变量数组声明 [英] Variable array declaration
问题描述
考虑以下C代码:
#include<stdio.h>
int main()
{int n,i;
scanf("%d",&n);
int a[n]; //Why doesn't compiler give an error here?
}
当编译器最初不知道时如何声明数组?
How can I declare an array when the compiler doesn't know initially?
推荐答案
当直到编译时才知道数组的确切大小时,需要使用动态内存分配。在C标准库中,有用于动态内存分配的函数:malloc,realloc,calloc和free。
When the the exact size of array is unknown until compile time, you need to use dynamic memory allocation. In the C standard library, there are functions for dynamic memory allocation: malloc, realloc, calloc and free.
这些函数可以在中找到。 < stdlib.h>
头文件。
These functions can be found in the <stdlib.h>
header file.
如果要创建数组,请执行以下操作:
If you want to create an array you do:
int array[10];
在动态内存分配中,您可以执行以下操作:
In dynamic memory allocation you do:
int *array = malloc(10 * sizeof(int));
您的情况是:
int *array = malloc(n * sizeof(int));
如果您分配了内存位置,请不要忘记取消分配:
If you allocate a memory position, never forget to deallocate:
if(array != NULL)
free(array);
内存分配是一个复杂的主题,我建议您搜索该主题,因为我的回答很简单。您可以从以下链接开始:
Memory allocation is a complex subject, I suggest you search for the subject because my answer was simple. You can start with this link:
https://www.programiz.com/c-programming/c-dynamic-memory-allocation
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