以下代码段的输出是什么？

Question

what's happening after clrscr?

#include<stdio.h>
#include<conio.h>
int *call();
void main()
{
int *ptr;
ptr=call();
clrscr();
printf("%d",*ptr);
getch();
}  

int*call()
{
int a=25;
a++;
return &a;
}

output:
-10

code works like this: call() is called, a=25, then a=26. let address of a be 65518. this address is returned to ptr. since return type is int, instead of 65518, (due to cyclic property) -18 is returned. so ptr=&a=-18. then clrscr clears it....but how *ptr is printed as output? i mean address cannot be negative(-18).

Answer 1

Returning a pointer to local is undefined behavior. Anything could happen - your program could crash, but more likely it is going to print some arbitrary number.

If you need to return a pointer from a C function, you need to either allocate a memory block in the dynamic storage, like this:

int*call()
{
    int *a=malloc(sizeof(int));
    *a = 25;
    *a++;
    return a;
}

or use a pointer to a statically allocated block, like this:

int* call()
{
    static int a=25;
    a++;
    return &a;
}

If you choose the dynamic allocation route, the caller must free the pointer returned by your function.

Answer 2

int*call()
{
int a=25; // <--- use malloc here i.e. int a = malloc(sizeof(int)); then you can set a value to a and return the pointer without any problemk, OTHERWISE => it will return an address of some random junks you don't want, its gonna be completely random
a++;
return &a;
}

Answer 3

When call() is called, a new stack frame is created with space for the local variable a , which has its lifetime during the execution of call() . When it returns, the stack frame is removed along with its local variable(s) and data. Trying to use this data outside the function is undefined, because it no longer exists, logically.

If you want to declare a inside a function and use it afterwards, you'll need to allocate it:

... 
int *a = malloc(sizeof int);
*a = 26;
return a;
... 

Remember to free() this pointer after you're finished using it.