Can not convert int to int *

[vishroxx]

#include "stdafx.h"
# include "stdio.h"
#include "conio.h"
void swap(int *x,int *y);

void main()
{
int a=10,b=20;
swap(a,b);
printf("value of a=%d and b=%d");
getch();
}

void swap(int *x,int *y)

{
  if(x!=y)
     {
      *x ^= *y;
         *y ^= *x;
         *x ^= *y;

     }
}

// I'm getting .. cann't convert int to int * ...

can anybody tell me why so . and how to solve it
regards.

hoping for quick and positive response.

[Char*Pntr]

#include "stdafx.h"
# include "stdio.h"
#include "conio.h"
void swap(int *x,int *y);

void main()
{
int a=10,b=20;
swap(a,b);
printf("value of a=%d and b=%d");
getch();
}

void swap(int *x,int *y)

{
  if(x!=y)
     {
      *x ^= *y;
         *y ^= *x;
         *x ^= *y;

     }
}

// I'm getting .. cann't convert int to int * ...

can anybody tell me why so . and how to solve it
regards.

hoping for quick and positive response.

Edit: The correct answer already provided below.

[ch4]

int main()
{
int a=10,b=20;
swap(&a,&b);
printf("value of a=%d and b=%d",a,b);
getch();
return 0;
}

Swap function needs two pointers (memory addresses) and you're passing raw integer values instead of their memory addresses by using & in front of them.

//Too Late for me

[bernt]

let's see... how many things are wrong here?

You are trying to return an int value in your function prototype, yet you declared it
to return "void".

When you call your function, you wont capture anything, even if it was written
correctly.

Also you dont need to use pointers in your function.

[STRIKETHROUGH]
I formally refute these statements.

-There are no returns anywhere, hence the declarations with void
-The "swap" function swaps two variables in-place. There's no need to return something.
-Pointers are necessary to swap variables like that (that is, in a function)
[/STRIKETHROUGH]

EDIT: Nevermind, it's gone now.

With that said,
1) void main() is evil. Do not use it. Please. See ch4's post for fixes.
2) Take a look at this. Read it twice, maybe three times, let it really sink in.
3) You're missing arguments for the printf function. Though it would have been easy to catch once you fixed the call to swap().

printf("value of a=%d and b=%d", a, b);

EDIT: Holy carpal, these edits make it hard to keep track of what's going on!

[kermit]

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

...

void swap(int *x,int *y)

{
  if(x!=y)
     {
      *x ^= *y;
         *y ^= *x;
         *x ^= *y;

     }
}

If you are using a good compiler, you are not gaining anything with that hacky xor swap. See here and here for more details.

[ch4]

#include "stdafx.h"
# include "stdio.h"
#include "conio.h"
void swap(int *x,int *y);

Quotes are used to include custom libraries.
For the default use <>

#include <stdafx.h>
# include <stdio.h>
#include <conio.h>
void swap(int *x,int *y);

[ch4]

EDIT: Holy carpal, these edits make it hard to keep track of what's going on!

Like goto

[Char*Pntr]

[QUOTE=bernt;968747][STRIKETHROUGH]
I formally refute these statements.

-There are no returns anywhere, hence the declarations with void
-The "swap" function swaps two variables in-place. There's no need to return something.
-Pointers are necessary to swap variables like that (that is, in a function)
[/STRIKETHROUGH]

EDIT: Nevermind, it's gone now.

I already deleted my suggestion.... like seconds after I posted it... surprised
that it hung on for so long.

[Char*Pntr]

Like goto

I took my first C class back in '92... the instructor who claimed he knew
Dennis Ritchie, told us that if we ever use a "goto" we would get
an "F" for the assignment.

He said the only reason the "goto" statement was included was because the ANSI C committee required it. Probably Ritchie was not too happy with that decision. But since I did not personally know him, maybe he used "goto's" for debugging or whatever. I've never used it, and I don't think I ever will.

That's pretty much all that remembered from the class since I haven't used
C since then.

[Salem]

> He said the only reason the "goto" statement was included was because the ANSI C committee required it.
This is a load of BS.
goto was part of the language long before ANSI got anywhere near it, It wasn't their decision to add or remove it.

[Char*Pntr]

> He said the only reason the "goto" statement was included was because the ANSI C committee required it.
This is a load of BS.
goto was part of the language long before ANSI got anywhere near it, It wasn't their decision to add or remove it.

I think I misquoted the statement that I heard over 18 years ago.... "ANSI C committee included the "goto" in the standard to make it backward compatible with existing C code?

Anyway, I'm off topic.

[cas]

the instructor ... told us that if we ever use a "goto" we would get
an "F" for the assignment.

This is probably good advice for a beginning class. I suspect a lot of students would have been exposed to BASIC or something similar, where the use of goto was normal, and would need to be taught how to think in another way.

He said the only reason the "goto" statement was included was because the ANSI C committee required it.

This can't be true. The Unix V4 kernel, dating from around 1973, has at least 90 cases of goto being used; both Ken Thompson and Dennis Ritchie used goto, if directory names are to be believed. The 1975 V6 system (kernel + userland) had at least 800 uses of goto. ANSI didn't even start looking at C until the early 1980s.

While goto can be abused, there are some legitimate uses for it. Common examples include breaking from inside nested loops, and for cleanup on error.

Edit: Late reply. Sorry.