C: 将具有数组的指针传递给函数时会出现 运行 时间错误。为什么？

Question

int add_up (int *a, int num_elements);

int main() 
{
  int orders[5] = {100, 220, 37, 16, 98};
  int *p;
  p=&orders[0];

  printf("Total orders is %d\n", add_up(*p, 5)); 

here If I pass pointer what is the problem? As pointer will point to no. 100 value. As 100 is an int value which will go to int *a. And the program should go well right? Edit 1: If I give p here in place of *p then will it be address of orders[0] and not value 100 right? But function int *a wants value 100 right. Correct me if I am wrong in this concept and thanks.

  return 0;
}

int add_up (int *a, int num_elements) 
{
  int total = 0;
  int k;

  for (k = 0; k < num_elements; k++) 
  {
    total += a[k];
  }
  return (total);
} 

Answer 1

用以下代码替换您的代码：

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

int add_up (int *a, int num_elements);

int main() 
{
  int orders[5] = {100, 220, 37, 16, 98};
  int *p;
  p=&orders[0];

  printf("Total orders is %d\n", add_up(p, 5)); 
  return 0;
}

int add_up (int *a, int num_elements) 
{
  int total = 0;
  int k;

  for (k = 0; k < num_elements; k++) 
  {
    total += a[k];
  }
  return (total);
}

您忘记包含 2 headers 和 add_up(*p, 5) 并删除了 *。