Question

我需要函数drop_balls来返回一个数组，以便我可以让我的下一个函数使用该数组。我需要它接受一个整数作为参数，然后返回一个int数组，以便它可以在我将要制作的另一个函数中使用。我一直在读它只能作为指针传递，但我完全不知道如何编码，有人可以帮助我。

#include <stdio.h>
#include <stdlib.h>
#include <time.h>


/* Prototype for drop_Balls, int parameter is number of balls being dropped */
int [] drop_balls(int); 

/* Gets the number of balls that need to be dropped by the user */
int get_num_balls(); 


int main()
{
    drop_balls(get_num_balls());
}


int get_num_balls()
{
    int num_balls;
    printf("How many balls should be dropped? ");
    scanf("%d", &num_balls);
    /* Ensure that it is atleast one ball */
    while(num_balls <= 0) 
    {
        printf("You have to drop at least one ball! \n ");
        printf("How many balls should be dropped? ");
        scanf("%d", &num_balls);
    }
    /* Return the number of balls that will be dropped */
    return num_balls; 
}



int [] drop_balls(int num_balls)
{
    /* Keeps track of how many balls landed where */
    int ball_count[21]; 
    /* What number ball are we on */
    int ball_num = 0; 
    /* Seed the generator */
    srand(time(NULL)); 
    /* Do the correct # of balls */
    for(ball_num = 0; ball_num < num_balls; ball_num++ ) 
    {
  /* Start at 10 since its the middle of 21 */
        int starting_point = 10; 
    /* Bounce each ball 10 times */
        for(starting_point = 10; starting_point > 0; starting_point--) 
        {
           int number;
       /* Create a random integer between 1-100 */
           number = rand() % 100; 
       /* If its less than 50 bounce to the right once */
           if(number >= 50) 
           {
               starting_point++;
           }
       /* If its greater than 50, bounce to the left once */
           else 
           {
               starting_point--;
           }
        }
    /* Add one to simulate one ball landing there */
        ball_count[starting_point]++;  
    }
    return ball_count;
}

Answer 1

它不完全确定你想要实现什么，但如果你想将一个整数数组balls传递给函数drop_balls，它将被声明为：

void drop_balls(int **balls);

int main ()
{
    int *balls; /* This is your array of balls*/

    /* Allocate memory for `number_of_balls` number of balls */
    balls = malloc(number_of_balls * sizeof(*balls));
    if (balls == NULL)
         printf("Error: failed to allocate memory");

    /* call the function by passing the pointer to your array of balls */
    drop_balls(&balls);

    free(balls);
    return 0;
}

Answer 2

您不能返回在函数中声明为局部变量的数组。它存在于堆栈中，一旦函数返回它将不再存在。你将返回数组的地址，这将是不好的。

相反，您需要使用malloc,在堆上创建数组并返回指向它的指针。您将需要检查此指针是否为NULL,，并且您需要调用free以在完成后释放内存。您还应该将函数的返回类型更改为int *。

Answer 3

返回数组的函数示例是malloc()，一个标准的库函数。您可以编写自己的函数来分配数组，对其执行某些操作并将其返回给调用者：

int *function_returning_an_array_of_size_n(int n)
{
   int *array = (int*)malloc(n * sizeof(int));
   // do something useful with the array
   // ...
   return array;
}

将数组作为参数的函数示例是free()，同样是标准库函数。

以下是编写接收数组作为参数的函数的方法。

void function_receiving_an_array_and_size_n(int *array, int n)
{
   // Do something with array. If accessing elements beware the array size violation.
}

我传递了一个额外的参数，数组大小。在不知道大小的情况下访问数组元素会很危险。

如何让我的数组返回一个int数组以用于另一个函数？

3 个答案: