C++ 使用“this”关键字时出现“错误：表达式必须具有指针类型”
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"Error: expression must have a pointer type" when using the "this" keyword
提问by user1779944
So I made a parent class, which i will call Parentthat has a Square*grid member variable. The grid variable is a pointer to a large array of Squares, which contain keymember variables. (Think of this project as a hashtable) The problem is I am making a function in the Parentclass that edits the key variables in the Squarearray, and getting an error. This line of code compiles:
所以我创建了一个父类，我称之为Parent具有Square*网格成员变量。grid 变量是一个指向包含key成员变量的大型 Squares 数组的指针。（将此项目视为哈希表）问题是我在Parent类中创建了一个函数来编辑Square数组中的关键变量，但出现错误。这行代码编译：
this->grid = new Square[row*col];
but this line does not compile:
但这行不编译：
this->grid[i*col + j]->key1 = j;
it underlines thisand says expression must have a pointer type. I was wondering if anyone had ideas to what i may be doing wrong?
它强调this并表示表达式必须具有指针类型。我想知道是否有人对我可能做错了什么有想法？
void Parent::initialize(int row,int col) {
this->grid = new Square[row*col];
for(int i = 0; i < row; i++) {
for(int j = 0;j < col; j++) {
this->grid[i*col + j]->key1 = j;
this->grid[i*col + j]->key2 = i;
}
}
回答by il_guru
You have to use
你必须使用
this->grid[i*col + j].key1
this->grid[i*col + j].key2
That is because even if it is true that your grid is a pointer, you have allocated in the are pointed by its memory an array of Squareobject. So when you use the [] operator you are obtaining an object of type Squareand not a Square*and for a Squareobject you hve to use the . operatorand not the -> operator.
那是因为即使您的网格确实是一个指针，您也已在其内存指向的Square对象中分配了一个对象数组。因此，当您使用 [] 运算符时，您将获得一个类型Square而不是 a的对象，Square*并且对于一个Square对象，您必须使用. 运算符而不是-> 运算符。
回答by PiotrNycz
I guess this->gridis of type Square*, so this->grid[0]is of type Square&and you must use .(dot) not ->(arrow) to access members from Square&. To use arrow for expression expression must have a pointer type...
我猜this->grid是 type Square*，所以this->grid[0]也是 type Square&，你必须使用.（点）而不是->（箭头）从Square&. 要使用箭头表达式表达式必须有一个指针类型...
this->grid[i*col + j]->key2
// ^^: this->grid[i*col + j] is not a pointer
this->grid[i*col + j].key2
// ^ use dot to access key2 and key1 too
