为什么可以在函数参数中指定数组的大小？

问题描述

我不明白为什么下面的示例编译和工作：

I don't understand why the following example compiles and works:

void printValues(int nums[3], int length) {
    for(int i = 0; i < length; i++)
        std::cout << nums[i] << " ";
    std::cout << '\n';
}

似乎3的大小完全被忽略，在编译错误。这是怎么回事？

It seems that the size of 3 is completely ignored but putting an invalid size results in a compile error. What is going on here?

推荐答案

在C ++（以及C）中，以数组类型声明的参数总是立即衰减到指针类型。以下三个声明是等效的

In C++ (as well as in C), parameters declared with array type always immediately decay to pointer type. The following three declarations are equivalent

void printValues(int nums[3], int length);
void printValues(int nums[], int length);
void printValues(int *nums, int length);

大小没有关系。但是，它并不意味着你可以使用一个无效的数组声明，例如，指定一个负或零大小是非法的。例如：

I.e. the size does not matter. Yet, it still does not mean that you can use an invalid array declaration there, i.e. it is illegal to specify a negative or zero size, for example.

（BTW，这同样适用于函数类型的参数 - 它立即衰减到指针到函数类型。）

(BTW, the same applies to parameters of function type - it immediately decays to pointer-to-function type.)

如果要强制参数和参数之间的数组大小匹配，请使用参数声明中的指针或数组类型

If you want to enforce array size matching between arguments and parameters, use pointer- or reference-to-array types in parameter declarations

void printValues(int (&nums)[3]);
void printValues(int (*nums)[3]);

当然，在这种情况下，大小将成为编译时常量， length 。

Of course, in this case the size will become a compile-time constant and there's no point of passing length anymore.

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