有让它工作。任何帮助表示赞赏。谢谢

Answer 1

函数参数需要一个指向 char 类型向量的指针，而调用函数传递的是 vector<char> 类型向量。您是否正在寻找函数中的以下变化？

//bits/stdc++.h is not a standard library and must not be included.
#include <iostream>
#include <vector> /* vector of strings */
using namespace std;

void solve_point(vector<vector <char>> &board){
    printf("solve_point\n");
    board[2][2] = 'c';
}

int main(){
    //dynamically allocate width and height
    int width = 7;
    int height = 9;
    //create 2d vector
    vector<vector<char>> boardExVector(width, vector<char>(height));
    boardExVector[1][2] = 'k';
    //pass to function by reference
    solve_point(boardExVector);
    //err: no suitable conversion function from "std::vector<std::vector<char, std::allocator<char>>, std::allocator<std::vector<char, std::allocator<char>>>>" to "std::vector<char, std::allocator<char>> *" exists
    printf("board[2][2] = %c\n", boardExVector[2][2]);
}

c ++将动态分配的二维向量传递给函数

c++ pass dynamically allocated 2d vector to function

c++

pointers

reference

vector

2d-vector