【c++】以函数调用的方式来书写矩阵乘法

/*m：A的行
* o：A的列，B的行
* n：B的列
* res：存储最终的结果
 */
void multi(double A0[21][21], double B0[21][21], int m, int o, int n, double res[21][21]) {
    double A[21][21];
    double B[21][21];
    double temp;
//防止对原矩阵进行修改
    for (int i = 0;i < 21;i++) {
        for (int j = 0;j < 21;j++) {
            A[i][j] = A0[i][j];
        }
    }
    for (int i = 0;i < 21;i++) {
        for (int j = 0;j < 21;j++) {
            B[i][j] = B0[i][j];
        }
    }
    for (int i = 0;i < m + 1;i++) {
        for (int j = 0;j < n + 1;j++) {
            temp = 0;
            for (int k = 0;k < o + 1;k++) {
                temp += A[i][k] * B[k][j];
            }
            res[i][j] = temp;
        }
    }
}

void test(double* arr) { //错误写法。

}

void test(double** arr) { //错误写法。二级指针用来存放一级指针的地址，而不是数组第一行的地址

}

void test(double* arr[2]) { //错误写法

}

void test(double (*arr)[2]) {//正确写法

}

#include <iostream>
#include <assert.h>

constexpr int row{ 2 };
constexpr int col{ 2 };


void multi(double (* A0)[col], double (* B0)[col], int row1, int col1, int row2, int col2, double (* res)[col]) {
    assert(col1 == row2);
    double temp;
    for (int i = 0;i ！= row1; ++i) {
        for (int j = 0;j ！= col2; ++j) {
            temp = 0;
            for (int k = 0;k ！= col1; ++k) {
                temp += *((*A0 + i) + k) * *((*B0 + k) + j);
                //temp += A0[i][k] * B0[k][j];
            }
            res[i][j] = temp;
        }
    }
}

#include <iostream>
#include <assert.h>
#include <vector>
#include <algorithm>
#include <array>

using vec = std::vector<std::vector<double>>;
using vecRow = std::vector<double>;

vec Multi(vec& A, vec& B) {
    int row1 = A.size();
    int col1 = A[0].size();
    int row2 = B.size();
    int col2 = B[0].size();
    vec res;
    vecRow temp;
    double sum;
    //double temp;
    assert(col1 == row2 && "第一个矩阵的列和第二个矩阵的行不相等");
    for (int i = 0;i != row1;++i) {
        for (int j = 0;j != col2;++j) {
            sum = 0;
            for (int k = 0;k != col1;++k) {
                sum += A[i][k] * B[k][j];
            }
            temp.push_back(sum);
        }
        res.push_back(temp);
        temp.erase(temp.begin(), temp.end());
    }
    return res;
}

int main() {
    vec A{ {1.0, 1.0}, { 2.0,2.0 } };
    vec B { {2.0, 2.0}, { 1.0,1.0 } };
    vec res;
    res = Multi(A, B);
    show(res);
}

vec multi(vec A, vec B) {
    int row1{ static_cast<int> (A.size()) };
    int col1{ static_cast<int>(A[0].size()) };
    int row2{ static_cast<int>(B.size()) };
    int col2{ static_cast<int>(B[0].size() )};
    vec res(row1, vecRow(col2));
    double temp{};
    assert(col1 == row2 && "第一个矩阵的列和第二个矩阵的行不相等");
    for (int i = 0;i != row1;++i) {
        for (int j = 0;j != col2;++j) {
            temp = 0;
            for (int k = 0;k != col1;++k) {
                temp += A[i][k] * B[k][j];
            }
            res[i][j] = temp;
        }
    }
    return res;
}

void show(vec& A) {
    for (vecRow i : A) {
        for (double j : i) {
            std::cout << j << " ";
       }
        std::cout << '\n';
    }
   /* for (int i = 0;i != A.size();++i) {
        for (int j = 0;j != A[0].size();++j) {
            std::cout << A[i][j] << "  ";
        }
        std::cout << "\n";
    }*/
}