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Feb 4th, 2007, 5:41 PM
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#1 |
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Newbie
Join Date: Feb 2007
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How do I swap pointers with a function?
I need help with the following problem about swapping pointers.
The following code is in main: main()
{
int x=1,y=2;
int *xp,*yp;
.
.
.
xp = &x;
yp = &y;
intpswap(<arg1>,<arg2>);
x=3;y=4;
printf("*xp = %d, *yp = %d\n",*xp,*yp);
}and thus printf prints "*xp = 4, *yp =3". I know how to swap int values by using a temp variable, but is it the same approach here? Any help or hints will be greatly appreciated!! Thank you!! Last edited by Thiengineer; Feb 4th, 2007 at 6:09 PM. |
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Feb 4th, 2007, 6:13 PM
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#2 |
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Resident Grouch
 
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Put one in a temp, put the second where the first was, and put the temp in the second.
__________________
Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers |
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Feb 4th, 2007, 8:19 PM
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#3 |
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Newbie
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This is what I have so far...am I on the right track? Please help!! You help is greatly appreciated!
void intpswap(int *xp, int *yp) { int t = *xp; *xp = *yp; *yp = t; } |
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Feb 4th, 2007, 10:23 PM
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#4 | |
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Resident Grouch
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Why would you declare t as an int? You're swapping int pointers. Since you want persistent effects (I imagine), you need to pass pointers to the pointers to swap, since copies will simply go out of scope when the function returns.
Note that a and b are not swapped, despite what the printf SAYS, only the pointers are swapped. #include <stdio.h>
void pSwap (int **a, int **b)
{
int *temp = *a;
*a = *b;
*b = temp;
}
int main ()
{
int a = 4;
int b = 6;
int *pA = &a;
int *pB = &b;
printf ("a: %d , b: %d\n", *pA, *pB);
pSwap (&pA, &pB);
printf ("(swapped) a: %d , b: %d\n", *pA, *pB);
}Quote:
__________________
Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers |
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Feb 6th, 2007, 3:02 PM
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#5 |
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Hobbyist Programmer
Join Date: Jun 2005
Location: New Mexico
Posts: 228
Rep Power: 4
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void swap( int *a, int *b)
{
int t=*a;
*a=*b;
*b=t;
}
int main()
{
int x=1;
int y=2;
int *a=&x;
int *b=&y;
printf("%d %d\n", *a, *b);
swap(a,b);
printf("%d %d\n", *a, *b);
return 0;
}
kcsdev:/home/jmcnama> cc t.c
kcsdev:/home/jmcnama> ./a.out
1 2
2 1FWIW. I think he meant to swap what the pointers were "aimed" at. Not the actual pointers themselves. But it doesn't matter.... |
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