C++ 通过引用传递数组
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/5724171/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Passing an array by reference
提问by John DB
How does passing a statically allocated array by reference work?
通过引用传递静态分配的数组如何工作？
void foo(int (&myArray)[100])
{
}
int main()
{
int a[100];
foo(a);
}
Does (&myArray)[100]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.
是否(&myArray)[100]有任何意义或它只是一个语法按引用传递任何阵列？我不明白单独的括号后面是大括号。谢谢。
回答by Erik
It's a syntax for array references - you need to use (&array)to clarify to the compiler that you want a reference to an array, rather than the (invalid) array of references int & array[100];.
这是数组引用的语法 - 您需要使用(&array)来向编译器说明您想要对数组的引用，而不是（无效的）引用数组int & array[100];。
EDIT: Some clarification.
编辑：一些澄清。
void foo(int * x);
void foo(int x[100]);
void foo(int x[]);
These three are different ways of declaring the same function. They're all treated as taking an int *parameter, you can pass any size array to them.
这三种是声明相同函数的不同方式。它们都被视为接受一个int *参数，您可以将任何大小的数组传递给它们。
void foo(int (&x)[100]);
This only accepts arrays of 100 integers. You can safely use sizeofon x
这仅接受 100 个整数的数组。您可以放心地使用sizeof上x
void foo(int & x[100]); // error
This is parsed as an "array of references" - which isn't legal.
这被解析为“引用数组”——这是不合法的。
回答by Martin York
It's just the required syntax:
这只是所需的语法：
void Func(int (&myArray)[100])
^ Pass array of 100 intby reference the parameters name is myArray;
^int通过引用传递 100 个数组，参数名称为myArray;
void Func(int* myArray)
^ Pass an array. Array decays to a pointer. Thus you lose size information.
^ 传递一个数组。数组衰减为指针。因此，您会丢失尺寸信息。
void Func(int (*myFunc)(double))
^ Pass a function pointer. The function returns an intand takes a double. The parameter name is myFunc.
^ 传递函数指针。该函数返回一个int并接受一个double。参数名称是myFunc.
回答by cpx
It is a syntax. In the function arguments int (&myArray)[100]parenthesis that enclose the &myArrayare necessary. if you don't use them, you will be passing an array of referencesand that is because the subscript operator []has higher precedence over the & operator.
它是一种语法。在函数参数int (&myArray)[100]中，括号&myArray是必要的。如果你不使用它们，你会传递一个array of references，就是因为subscript operator []拥有更高的优先级& operator。
E.g. int &myArray[100] // array of references
例如 int &myArray[100] // array of references
So, by using type construction ()you tell the compiler that you want a reference to an array of 100 integers.
因此，通过使用，type construction ()您可以告诉编译器您想要一个包含 100 个整数的数组的引用。
E.g int (&myArray)[100] // reference of an array of 100 ints
例如 int (&myArray)[100] // reference of an array of 100 ints
回答by Eduardo A. Fernández Díaz
Arrays are default passed by pointers. You can try modifying an array inside a function call for better understanding.
数组默认通过指针传递。您可以尝试在函数调用中修改数组以更好地理解。
