Using pointers to swap int array values

Question

I am supposed to use pointers to swap ints in an array. It compiles with no errors or warnings and runs but does not swap the ints. Any suggestions would be helpful!!!

Here is the tester:

#import <stdio.h>

void swap( int ary[] );

int main(  int argc, char*argv[] )
{
    int ary[] = { 25, 50 };
    printf( "The array values are: %i and %i \n", ary[0], ary[1] );
    swap( ary );
    printf( "After swaping the values are: %i and %i \n", ary[0], ary[1] );

    return 0;
}

Here is the swap function:

void swap( int ary[] )
{
    int temp = *ary;
    *ary = *(ary + 1);
    *ary = temp;
}

This is what is displayed after running:

The array values are: 25 and 50
After swaping the values are: 25 and 50

Ron Warholic · Accepted Answer · 2009-11-03 23:28:59Z

21

I hate spoiling this but it looks like a typo more than anything.

In your swap function:

*ary = temp;

should be:

*(ary + 1) = temp;

edit: Is there a reason you're not using array notation? I think it's a bit clearer for things like this:

int temp = ary[0];
ary[0] = ary[1];
ary[1] = temp;

answered Nov 3, 2009 at 23:28

Ron Warholic

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6 Comments

Ron Warholic Over a year ago

I think we've all been there.

GManNickG Over a year ago

Indeed, use [], it's much cleaner.

paxdiablo Over a year ago

It should also be include, not import.

Ed Swangren Over a year ago

@paxdiablo: Perhaps he is running in XCode and is actually compiling as Objective-C (.m suffix insteaD of .c)?

Chris Lutz Over a year ago

@Josh - Why? Both are mathematically and machine-code-ally equivalent.

|

Lara Bailey · Accepted Answer · 2009-11-03 23:28:56Z

6

Examine your swap function more carefully:

void swap( int ary[] )
{
    int temp = *ary;
    *ary = *(ary + 1);
    *ary = temp;
}

When does *(ary + 1) get assigned to?

answered Nov 3, 2009 at 23:28

Lara Bailey

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Comments

theycallmemorty · Accepted Answer · 2009-11-03 23:29:18Z

4

You move the second value into the first spot, and then move the first value back into the first spot.

answered Nov 3, 2009 at 23:29

theycallmemorty

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Comments

John La Rooy · Accepted Answer · 2009-11-04 00:36:22Z

4

just for fun; It's also possible to swap without using a temporary value

void swap( int ary[] )
{
    *ary ^= *(ary + 1);
    *(ary + 1) ^= *ary;
    *ary ^= *(ary + 1);
}

As GMan points out, this code obscures your intent from the compiler and the processor, so the performance may be worse than using a temp variable, especially on a modern CPU.

edited Nov 4, 2009 at 0:36

answered Nov 3, 2009 at 23:38

John La Rooy

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4 Comments

Christopher Tarquini Over a year ago

What does the '^=' operator do here? Never seen it used before

John La Rooy Over a year ago

^ is exclusive or, so x^=y means x=x^y

f0b0s Over a year ago

^ is a bit XOR. 0 ^ 0 = 0 1 ^ 1 = 0 1 ^ 0 = 0 0 ^ 1 = 0

GManNickG Over a year ago

You should mention this probably isn't worth it. Modern CPU's execute instructions in parallel, and this causes a pipe-line stall. Not to mention it's ugly.

user167864 · Accepted Answer · 2009-11-04 13:38:23Z

2

You can also swap the values without any temporary variable:

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

then call:

swap(&ary[0], &ary[1]);

answered Nov 4, 2009 at 13:38

user167864

Comments

Peter Rowell · Accepted Answer · 2009-11-03 23:31:54Z

0

Try this instead:

void swap( int ary[] )
{
    int temp = ary[0];
    ary[0] = ary[1];
    ary[1] = temp;
}

answered Nov 3, 2009 at 23:31

Peter Rowell

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4 Comments

Josh Curren Over a year ago

I know how to do it with array notation but we were required to use pointers.

f0b0s Over a year ago

@Josg i'll open a little secret to you. You can take [] notation, and make the thing, that compilers do: a[x] = *(a + x) = x[a]. YEAH! it's a little c-trick about a[x] = x[a], cause compilers don't use []-method.

f0b0s Over a year ago

i mean you can take the Peter's code and just edit it little bit: arr[1] -> *(arr+1)

Peter Rowell Over a year ago

What f0b0s is saying (and I agree) is that for all intents and purposes there is no difference between an array name and a pointer (ignoring doing foo++); the only questions are where is the space allocated? (stack, heap, or 'global space'), and how many elements are in the array? Both of these questions also should send up red flags because it means a) you don't know how far you can meaningfully index into the array, and b) you don't know if you can/should deallocate anything. Gack. Give me Python.

f0b0s · Accepted Answer · 2009-11-04 00:24:57Z

0

your swap function will work only for 2-ints array, so show it to your compiler (it won't change anything, but make code cleaner)

void swap( int ary[2] )

answered Nov 4, 2009 at 0:24

f0b0s

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1 Comment

Chris Lutz Over a year ago

-1 this doesn't accomplish anything, and really doesn't help anyone in the long run. Your "clearer code" to try to help us understand your intent is no substitute for decent documentation.

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