在C中传递数组:方括号与指针 [英] Passing arrays in C: square brackets vs. pointer

问题描述

我想将数组传递给函数.从我所看到的，有两种方法可以做到这一点:

I'm wanting to pass an array into a function. From what I can see, there are 2 ways of doing this:

1.

void f (int array[]) {
    // Taking an array with square brackets
}

2.

void f (int *array) {
    // Taking a pointer
}

每个呼叫者是:

int array[] = {0, 1, 2, 3, 4, 5};
f (array);

这两种方法之间有什么实际区别吗?

Is there any actual difference between these 2 approaches?

推荐答案

在您的特定示例中没有区别.

In your specific example there is no difference.

在更一般的情况下，这两种方法之间的区别在于，在 [] 语法的情况下，该语言对数组声明的正确性执行常规"检查.例如，当使用 [] 语法时，数组元素类型必须完整.指针语法没有这种要求

In more general case one difference between these two approaches stems from the fact that in case of [] syntax the language performs "usual" checks for correctness of array declaration. For example, when the [] syntax is used, the array element type must be complete. There's no such requirement for pointer syntax

struct S;
void foo(struct S *a); // OK
void bar(struct S a[]); // ERROR

此规则的特定副作用是您可以将 void * 参数声明为 void [] 参数.

A specific side-effect of this rule is that you canon declare void * parameters as void [] parameters.

如果您指定数组大小，则它必须为正数(即使之后会被忽略).

And if you specify array size, it has to be positive (even though it is ignored afterwards).

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