hi , i'm having a problem with pointers in C ...

blackslither · 10-25-2008, 04:53 PM

I will give a simple example inspired from my problem :

Code:

void change(int *x, int *y){x=y;}

int main(){
int *xx,*yy;

yy = (int*)malloc(sizeof(int));
change(xx,yy);

if(xx == NULL)printf("is null\n");
else printf("---- %d ----\n",*xx);

.................................................. .........

xx doesn't point to the value of yy (*yy) , the program prints "is null" . Why ?

xx and yy must be declared int* , it's something that i must do in my program . and xx =yy must be done in a function , not in main .

10x

matsp · 10-25-2008, 05:06 PM

Since your swap function only changes the local variables inside the function change, you will not see any effect in main. To modify a pointer, you need to pass a pointer to pointer - this applies to all of C: to change something, you need to know it's location, so you need a pointer.

Considering that xx could be just about any value in the world that can be held in a pointer, who knows what is going to happen. Only yy is initialized.

It's exactly the same as:

Code:

int x;
printf("%d\n", x);

x could be all values that are valid for an integer.

Unless a local variable is actually given a value, it will have whatever value happens to be in it's location in memory.

--
Mats

Daper · 10-26-2008, 03:57 AM

So your code that changes the pointer should look like:

Code:

void change(int **x, int **y)
{
  *x=*y;
}

// ...

change (&xx, &yy);

--
Daper
http://www.linuxprogrammingblog.com

10-25-2008, 04:53 PM	#1
blackslither Registered User Join Date: Oct 2008 Posts: 4	hi , i'm having a problem with pointers in C ... I will give a simple example inspired from my problem : Code: void change(int x, int y){x=y;} int main(){ int xx,yy; yy = (int)malloc(sizeof(int)); change(xx,yy); if(xx == NULL)printf("is null\n"); else printf("---- %d ----\n",xx); .................................................. ......... xx doesn't point to the value of yy (yy) , the program prints "is null" . Why ? xx and yy must be declared int , it's something that i must do in my program . and xx =yy must be done in a function , not in main . 10x
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10-25-2008, 05:06 PM	#2
matsp Kernel hacker Join Date: Jul 2007 Location: Farncombe, Surrey, England Posts: 15,686	Since your swap function only changes the local variables inside the function change, you will not see any effect in main. To modify a pointer, you need to pass a pointer to pointer - this applies to all of C: to change something, you need to know it's location, so you need a pointer. Considering that xx could be just about any value in the world that can be held in a pointer, who knows what is going to happen. Only yy is initialized. It's exactly the same as: Code: int x; printf("%d\n", x); x could be all values that are valid for an integer. Unless a local variable is actually given a value, it will have whatever value happens to be in it's location in memory. -- Mats __________________ Compilers can produce warnings - make the compiler programmers happy: Use them! Please don't PM me for help - and no, I don't do help over instant messengers.
matsp is offline

10-26-2008, 03:57 AM	#3
Daper Registered User Join Date: Oct 2008 Posts: 2	So your code that changes the pointer should look like: Code: void change(int x, int y) { x=y; } // ... change (&xx, &yy); -- Daper http://www.linuxprogrammingblog.com
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