为什么在C＆＃x2B;＆＃x2B;中编写比较器时，我们会添加括号？？

发布于 2025-02-03 08:23:25 字数 253 浏览 3 评论 0原文

这是一个解释我的意思的代码。


static bool comparator(int a, int b) {
    if(a > b) return false;
    return true;
}

sort(arr.begin(), arr.end(), comparator); // why don't we write comparator()

原文

Here is a code explaining what I mean.


static bool comparator(int a, int b) {
    if(a > b) return false;
    return true;
}

sort(arr.begin(), arr.end(), comparator); // why don't we write comparator()

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冬天旳寂寞 2025-02-10 08:23:25

为什么在c ++？
中编写时，我们不添加括号

因为我们不是传达调用该功能的结果，而是指向该功能的指针。特别是，第三个参数名为比较器 to std :: sort will 隐式衰减 指向该功能类型由于类型衰减。您可能已经熟悉内置阵列的类型衰减，这也会自动衰减到指向其第一个元素的指针。此（免费功能指针的功能）只是类型衰减的另一个实例。

如果我们要通过comaparator（），那么我们将传递一个类型bool的值，这不是预期的。此外，由于比较器需要2 int参数，因此，如果我们写comparator（），那将是无效的，因为我们不会传递任何传递任何争论它。

它是，好像您写了：

//---------------------------v--------------->note the & operator used here which is optional
sort(arr.begin(), arr.end(), &comparator);   //this statement is equivalent to the statement that you have in your question

上面修改的语句等同于您在示例中的陈述。这里唯一的句法差异是，这里我们有明确使用＆amp;来指示我们将指针传递给了一个函数，代码>比较器。

Why don't we add parenthesis when writing comparator in c++?

Because we're not passing the result of calling that function but instead a pointer to that function. In particular, the third argument named comparator to std::sort will implicitly decay to a pointer to that function type due to type decay. You might already be familiar with type decay for built in arrays which also automatically decay to a pointer to their first element. This(a free function to function pointer) is just another instance of type decay.

If we were to pass comaparator() then we would be passing a value of type bool which is not what is expected. Additionally since comparator takes 2 int arguments and so if we'were to write comparator() it would be invalid because we're not passing any argument to it.

It is as if you wrote:

//---------------------------v--------------->note the & operator used here which is optional
sort(arr.begin(), arr.end(), &comparator);   //this statement is equivalent to the statement that you have in your question

The above modified statement is equivalent to the statement that you have in your example. The only syntactic difference here is that here we have explicitly used the & to indicate that we're passing a pointer to the function named comparator.

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想你只要分分秒秒 2025-02-10 08:23:25

如果您要编写

sort(arr.begin(), arr.end(), comparator());

，则意味着必须评估函数std :: sort code> comparator（）的参数表达式。但是，这两个参数都不提供给函数调用comparator（）。因此，编译器将发出错误消息。

另一方面，如果您要编写，

sort(arr.begin(), arr.end(), comparator);

则功能指示器比较器将隐式转换为指向函数的指针，并且内部内部的函数std :: sort本身将调用该功能使用指针并为其提供两个参数。

If you will write

sort(arr.begin(), arr.end(), comparator());

then it means that the argument expression of the function std::sort comparator() must be evaluated. But neither arguments are supplied to the function call comparator(). So the compiler will issue an error message.

On the other hand, if you will write

sort(arr.begin(), arr.end(), comparator);

then the function designator comparator is implicitly converted to a pointer to the function and the function std::sort within itself will call the function using the pointer and supplying two arguments to it.

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