通过引用传递数组 [英] Passing an array by reference

问题描述

如何通过引用传递静态分配的数组?

How does passing a statically allocated array by reference work?

void foo(int (&myArray)[100])
{
}

int main()
{
    int a[100];
    foo(a);
}

(&myArray)[100] 是否有任何意义，或者它只是一种通过引用传递任何数组的语法?我不明白这里的单独括号后跟大括号.谢谢.

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.

推荐答案

这是数组引用的语法 - 您需要使用 (&array) 向编译器说明您需要引用到一个数组，而不是(无效的)引用数组 int &数组[100];.

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];.

一些说明.

void foo(int * x);
void foo(int x[100]);
void foo(int x[]);

这三种是声明同一个函数的不同方式.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.

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.

void foo(int (&x)[100]);

这仅接受 100 个整数的数组.您可以安全地在 x

This only accepts arrays of 100 integers. You can safely use sizeof on x

void foo(int & x[100]); // error

这被解析为引用数组"——这是不合法的.

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

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