free(ptr); question?

Hello, once again :) today is devoted to the pointers lesson! Now I have compiled everything successfully on the lesson so far except for the function

Code:

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free(ptr);

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here is what I have so far that compiles.

Code:

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

main()
{
 int x,*p=malloc(sizeof(*p));
 p=&x;
 printf("Enter a value for x:");
 scanf("%d",&x);
 printf("this is the output of *p:%d\n",*p);
 free(p);
 getchar();
}

---

but when I run the program I get this

Code:

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beatzz@AMD2500:~/c$ gcc -o pointers pointers.c
beatzz@AMD2500:~/c$ ./pointers
Enter a value for x:3
this is the output of *p:3
*** glibc detected *** ./pointers: free(): invalid pointer: 0xbfb8e150 ***
======= Backtrace: =========
/lib/tls/i686/cmov/libc.so.6[0xb7e16a85]
/lib/tls/i686/cmov/libc.so.6(cfree+0x90)[0xb7e1a4f0]
./pointers[0x80484a9]
/lib/tls/i686/cmov/libc.so.6(__libc_start_main+0xe0)[0xb7dc1450]
./pointers[0x80483e1]
======= Memory map: ========
08048000-08049000 r-xp 00000000 08:01 3801091    /home/beatzz/c/pointers
08049000-0804a000 rw-p 00000000 08:01 3801091    /home/beatzz/c/pointers
0804a000-0806b000 rw-p 0804a000 00:00 0          [heap]
b7c00000-b7c21000 rw-p b7c00000 00:00 0
b7c21000-b7d00000 ---p b7c21000 00:00 0
b7daa000-b7dab000 rw-p b7daa000 00:00 0
b7dab000-b7ef4000 r-xp 00000000 08:01 12289040  /lib/tls/i686/cmov/libc-2.7.so
b7ef4000-b7ef5000 r--p 00149000 08:01 12289040  /lib/tls/i686/cmov/libc-2.7.so
b7ef5000-b7ef7000 rw-p 0014a000 08:01 12289040  /lib/tls/i686/cmov/libc-2.7.so
b7ef7000-b7efa000 rw-p b7ef7000 00:00 0
b7efd000-b7f07000 r-xp 00000000 08:01 12271638  /lib/libgcc_s.so.1
b7f07000-b7f08000 rw-p 0000a000 08:01 12271638  /lib/libgcc_s.so.1
b7f08000-b7f0c000 rw-p b7f08000 00:00 0
b7f0c000-b7f0d000 r-xp b7f0c000 00:00 0          [vdso]
b7f0d000-b7f27000 r-xp 00000000 08:01 12271629  /lib/ld-2.7.so
b7f27000-b7f29000 rw-p 00019000 08:01 12271629  /lib/ld-2.7.so
bfb7b000-bfb90000 rw-p bffeb000 00:00 0          [stack]
Aborted

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Did I make a mistake? or am I missing libraries?

it should crash

p=&x;

address returned by malloc is overwritten with the address of x

when you call free(p) - you pass the address malloc has no idea about. So you get a crash here

so how dose one appropriately use free(); ?

Quote:

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Originally Posted by elsheepo

so how dose one appropriately use free(); ?

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give to free exactly the same address as received from malloc

If you want a pointer to point to a variable on the stack, don't allocate any space:

Code:

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int n;
int* p;
p = &n;
/*no free*/

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I understand what anon is saying, just leave the malloc out, and it would work fine, but I'm trying to use it just for the tutorials sake.

Code:

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

main()
{
 int x,*p=malloc(sizeof(*p));
 p=&x;
 printf("Enter a value for x:");
 scanf("%d",&x);
 printf("this is the output of *p:%d\n",*p);
 free(p);
 getchar();
}

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by "give to free exactly the same address as received from malloc" do you mean by changing this, free(p); to this free(*p); because I tried that and it wouldn't even compile.

No, more like:

Code:

---

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

int main()
{
    int x;
    int *p = malloc(sizeof(*p));
    printf("Enter a value for x:");
    scanf("%d", &x);
    *p = x; /* Assume that malloc() did not return a null pointer. */
    printf("this is the output of *p: %d\n", *p);
    free(p);
    getchar();
    return 0;
}

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Though in this case the dynamic memory allocation is not useful since just using x will suffice.

