C语言 C 编程，错误：被调用的对象不是函数或函数指针
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C programming, error: called object is not a function or function pointer
提问by Freddie Chopin
I am trying to write a program which implements the Pop and Push functions. The problem is, I am trying to pass the pointer that points to integer Top to the function, so that this integer keeps changing, but when I try to compile I always get this line:
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，以便这个整数不断变化，但是当我尝试编译时，我总是得到这一行：
**error: called object is not a function or function pointer (*t)--
**错误：被调用的对象不是函数或函数指针 (*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 popped;
if((*t) == -1)
{
return(0);
}
else
{
popped = stac[*t]
(*t)--;
return popped;
}
}
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 popped is: %d\n", decision);
}
}
}while(choice != 3);
printf("Top is %d\n", top);
}
回答by Freddie Chopin
You missed one semicolon just before that line with error:
您在该行之前错过了一个分号并出现错误：
poped = stac[*t] <----- here
(*t)--;
The reason for this strange error is that compiler saw sth like that:
这个奇怪错误的原因是编译器看到了这样的东西：
poped = stac[*t](*t)--;
Which it could interpret as a call to a function pointer coming from a table, but this obviously makes no sense, because stac is an array of ints, not an array of function pointers.
它可以解释为对来自表的函数指针的调用，但这显然没有意义，因为 stac 是一个整数数组，而不是一个函数指针数组。
