blocks|key|263362|text|作为论据传递时，&function和function之间没有区别|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|263363|但是，您的typedefs之间存在差异。我不知道官方的解释，也就是说到底有什么不同，但根据我的记忆|263364|typedef+void+(*name1)(void);|code-block|syntax|javascript|263365|和|263366|typedef+void(name2)(void);|263367|是不同的：|263368|name1是指向不带参数且不返回任何内容的函数的指针|263369|name2是一个不带参数且不返回任何内容的函数|263370|你可以通过编译来测试它：|263371|typedef+void+(*pointer)(void);
typedef+void+(function)(void);

void+foo(void){}

int+main()
{
++++pointer+p;
++++function+f;

++++p+=+foo;+//compiles
++++p();

++++f+=+foo;+//does+not+compile
++++f();
}|263372|再说一次，我不适合解释这种行为的确切原因，但我相信，如果你看看标准，你会在某个地方找到解释。|263373|entityMap^0|8|9|I|8|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|19|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1B|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1D|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1E|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1F|8|@]|D|@]|E|$]]|$1|W|3|X|5|6|7|1G|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|J|7|1H|8|@]|D|@]|E|$K|L]]|$1|10|3|11|5|6|7|1I|8|@]|D|@]|E|$]]|$1|12|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|13|$]]

there is no difference between <code>&amp;function</code> and <code>function</code> when passing as arguement

however there is a difference between your typedefs. I do not know the official explanation, i.e what exactly the difference, but from what i remember

<pre><code>typedef void (*name1)(void);
</code></pre>

and

<pre><code>typedef void(name2)(void);
</code></pre>

are different:

name1 is a pointer to a function that takes no paramter and returns nothing

name2 is a function that takes no paramter and returns nothing

you can test it by compiling:

<pre><code>typedef void (*pointer)(void);
typedef void (function)(void);

void foo(void){}

int main()
{
 pointer p;
 function f;

 p = foo; //compiles
 p();

 f = foo; //does not compile
 f();
}
</code></pre>

again, i am not the right person to explain exact reason of this behavior but i believe if you take a look at standards you will find the explanation somewhere there

blocks|key|2960242|text|&function和function没有区别--它们都是地址。您可以通过同时输出这两个命令来查看：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2960243|function+bar();+

....+
printf("addr+bar+is+0x%25d\n",+&bar);
printf("bar+is+0x%25d\n",+bar);|code-block|syntax|javascript|2960244|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

There is no difference between &amp;function and function - they're both addresses. You can see this by printing them both: 

<pre><code>function bar(); 

.... 
printf("addr bar is 0x%d\n", &amp;bar);
printf("bar is 0x%d\n", bar);
</code></pre>

blocks|key|265550|text|这是没有区别的。有关证据，请参阅C99+specification+(第6.7.5.3.8节)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|265551|“将参数声明为”“函数返回类型”“应调整为”“指向函数返回类型的指针”“，如6.3.2.1中所示。”|265552|entityMap|0|LINK|mutability|MUTABLE|url|http://www.open-std.org/jtc1/sc22/WG14/www/docs/n1256.pdf^0|G|H|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]]|C|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]]]

There is no difference. For evidence see the <a href="http://www.open-std.org/jtc1/sc22/WG14/www/docs/n1256.pdf" rel="noreferrer">C99 specification</a> (section 6.7.5.3.8).

"A declaration of a parameter as ‘‘function returning type’’ shall be adjusted to ‘‘pointer to
function returning type’’, as in 6.3.2.1."

blocks|key|266990|text|区别只是风格上的不同。使用函数指针时也有同样的场景：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|266991|void+func+(void);|code-block|syntax|javascript|266992|..。|266993|void(*func_ptr)(void)+=+func;

func_ptr();++++//+call+func
(*func_ptr)();+//+call+func

printf("%25d\n",+ptr);+
printf("%25d\n",+*ptr);|266994|有一些人说(*func_ptr)()语法是首选的，以清楚地表明函数调用是通过函数指针完成的。其他人认为*的风格更清晰。|266995|像往常一样，可能没有科学研究证明这两种形式中的任何一种都比另一种更好，所以只选择一种风格并坚持下去。|266996|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The difference is only stylistic. You have the same scenario when using function pointers:

<pre><code>void func (void);
</code></pre>

...

<pre><code>void(*func_ptr)(void) = func;

func_ptr(); // call func
(*func_ptr)(); // call func

printf("%d\n", ptr); 
printf("%d\n", *ptr);
</code></pre>

There are some who say that the (*func_ptr)() syntax is to prefer, to make it clear that the function call is done through a function pointer. Others believe that the style with the * is clearer. 

As usual, there are likely no scientific studies proving that either form is better than the other, so just pick one style and stick to it.

For example:

<pre><code>#include &lt;stdio.h&gt;

typedef void (* proto_1)();
typedef void proto_2();

void my_function(int j){
 printf("hello from function. I got %d.\n",j);
}

void call_arg_1(proto_1 arg){
 arg(5);
}
void call_arg_2(proto_2 arg){
 arg(5);
}
void main(){
 call_arg_1(&amp;my_function);
 call_arg_1(my_function);
 call_arg_2(&amp;my_function);
 call_arg_2(my_function);
}
</code></pre>

Running this I get the following:

<pre><code>&gt; tcc -run try.c
hello from function. I got 5.
hello from function. I got 5.
hello from function. I got 5.
hello from function. I got 5.
</code></pre>

My two questions are:

<ul>
<li>What is the difference between a function prototype defined with <code>(* proto)</code> and one defined without?</li>
<li>What is the difference between calling a function with the reference operator (<code>&amp;</code>) and without?</li>
</ul>

In C, what is the difference between `&function` and `function` when passed as arguments?

例如：#include <stdio.h>typedef void (* proto_1)();typedef void proto_2();void my_function(int j){ printf("hello from function. I got %d.\n",j);}void call_arg_1...

问在C中，当作为参数传递时，`&function`和`function`有什么区别？
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问在C中，当作为参数传递时，`&function`和`function`有什么区别？
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