C++ 而循环正在更改每个对象，而不是遍历

这就是实现：

 result[j]->coordinate[i] = (*tmp[i] * *v.columns[j])

for（自动i=0；idimension.rows；i++）{
for（auto j=0；jdimension.rows；k++）{
结果[j]->坐标[i]+=this->columns[i]->坐标[k]*v.columns[k]->坐标[j]；
printf（“j=%d，%s和%s\n”，j，结果[j]->printV（），结果[0]->printV（））；
}
}
}   

当j=1时，结果[0]也会改变，我不知道为什么。这是输出

for(auto i=0; i < this->dimension.rows;i++){
            for(auto j=0; j<v.dimension.columns;j++){
                for(auto k=0; k< this->dimension.rows;k++){
                    result[j]->coordinate[i] += this->columns[i]->coordinate[k] * v.columns[k]->coordinate[j];
                    printf("j = %d, %s and %s\n",j,result[j]->printV(), result[0]->printV());
                }
            }
        }   

j=0，且
j=0，且
j=1，且
j=1，且
j=2，且
j=2，且
j=0，且
j=0，且
j=1，且
j=1，且
j=2，且
j=2，且

矩阵乘法的定义不同。它需要3个嵌套for循环，而不仅仅是两个：

对于每对

（i，j）

，需要迭代

k

的所有值，并执行以下伪代码：

j = 0, <2.000000,0.000000> and <2.000000,0.000000>
j = 0, <2.000000,0.000000> and <2.000000,0.000000>
j = 1, <2.000000,0.000000> and <2.000000,0.000000>
j = 1, <1.000000,0.000000> and <1.000000,0.000000>
j = 2, <1.000000,0.000000> and <1.000000,0.000000>
j = 2, <1.000000,0.000000> and <1.000000,0.000000>
j = 0, <1.000000,-1.000000> and <1.000000,-1.000000>
j = 0, <1.000000,-1.000000> and <1.000000,-1.000000>
j = 1, <1.000000,-1.000000> and <1.000000,-1.000000>
j = 1, <1.000000,1.000000> and <1.000000,1.000000>
j = 2, <1.000000,1.000000> and <1.000000,1.000000>
j = 2, <1.000000,1.000000> and <1.000000,1.000000> 

---

printV函数在哪里？？？

strncat（s

导致UB，

strncat

需要以字符串开头

snprintf

被调用时长度参数大于缓冲区大小（尽管在这种特殊情况下不太可能溢出）

strncat

3参数在每次调用中都是不正确的；所有这些参数都可能是

strcat

（尽管

snprintf

会更高效、更整洁）

操作符*

（以及其他操作符）应该按值返回；您现在这样做会导致内存泄漏。根本不要使用

new

。我已经编辑了原始帖子来实现您的伪代码。它仍然不起作用。我犯了一个错误。现在编辑了我的帖子。它仍然会给出相同的问题，即向量结果的第一个坐标[0]正在变化。我想你把结果（I，j）和结果（j，I）混淆了。当j=1和k=1时似乎会发生这种情况。结果（I，j）是指第I行和第j列？

for(auto i=0; i < this->dimension.rows;i++){
            for(auto j=0; j<v.dimension.columns;j++){
                for(auto k=0; k< this->dimension.rows;k++){
                    result[j]->coordinate[i] += this->columns[i]->coordinate[k] * v.columns[k]->coordinate[j];
                    printf("j = %d, %s and %s\n",j,result[j]->printV(), result[0]->printV());
                }
            }
        }   

j = 0, <2.000000,0.000000> and <2.000000,0.000000>
j = 0, <2.000000,0.000000> and <2.000000,0.000000>
j = 1, <2.000000,0.000000> and <2.000000,0.000000>
j = 1, <1.000000,0.000000> and <1.000000,0.000000>
j = 2, <1.000000,0.000000> and <1.000000,0.000000>
j = 2, <1.000000,0.000000> and <1.000000,0.000000>
j = 0, <1.000000,-1.000000> and <1.000000,-1.000000>
j = 0, <1.000000,-1.000000> and <1.000000,-1.000000>
j = 1, <1.000000,-1.000000> and <1.000000,-1.000000>
j = 1, <1.000000,1.000000> and <1.000000,1.000000>
j = 2, <1.000000,1.000000> and <1.000000,1.000000>
j = 2, <1.000000,1.000000> and <1.000000,1.000000> 

result = zeros(a.rows,b.cols)
for i = 0:a.rows-1
   for j = 0:b.cols-1
      for k = 0:a.cols-1
         result(i,j) += a(i,k) * b(k,j)