Thread: Need help to understanding the code

Hi Guys,

I am working on this code, and it is working fine. However, I would like to how it actually works.

Here is the code:

Code:

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

/*
 * 
 */

void swap(void *,void*,int);

int main(int argc, char** argv) {

   
    char *husband = "Fred";
    char *wife = "Wilma";

    swap(&husband, &wife,sizeof(char *));
    

    printf("The husband is: %s\n", husband);
    printf("The wife is: %s\n", wife);

    return (EXIT_SUCCESS);
}
void swap(void *vp1, void *vp2, int size){

    char buf[size];
    
    //copy the content of vp1 to buff
    memcpy(buf,vp1,size);
    memcpy(vp1,vp2,size);
    memcpy(vp2,buf,size);

}

I understand that - The buf should have a pattern of whatever in vp1. But I can not confirmed it. I am assuming that - the buf should have a copy of the address of whatever in vp1. But how can I print the content of the buf out ?

I have tried:

printf("buf has: %p", buf) and I am getting: 0x22cce0
printf("vp1 has: %p",vp1) and I am getting: 0x22cd34

I thought: the buf and the vp1 should have the same bit pattern in them.

I have tried:

printf("buf has: %p", *buf); and I am getting: 0x20

Would some one help ?

Thanks,

z

You're dealing with strings, print them out using %s in your printf format.

"yeah... yeah.. YEAHHH!!! i mean i'm just an IT that doesn't really know any of the real science and technical details and mostly only work on plugging this to this and i work in well with linux. so i should know that this guy's crackpot, right!! yeah!!"

"i mean it's not like he's writing code for matrix arithmetic, right??? yeah!! coz like, it's the hardest arithmetic code i can think of" <-(by that last sentence i really am speaking from their point of view... it is the hardest calculation that they know to do... i wonder how they'll fare in even reading my code)

Originally Posted by renzokuken01

"i mean it's not like he's writing code for matrix arithmetic, right??? yeah!! coz like, it's the hardest arithmetic code i can think of" <-(by that last sentence i really am speaking from their point of view... it is the hardest calculation that they know to do... i wonder how they'll fare in even reading my code)

Hey man... have you even seen the travel folders for reality?

Really this is totally unhelpful behavior and it's not going to make you any friends.

Originally Posted by zbonzbon

Hi Guys,

I am working on this code, and it is working fine. However, I would like to how it actually works.

Here is the code:

Code:

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

void swap(void *,void*,int);

int main(int argc, char** argv) {

   
    char *husband = "Fred";
    char *wife = "Wilma";

    swap(&husband, &wife,sizeof(char *));
    

    printf("The husband is: %s\n", husband);
    printf("The wife is: %s\n", wife);

    return (EXIT_SUCCESS);
}


void swap(void *vp1, void *vp2, int size){

    char buf[size];
    
    //copy the content of vp1 to buff
    memcpy(buf,vp1,size);
    memcpy(vp1,vp2,size);
    memcpy(vp2,buf,size);

}

Right off you are dealing with buffer overflows... Fred and Wilma are different lengths. When you used them to intialize the string pointers husband and wife you got 5 and 6 bytes respectively... when you try to reverse them you can put Fred where Wilma came from, because it's shorter. You are not going to put Wilma where Fred came from without causing a buffer overflow because it's longer.

Secondly when you are sending in the size of the buffers to memcpy, you are making the mistake of telling it to copy only 4 characters... sizeof (char*)... instead of the size of the actual strings... for which you need strlen(). At most it's going to swap the first 4 characters of of each string. (a char* pointer is 4 bytes on most systems)

I understand that - The buf should have a pattern of whatever in vp1. But I can not confirmed it. I am assuming that - the buf should have a copy of the address of whatever in vp1. But how can I print the content of the buf out ?

I have tried:

printf("buf has: %p", buf) and I am getting: 0x22cce0
printf("vp1 has: %p",vp1) and I am getting: 0x22cd34

You need to print the CONTENT not the pointer.

Code:

printf("buf has: %s",buf);
printf("vp1 has: %s",vp1);

Originally Posted by Subsonics

I think the point here is to swap the addresses that the pointers contain, hence the sizeof(char *), and the & in the swap call.

So if wife points to "wilma" and husband points to "fred"

after the action:

wife points to "fred" and husband points to "wilma"

He's using memcpy()... so I went from there.

Yes, in this case swapping pointers makes more sense.

Originally Posted by CommonTater

He's using memcpy()... so I went from there.

Yes, in this case swapping pointers makes more sense.

Yes, that is probably since the swap function uses void pointers.

Subsonics are correct. What the program doing is that - It is actually swapping the address that contain in vp1 and vp2. And it uses the char buf[size] as a storage for copying the address of the vp1 into it first and then copy the vp2 to vp1 and then finally, it copied the buf back to vp2. Effectively, it is swapping the pointer that pointed to the string Wilma and Fred.

That logic is very cooled. However, I would like to confirm it by:

Try to print out the content of the buf to see if it is actually containing the address of Fred. But I could not.

Regards,

z