原型变长数组 [英] Prototype for variable-length arrays

问题描述

我想编写一个函数，它接受一个可变大小的数组在C。

I am trying to write a function that takes an array of an variable size in c.

void sort(int s, int e, int arr[*]){
    ...
}

它说，对可变长度数组，它需要在函数声明有界。这意味着什么？我用x code 4.0，与LLVM编译器2.0。

It says that for variable length arrays, it needs to be bounded in the function declaration. What does that mean? I am using xcode 4.0, with the LLVM compiler 2.0.

感谢您的帮助。

推荐答案

如果你不使用的C99变长数组，通常的解决方法是在一个指针传递到第一个元素，任何你想要的索引沿用来访问的元素。

If you're not using the C99 variable length arrays, the usual solution is to pass in a pointer to the first element, along with any indexes you want to use for accessing the elements.

下面是一张code的打印出一系列的数组，类似于你想与做你的排序。

Here's a piece of code that prints out a range of an array, similar to what you're trying to do with your sort.

#include <stdio.h>

static void fn (int *arr, size_t start, size_t end) {
    size_t idx;
    for (idx = start; idx <= end; idx++) {
        printf ("%d ", arr[idx]);
    }
    putchar ('\n');
}

int main (void) {
    int my_array[] = {9, 8, 7, 6, 5, 4, 3, 2, 1, 0};
    fn (my_array, 4, 6);
    return 0;
}

这个输出要素四至六，包容性（从零开始），赠送：

This outputs elements four through six inclusive (zero-based), giving:

5 4 3

有几点要注意。

• 使用 my_array 在函数调用 FN 自动衰变的数组指针它的第一个元素。这实际上在大多数（不是全部）情况发生在你使用数组，所以你不必明确说明＆放大器;（my_array [0]）

• Using my_array in that function call to fn automatically "decays" the array into a pointer to its first element. This actually happens under most (not all) circumstances when you use arrays, so you don't have to explicitly state &(my_array[0]).

C处已经的的的一个很好的排序功能内置于标准库，名为的qsort 。在许多情况下，这是你应该使用什么（除非要么你有一个特定的算法要用于排序，或者你正在做的一门功课/自我教育运动）。

C already has a very good sort function built in to the standard library, called qsort. In many cases, that's what you should be using (unless either you have a specific algorithm you want to use for sorting, or you're doing a homework/self-education exercise).

这篇关于原型变长数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！