A pointer can refer to an int on the stack (int x) or it can point to a int that is allocated dynamically. With malloc you ask for enough memory to store an int in, and p is the only means that allows you to access it.

Code:

---

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

main()
{
 int *p=malloc(sizeof(*p));
 printf("Enter a value to be stored:");
 scanf("%d",p);
 printf("this is the output of *p:%d\n",*p);
 free(p);
 getchar();
}

---

okay, I got the program to work the way I wanted it to at least. I'm not sure if I fully understand pointers at the moment, but I supose with time I will. Thanks for the help guys

Code:

---

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

main()
{
 int x,*p=malloc(sizeof(*p));
 printf("Enter a value for x:");
 scanf("%d",&x);
 *p=x;
 printf("this is the output of *p:%d\n",*p);
 free(p);
 getchar();
}

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for some reason the program works just the same without

Code:

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return 0;

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but thanks lazerlight for your example, it worked :)

Quote:

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Originally Posted by elsheepo

for some reason the program works just the same without

Code:

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return 0;

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but thanks lazerlight for your example, it worked :)

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Yes, it would do - except that any program askign your program whether it was successful or not would get an answer that may be just about anything, as the return value from a function that doesn't have a return statement will be whatever happens to be in the register that the compiler designates "the return value" - in your case, it would probably be the return value from getchar(), but that is not guaranteed.

--
Mats

so is it a good idea to have

Code:

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return 0;

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in all functions?

Quote:

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Originally Posted by elsheepo

so is it a good idea to have

Code:

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return 0;

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in all functions?

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Yes. Unless of course something went wrong (e.g. the application could not open a file, ran out of memory, or the user input was invalid), in which case you may want to return some other number to indicate "not a success". By convention, successfull execution of the program should return 0 - other times, return non-zero. You may even return different things to indicate WHY it went wrong.

--
Mats

To add to that, you really should explicitly specify the return type of all functions, including main. Omitting return types is an error in C99 and C++ (and deprecated in C90).
And as for pointers, remember this:
A pointer is a variable that stores memory addresses. Thus, when assigning a value to it, you are overwriting its contents--its address.
* is the dereference operator. It basically tells the compiler "go to the address stored inside this variable and do x"

The function free relies on that the address passed in is an address that was returned by malloc. If it is not, you can expect bad consequences.

Quote:

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Originally Posted by Elysia

you really should explicitly specify the return type of all functions,

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1. Dose that mean when declaring a function such as "main()" I should write it "int main()"?
2. Is "free()" only used in conjunction with"malloc(sizeof())"?
3. Can someone please explain these steps that were taken to fix the program?

Bugged code

Code:

---

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

main()
{
 int x,*p=malloc(sizeof(*p));
 p=&x;
 printf("Enter a value for x:");
 scanf("%d",&x);
 printf("this is the output of *p:%d\n",*p);
 free(p);
 getchar();
}

---

Fixed code

Code:

---

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

int main()
{
    int x;
    int *p = malloc(sizeof(*p));
    printf("Enter a value for x:");
    scanf("%d", &x);
    *p = x; /* Assume that malloc() did not return a null pointer. */
    printf("this is the output of *p: %d\n", *p);
    free(p);
    getchar();
    return 0;
}

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3a. "main()" was changed to "int main()"

3b. Notice she separated the variables during declaration, even though they are both integers. Is this necessary or just personal preference?

3c. "p=&x;" was changed to "*p=x;" also moved below scanf why?

3d. "return 0;" was used, I think I understand that one now.

Quote:

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Originally Posted by elsheepo

1. Dose that mean when declaring a function such as "main()" I should write it "int main()"?

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Yes. Of course, if the function returns an object of some other type then you should declare it with the appropriate return type, or void if the function does not return any value at all.

Quote:

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Originally Posted by elsheepo

2. Is "free()" only used in conjunction with"malloc(sizeof())"?

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It can be used in conjunction with realloc() as well. The sizeof is useful but not essential here.

Quote:

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Originally Posted by elsheepo

3a. Notice she separated the variables during declaration, even though they are both integers. Is this nessisary or just personal preference?

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Personal preference. It can be clearer to have declarations, especially those involving pointers, on their own lines.

Quote:

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Originally Posted by elsheepo

3b. p=&x; was changed to *p=x; also moved below scanf why?

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The change from p = &x; to *p = x; is the "fix". The move to just below scanf() is not important.