C编程,错误：被调用的对象不是函数或函数指针[已关闭]

我正在尝试编写一个实现Pop和Push功能的程序.问题是,我试图将指向整数Top的指针传递给函数,这样整数不断变化,但是当我尝试编译时,我总是得到这个行：

**error: called object is not a function or function pointer (*t)–

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

#define MAX 10
int push(int stac[],int *v,int *t)
{
  if((*t) == MAX-1)
  {
      return(0);
  }
  else
  {
      (*t)++;
      stac[*t] = *v;
      return *v;
   }
}

int pop(int stac[],int *t)
{
 int poped;
 if((*t) == -1)
 {
      return(0);
 }
 else
 {
     poped = stac[*t]
     (*t)--;
     return poped;
 } 
}
int main()
{
int stack[MAX];
int value;
int choice;
int decision;
int top;
top = -1;
do{
   printf("Enter 1 to push the value\n");
   printf("Enter 2 to pop the value\n");
   printf("Enter 3 to exit\n");
   scanf("%d",&choice);
   if(choice == 1)
   {
       printf("Enter the value to be pushed\n");
       scanf("%d",&value);
       decision = push(stack,&value,&top);
       if(decision == 0)
       { 
           printf("Sorry,but the stack is full\n");  
       }
       else
       {
           printf("The value which is pushed is: %d\n",decision);
       }
   }
   else if(choice == 2)
    {
         decision = pop(stack,&top);
        if(decision == 0)
          {
               printf("The stack is empty\n");
          }
         else
          {
              printf("The value which is poped is: %d\n",decision);
          }

    }
 }while(choice != 3);
 printf("Top is %d\n",top);

}

解决方法

你错过了一行之前错过了一个分号：

poped = stac[*t] <----- here
 (*t)--;

这个奇怪的错误的原因是编译器看到这样的：

poped = stac[*t](*t)--;

它可以解释为来自表的函数指针的调用,但这显然没有任何意义,因为stac是一个int数组,而不是一个函数指针数组.