如何在C中声明运行时数组的大小？

Question

我基本上想要相当于C的C(好吧,只是数组的部分,我不需要类和字符串解析以及所有这些):

public class Example
{
    static int[] foo;
    public static void main(String[] args)
    {
        int size = Integer.parseInt(args[0]);
        foo = new int[size]; // This part
    }
}

请原谅我的无知.我被java损坏了;)

Answer 1

/* We include the following to get the prototypes for:
 * malloc -- allocates memory on the freestore
 * free   -- releases memory allocated via above
 * atoi   -- convert a C-style string to an integer
 * strtoul -- is strongly suggested though as a replacement
*/
#include <stdlib.h>
static int *foo;
int main(int argc, char *argv[]) {
    size_t size = atoi(argv[ 1 ]); /*argv[ 0 ] is the executable's name */
    foo = malloc(size * sizeof *foo); /* create an array of size `size` */
    if (foo) {  /* allocation succeeded */
      /* do something with foo */
      free(foo); /* release the memory */
    }
    return 0;
}

警告:关闭袖口,没有任何错误检查.

Answer 2

在C中,如果忽略错误检查,则可以使用此方法:

#include <stdlib.h>
static int *foo;

int main(int argc, char **argv)
{
     int size = atoi(argv[1]);
     foo = malloc(size * sizeof(*foo));
     ...
}

如果您不想使用全局变量并且使用的是C99,则可以执行以下操作:

int main(int argc, char **argv)
{
    int size = atoi(argv[1]);
    int foo[size];
    ...
}

这使用VLA - 可变长度数组.