blocks|key|146208|text|第一种在某种程度上是第二种的语法糖。第二种方法很明显是通过指针进行调用，它主要用于函数指针而不是常规函数，以使区别更加明显。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|146209|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The first one is somewhat of a syntactic sugar for the second one. The second one makes it obvious that you're making the call through a pointer, and it's used mainly with function pointers rather than regular functions, to make the distinction more obvious.

blocks|key|145532|text|就像数组类型几乎完全等效于对应的指针到元素类型一样，函数类型完全等效于对应的指针到函数类型：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|145533|void+(*func1)()+=+f;++//+function+type+->+pointer-to-function+type
void+(*func2)()+=+&f;+//+pointer-to-function+type+->+pointer-to-function+type|code-block|syntax|javascript|145534|而且还|145535|void+(*func)()+=+...;
func();+++++++++++++++//+pointer-to-function+type+%2B+function-call+operator
(*func)();++++++++++++//+function+type+%2B+function-call+operator|145536|所以，在|145537|(*f)();|145538|您正在取消引用f+(隐式转换为&f)，然后应用函数调用操作符。|offset|length|style|CODE|145539|entityMap^0|0|0|0|0|0|0|7|1|F|2|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@$Q|13|R|14|S|T]|$Q|15|R|16|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|V|$]]

Just as an array type is almost entirely equivalent to the corresponding pointer-to-element type, a function type is entirely equivalent to the corresponding pointer-to-function type:

<pre><code>void (*func1)() = f; // function type -&gt; pointer-to-function type
void (*func2)() = &amp;f; // pointer-to-function type -&gt; pointer-to-function type
</code></pre>

and also

<pre><code>void (*func)() = ...;
func(); // pointer-to-function type + function-call operator
(*func)(); // function type + function-call operator
</code></pre>

So, in

<pre><code>(*f)();
</code></pre>

you're dereferencing <code>f</code> (implicitly converted to <code>&amp;f</code>), and then apply the function-call operator.

blocks|key|1305825|text|在C%2B%2B中有两种调用函数的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1305826|按姓名：|1305827|f();|code-block|syntax|javascript|1305828|通过函数指针：|1305829|typedef+void+(*fptr_t)();
fptr_t+fptr+=+&f;
(*fptr)();|1305830|现在，在函数名(&f)上使用address-of操作符显然会创建一个函数指针.但是，当满足某些条件时，函数名可以隐式转换为函数指针。因此，上面的代码可以写成：|offset|length|style|CODE|1305831|typedef+void+(*fptr_t)();
fptr_t+fptr+=+f;+//+no+address-of+operator,+implicit+conversion
(*fptr)();|1305832|第二个例子正是这样做的，但是使用一个临时变量来保存函数指针，而不是使用一个命名的局部变量。|1305833|我更喜欢使用address-of，在创建函数指针时，含义更加清晰。|1305834|相关注意:如果提供函数指针，函数调用操作符将自动取消对函数指针的引用。所以这也是合法的：|1305835|typedef+void+(*fptr_t)();
fptr_t+fptr+=+&f;
fptr();|1305836|这对于模板非常有用，因为无论函数指针还是函子(实现operator()的对象)传入，相同的语法都能工作。|1305837|而且两种快捷方式都不适用于指针到成员，在那里您需要显式的地址和取消引用操作符。|1305838|1305839|在C，@Mehrdad+explains+that+all+function+calls+use+a+function+pointer.中|1305840|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/541686/mehrdad^0|0|0|0|0|0|8|2|0|0|0|0|0|0|P|A|0|0|0|3|1T|0|0^^$0|@$1|2|3|4|5|6|7|1H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1I|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1J|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1K|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|1L|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|1M|8|@$O|1N|P|1O|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|F|7|1P|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1R|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1S|8|@]|9|@]|A|$]]|$1|10|3|11|5|F|7|1T|8|@]|9|@]|A|$G|H]]|$1|12|3|13|5|6|7|1U|8|@$O|1V|P|1W|Q|R]]|9|@]|A|$]]|$1|14|3|15|5|6|7|1X|8|@]|9|@]|A|$]]|$1|16|3|-4|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|17|3|18|5|6|7|1Z|8|@]|9|@$O|20|P|21|1|22]]|A|$]]|$1|19|3|-4|5|6|7|23|8|@]|9|@]|A|$]]]|1A|$1B|$5|1C|1D|1E|A|$1F|1G]]]]

There are two ways to call a function in C++:

By name:

<pre><code>f();
</code></pre>

Through a function pointer:

<pre><code>typedef void (*fptr_t)();
fptr_t fptr = &amp;f;
(*fptr)();
</code></pre>

Now, using the address-of operator on a function name (<code>&amp;f</code>) obviously creates a function pointer. But the function name can implicitly convert to a function pointer when certain conditions are met. So the above code could be written as:

<pre><code>typedef void (*fptr_t)();
fptr_t fptr = f; // no address-of operator, implicit conversion
(*fptr)();
</code></pre>

Your second example is doing exactly this, but using a temporary variable to hold the function pointer instead of a named local variable.

I prefer to use address-of when creating function pointers, the meaning is much clearer.

A related note: The function call operator will automatically dereference a function pointer if one is provided. So this also is legal:

<pre><code>typedef void (*fptr_t)();
fptr_t fptr = &amp;f;
fptr();
</code></pre>

That's pretty useful with templates, because the same syntax works whether you have a function pointer or a functor (object implementing <code>operator()</code>) passed in.

And neither shortcut works with pointer-to-members, there you NEED explicit address-of and dereference operators.

<hr>

In C, <a href="https://stackoverflow.com/users/541686/mehrdad">@Mehrdad explains that all function calls use a function pointer.</a>

<blockquote>
 Possible Duplicate: 
 <a href="https://stackoverflow.com/questions/2795575/how-does-dereferencing-of-a-function-pointer-happen">How does dereferencing of a function pointer happen?</a> 
</blockquote>



If we have

<pre><code>void f() {
 printf("called");
}
</code></pre>

Then the following code will result in output of "calledcalled":

<pre><code>f();
(*f)();
</code></pre>

I don't really understand how this works…what is the difference between <code>*f</code> and <code>f</code>? And why would you call a function using the latter syntax?

value of function in c/C++

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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 可能重复：如果我们有void f() {    printf("called");}然后，以下代码将产生“被调用”的输出：f();(*f)();我不太明白这是怎么回事，…*f和f有什么区别？为什么要使用后一种语法调用函数呢？

问c/C++中的函数值
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