小编sar*_*mdy的帖子

我不明白为什么这段代码给我乘法、除法和减法的错误结果？它给我任何乘法运算的结果为零，任何除法错误的结果最终减法结果始终为 1

#include <stdio.h>
long operations(char x,int num1 ,int num2 );

int main(void) {
    char x;
    int num1 , num2;
    long result;
    setbuf(stdout,NULL);
    result=operations(x,num1 ,num2 );

}
long operations(char x,int num1 ,int num2 )
{
    printf("please enter your operation sum (1) ,subtrcation(2),mul(3),div(4) :");
    scanf("%d",&x);

    if (x==(1||2||3||4))
    {
    printf("please enter the first and second numbers : \n ");
    scanf ("%d%d" ,&num1 ,&num2);
    }
    else
    {
        printf("error ") ;
    }

    switch (x)
    {
    case 1:
    {
        return  printf("the result  sum is  :%d …

（我想编写一个代码来计算数组元素的总和和平均值）我想找出错误平均结果的原因，其中平均值可能是浮点数，所以我进行了强制转换以使用以下方法将结果转换为浮点数(float)sum)/size.

注意：我知道如果我确定变量平均值这个问题将得到解决

float average = (float)sum/arr_size;
printf("The average of the array elements is: %f\n", average);

但我需要分析我错误的平均输出

#include<stdio.h>
#define size 5
int main(void)
{ 
  int i,sum=0;
  int arr[size];
  printf("please enter array elements: \n");
  // for loop to scant element elemnt of array
  for(i=0;i<size;i++)
  {
      scanf("%d",&arr[i]);
  }
  // sum of array
  for(i=0;i<size;i++)
  {
      sum+=arr[i];
  }
  // print sum after fishing
  printf("the summation is : %d \n",sum);

  // average of array
  printf("The average of array : %d",(((float)sum)/size));

  return 0;
}

我的代码在我想运行操作结果的地方没有按预期工作有什么问题？运行结果告诉我“编译器错误”，但我找不到错误

int main(void) {

int operation,num1,num2;

printf("please enter operation (sum,sub,mul,div) : ");
scanf("%d",& operation);
printf("please enter two number: ");
scanf("%d%d",& num1,& num2);



switch (operation)
?{
    case sum :
      printf("The summation is : %d", num1+num2);
      break; 
    case sub :
      printf("The submition is : %d", num1-num2);
      break;

    case mul :
      printf("The multiplication n is : %d", num1*num2);
      break;

    case div :
      printf("The division is : %d", num1/num2);
      break;

    default:
      printf("Not valid");
}

  return 0;
}