blocks|key|708967|text|这是一种数组引用的语法--您需要使用(&array)向编译器说明您需要的是对数组的引用，而不是引用int+&+array[100];的(无效)数组。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|708968|编辑:一些说明。|708969|void+foo(int+*+x);
void+foo(int+x[100]);
void+foo(int+x[]);|code-block|syntax|javascript|708970|这三种是声明相同函数的不同方式。它们都被视为接受int+*参数，您可以将任意大小的数组传递给它们。|708971|void+foo(int+(&x)[100]);|708972|这只接受100个整数的数组。您可以在x上安全地使用sizeof|708973|void+foo(int+&+x[100]);+//+error|708974|这被解析为“引用数组”--这是不合法的。|708975|entityMap^0|I|8|1D|H|0|0|0|O|5|0|0|I|1|P|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|S|3|T|5|J|7|1E|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1F|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|X|$]]

It's a syntax for array references - you need to use <code>(&amp;array)</code> to clarify to the compiler that you want a reference to an array, rather than the (invalid) array of references <code>int &amp; array[100];</code>. 

EDIT: Some clarification.

<pre><code>void foo(int * x);
void foo(int x[100]);
void foo(int x[]);
</code></pre>

These three are different ways of declaring the same function. They're all treated as taking an <code>int *</code> parameter, you can pass any size array to them.

<pre><code>void foo(int (&amp;x)[100]);
</code></pre>

This only accepts arrays of 100 integers. You can safely use <code>sizeof</code> on <code>x</code>

<pre><code>void foo(int &amp; x[100]); // error
</code></pre>

This is parsed as an "array of references" - which isn't legal.

blocks|key|709099|text|这只是必需的语法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|709100|void+Func(int+(&myArray)[100])|code-block|syntax|javascript|709101|%5E引用传递100+int数组，参数名为myArray；|offset|length|style|CODE|709102|void+Func(int*+myArray)|709103|%5E传递数组。数组衰减为指针。因此，您将丢失大小信息。|709104|void+Func(int+(*myFunc)(double))|709105|%5E传递函数指针。该函数返回一个int并接受一个double。参数名称为myFunc。|709106|entityMap^0|0|0|9|3|J|7|0|0|0|0|F|3|N|6|Z|6|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]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|16|8|@$I|17|J|18|K|L]|$I|19|J|1A|K|L]|$I|1B|J|1C|K|L]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$]]

It's just the required syntax:

<pre><code>void Func(int (&amp;myArray)[100])
</code></pre>

^ Pass array of 100 <code>int</code> by reference the parameters name is <code>myArray</code>;

<pre><code>void Func(int* myArray)
</code></pre>

^ Pass an array. Array decays to a pointer. Thus you lose size information.

<pre><code>void Func(int (*myFunc)(double))
</code></pre>

^ Pass a function pointer. The function returns an <code>int</code> and takes a <code>double</code>. The parameter name is <code>myFunc</code>.

blocks|key|925437|text|它是一种语法。在函数参数中，必须使用包含&myArray的int+(&myArray)[100]括号。如果不使用它们，您将传递一个array+of+references，这是因为subscript+operator+[]比&+operator具有更高的优先级。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|925438|例如，int+&myArray[100]+//+array+of+references|925439|因此，通过使用type+construction+()，您告诉编译器您希望引用一个由100个整数组成的数组。|925440|例如int+(&myArray)[100]+//+reference+of+an+array+of+100+ints|925441|entityMap^0|K|8|T|J|1T|J|2H|L|33|A|0|3|14|0|7|K|0|2|1K|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|L|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|M|$]]

It is a syntax. In the function arguments <code>int (&amp;myArray)[100]</code> parenthesis that enclose the <code>&amp;myArray</code> are necessary. if you don't use them, you will be passing an <code>array of references</code> and that is because the <code>subscript operator []</code> has higher precedence over the <code>&amp; operator</code>.

E.g. <code>int &amp;myArray[100] // array of references</code>

So, by using <code>type construction ()</code> you tell the compiler that you want a reference to an array of 100 integers.

E.g <code>int (&amp;myArray)[100] // reference of an array of 100 ints</code>

blocks|key|925486|text|默认情况下，数组由指针传递。您可以尝试修改函数调用中的数组，以便更好地理解。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|925487|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|$]]

Arrays are default passed by pointers. You can try modifying an array inside a function call for better understanding.

blocks|key|709273|text|下面创建一个泛型函数，通过引用获取任意大小和任意类型的数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|709274|template<typename+T,+std::size_t+S>
void+my_func(T+(&arr)[S])+{
+++//+do+stuff
}|code-block|syntax|javascript|709275|play+with+the+code.|offset|length|709276|entityMap|0|LINK|mutability|MUTABLE|url|https://godbolt.org/z/nPqUXY^0|0|0|0|J|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

The following creates a generic function, taking an array of any size and of any type by reference:

<pre><code>template&lt;typename T, std::size_t S&gt;
void my_func(T (&amp;arr)[S]) {
 // do stuff
}
</code></pre>

<a href="https://godbolt.org/z/nPqUXY" rel="noreferrer">play with the code.</a>

How does passing a statically allocated array by reference work?

<pre><code>void foo(int (&amp;myArray)[100])
{
}

int main()
{
 int a[100];
 foo(a);
}
</code></pre>

Does <code>(&amp;myArray)[100]</code> have any meaning or its just a syntax to pass any array by reference?
I don't understand separate parenthesis followed by big brackets here. Thanks.

Passing an array by reference

编译器

通过引用传递静态分配的数组是如何工作的？void foo(int (&myArray)[100]){}int main(){    int a[100];    foo(a);}(&myArray)[100]有什么意义吗?或者它只是一种通过引用传递数组的语法？我不明白这里后面跟着大括号的单独括号。谢谢。

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