Local variable address used as errno

Question

#include<stdio.h>

struct a
{
    void *ptr;
    unsigned long val;
};

void main()
{
    unsigned char errno;

    struct a *id;
    id = malloc(sizeof(*id));
    func2(id);
    printf("After changing %d\n", id->val);
}


void func2(struct a *id)
{
    unsigned char errno;
    func(id,&errno);
}

void func(struct a *id,void *ptr)
{   
    memset(id, 0, sizeof(*id));

    id->ptr = ptr;
    if (sizeof(id->val) >= sizeof(id->ptr))
    {   
        id->val = (unsigned long)id->ptr;
        return;
    }

}

when i am printing id->val in main function it is printing -1075050593 . But i am trying to access a invalid address. Please explain. I am very new to c programming.

Also, you should turn on compiler warning messages. You will find that this code doesn't compile cleanly. You should fix all compilation problems first, and then ask the question if it still doesn't work as expected. — Oliver Charlesworth, Jul 12 '12 at 11:35
possible duplicate of [Pointer to local variable](http://stackoverflow.com/questions/4570366/pointer-to-local-variable) — Eran, Jul 12 '12 at 11:40
`func2(&id);` has an ampersand too many. And main() should return int. sigh. — wildplasser, Jul 12 '12 at 11:41
Beware that `errno` has a fixed meaning in the context of C. — Jens Gustedt, Jul 12 '12 at 11:43
Have you tried printing the value of `errno` in `func2()`. It probably gets initialised to 0 and then 0 gets assigned to `id->val` in `func()`. — Wernsey, Jul 12 '12 at 11:44
@shunty have a look at https://en.wikipedia.org/wiki/Errno.h — Jens Gustedt, Jul 12 '12 at 11:45

score 1 · Answer 1 · answered Jul 12 '12 at 14:50

Usually, what is "accessing an invalid address"? It consists of 2 parts. 1, accessing: it includes read/write/execute. 2, invalid address: the kernel space and the un-malloced heap are the invalid address for the user application.

In this case, the address (&errno) belongs to the stack, so it is not an invalid address. And you doesn't read/write/execute the content in this address. So you are not accessing the invalid address.

BTW, with the "%d" placeholder in the printf() call, the "id->val" will be explained as a signed int type, it is why you got a negative value. Please use "%p" for pointers and "%u" for unsigned int.

so if i tries to access *id->val what will be the behavior? – shunty Jul 12 '12 at 23:33 — shunty, Jul 12 '12 at 23:33

hmjd · Accepted Answer · 2012-07-12T11:59:13.577

0

The following statement is not attempting to access an invalid address, it is just printing the address:

printf("After changing %d\n", id->val);

what value it will hold is unknown as id is being incorrectly passed to func2().

Some problems with the code (some already mentioned in the comments):

incorrectly taking address of id when passing to func2() as id is already a struct a*
implicit declarations of memset() and malloc() as missing include directives
implicit declarations of func() and func2()
return type of main() should be int

edited Jul 12 '12 at 11:59

answered Jul 12 '12 at 11:40

hmjd

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i made an edit instead of address of structure only pointer should be passed. printf("After changing %d\n", id->val); it trying to print a address which is invalid. – shunty Jul 12 '12 at 11:51
1

@shunty, it is just _printing_ the address value and is not attempting to access it which is fine. See http://ideone.com/ZH3Ws – hmjd Jul 12 '12 at 11:52

Local variable address used as errno

2 Answers2