别名向量正确 [英] Aliasing vector correctly

问题描述

我在其他地方找不到答案，所以我想我只需要问这个：我试图得到一个向量的别名（其中存储int指针），如下：

I have not been able to find the answer elsewhere, so I guess I just have to ask this one: I am trying to get an alias for a vector (in which int pointers are stored), as below:

    void conversion(Engine * ENGINES) 
    {//The Engine class has a vector of int* as a public data member called SITE
        for (int i = 0; i < 3; i++)
        {
            vector <int*>* current = &(ENGINES[i].SITE);//the problematic line
            int j_max = current -> size();
            cout << j_max << endl;
            for (int j = 0; j < j_max; j++)
            {
                for (int k = 0; k < 3; k++)
                {
                     if (*current[j][k] == 2)
                    *current[j][k] = 1;
                     if (*current[j][k] == -1)
                    *current[j][k] = 0;
                }
            }
        }
    }

问题是似乎存在电流[a] [b]的指数的反转。我想要能够使用当前作为一个普通的向量，但现在的索引是相反的：

The problem is that there seems to be an inversion of the indices for the *current[a][b]. I want to be able to use current as a normal vector, but now the indexing is reversed compared to:

vector <int*> current1 = ENGINES[1].SITE;

，所以 * current [i] [j] = current1 [j] [i] 。我的语法有错误吗？

so that *current[i][j] = current1[j][i] for some reason. Is there a mistake in my syntax?

推荐答案

我相信你的问题是 [] 的优先级高于一元的 * 。所以你得到 *（current [j] [k]）而不是（* current）[j] [k] ，这是你想要的。

I believe your problem is that [] has higher precedence than unary *. So you're getting *(current[j][k]) instead of (*current)[j][k], which is what you want.

但是你可以通过引用而不是一个指针来消除这个问题：

However you could eliminate that problem by just taking a reference rather than a pointer:

vector< int *>& current =（ENGINES [i] .SITE）; 然后只需删除您的额外负载 * 。

vector <int*>& current = (ENGINES[i].SITE); and then just remove your extra loading * operators on access to current.

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