难以用C理解变长数组 [英] Difficulty in understanding variable-length arrays in C

问题描述

我在读一本书时，我发现数组的大小必须在申报时给予或使用malloc在runtime.I堆中分配在C写这个程序：

I was reading a book when I found that array size must be given at time of declaration or allocated from heap using malloc at runtime.I wrote this program in C :

#include<stdio.h>

int main() {
  int n, i;
  scanf("%d", &n);
  int a[n];
  for (i=0; i<n; i++) {
    scanf("%d", &a[i]);
  }
  for (i=0; i<n; i++) {
    printf("%d ", a[i]);
  }
  return 0;
}

这code正常工作。

我的问题是如何code可以工作correctly.Isn't它侵犯的C基本概念的数组大小必须在运行之前进行声明或使用malloc（）分配它在runtime.I'm没有做任何这两个东西，那么为什么它的工作是否正常？

My question is how this code can work correctly.Isn't it the violation of basic concept of C that array size must be declared before runtime or allocate it using malloc() at runtime.I'm not doing any of these two things,then why it it working properly ?

解决我的问题是这是在C99支持变长数组，但如果我打aroundmy code，并把语句int类型的[N];上述scanf函数（％D，和放大器; N）;那么它停止工作为什么so.if变长数组在C的支持

Solution to my question is variable length arrays which are supported in C99 but if I play aroundmy code and put the statement int a[n]; above scanf("%d,&n); then it's stops working Why is it so.if variable length arrays are supported in C ?

推荐答案

C99标准支持变长数组 。这些阵列的长度是在运行时确定的。

The C99 standard supports variable length arrays. The length of these arrays is determined at runtime.

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