Question

存在问题：＆＃34;实现一个函数来接收表示整数的正方形数组的向量向量并返回指向包含上对角元素的副本（逐行）的1-D运行时数组的指针。＆＃34; 以下是我到目前为止所做的事情，请原谅我第一次在这里发布的任何错误：

#include<iostream>
#include<vector>
#include<cstdlib>
using namespace std;
const int N = 4;
int * pointer(vector<vector<int>>amatrix)
{
    int L = (N*N - N) / 2;
    int * b;
    b = new int[L];
        for (int i = 0; i < N-1; i++)
            for (int j = 1; j < N; j++)
            {
                int c = amatrix[i][j];
                for (int k = 0; k < (N*N - N) / 2; k++)
                    b[k] = c;
            }
    return b;
}
void main()
{
    vector<vector<int>>amatrix(N, vector<int>(N));
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            amatrix[i][j] = rand() % 10;
            cout << amatrix[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
    for (int i = 0; i < (N*N - N) / 2; i++)
        cout << pointer(amatrix)[i] << " ";

    cout << endl;
    system("pause");
}

Answer 1

捕捉！：）

#include <iostream>
#include <vector>
#include <cstdlib>
#include <ctime>

int * pointer( const std::vector<std::vector<int>> &amatrix )
{
    size_t n = amatrix.size();
    size_t size = ( n * n - n ) / 2;

    int *p = new int[size];
    int *q = p;

    for ( size_t i = 0; i < n; i++ )
    {
        for ( size_t j = i + 1; j < n; j++ ) *q++ = amatrix[i][j];
    }        

    return p;
}

int main()
{
    const size_t N = 4;

    std::vector<std::vector<int>> amatrix( N, std::vector<int>( N ) );

    std::srand( ( unsigned int )std::time( 0 ) );

    for ( size_t i = 0; i < N; i++ )
    {
        for ( size_t j = 0; j < N; j++ ) amatrix[i][j] = std::rand() % 10;
    }

    for ( size_t i = 0; i < N; i++ )
    {
        for ( size_t j = 0; j < N; j++ ) std::cout << amatrix[i][j] << ' ';
        std::cout << std::endl;
    }
    std::cout << std::endl;

    int *p = pointer( amatrix );

    for ( size_t i = 0; i < ( N * N - N ) / 2; i++ ) std::cout << p[i] << ' ';
    std::cout << std::endl;

    delete []p;
}

程序输出可能看起来像

6 3 4 7 
6 0 1 2 
9 9 7 8 
8 8 4 4 

3 4 7 1 2 8

至于你的代码，然后在函数

中循环

            for (int k = 0; k < (N*N - N) / 2; k++)
                b[k] = c;

没有意义。

同样没有意义为循环的每次迭代调用函数

for (int i = 0; i < (N*N - N) / 2; i++)
    cout << pointer(amatrix)[i] << " ";

因为会有内存泄漏，因为每次调用它时都会分配一个新数组。

从向量返回指向数组的指针

1 个答案: