Question

我正在尝试通过c ++中的引用将动态分配的2d向量传递给函数。

最初，我尝试使用2d数组来执行此操作，但是却被告知要尝试使用2d向量。由于转换错误，我下面的代码在solve_point（boardExVector）行处失败。

#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]);
}

我只是重新使用C ++，所以我正在努力改进指针和引用，我在网上寻找了解决方案，并且已经尝试了一些方法，这些方法通常涉及更改resolve_point函数标头以包括*或＆，但我还没有开始工作。任何帮助表示赞赏。谢谢

Answer 1

函数参数需要一个指向char类型的向量的指针，而调用方函数正在传递vector<char>类型的向量。您是否正在寻找功能上的以下变化？

//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2d向量传递给函数

1 个答案: