“此" C语言中的指针(不是C ++) [英] "this" pointer in C (not C++)

问题描述

我正尝试在C中创建一个堆栈，这很有趣，并且想到了使用struct表示堆栈的想法.然后，将函数指针添加到针对push()和pop()操作的结构.

I'm trying to create a stack in C for fun, and came up with the idea of using struct to represent the stack. Then I add function pointers to the struct for push() and pop() operations.

到目前为止，看起来一切都很好，但是，对于push()和pop()函数的实现，我需要以某种方式引用*.那怎么办(可以吗?)?

So far all is good it seems, but, for the implementation of the push() and pop() functions I need to refer to *this somehow. How can that (can it?) be done?

这是我的结构

struct Stack {
    int *data;
    int current_size;
    int max_size;
    int (*push)(int);
    int (*pop)();
};

以示例为例，

int push(int val) {
    if(current_size == max_size -1)
            return 0;

    data[current_size] = val;
    current_size++;

    return 1;
}

您可以想象，编译器不知道current_size是什么，因为它会像stack->current_size这样.

As you can imagine, the compiler has no idea what current_size is, as it would expect something like stack->current_size.

这有可能克服吗?

推荐答案

C中没有隐式的this.使其明确:

There's no implicit this in C. Make it explicit:

int push(Stack* self, int val) {
    if(self->current_size == self->max_size - 1)
            return 0;

    self->data[self->current_size] = val;
    (self->current_size)++;

    return 1;
}

您当然必须在每次调用push和类似方法的过程中都将指向该结构的指针传递给我们.

You will of course have to pass the pointer to the struct into every call to push and similar methods.

当您将Stack定义为类而将push等定义为方法时，这基本上就是C ++编译器为您所做的事情.

This is essentially what the C++ compiler is doing for you when you define Stack as a class and push et al as methods.

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