[SOLVED] Programming help

Ajit Gunge · 01-05-2011, 03:13 PM

HI All,
I have written a simple C program in which I dont feel anything is wrong but still the GCC compiler is giving errors and warning.Can someone help me find out whats exactly going wrong in here.
The errors are

Quote:

prog2.c:3: error: syntax error before '*' token
prog2.c: In function `swap':
prog2.c:4: error: number of arguments doesn't match prototype
prog2.c:2: error: prototype declaration
prog2.c:6: error: `x' undeclared (first use in this function)
prog2.c:6: error: (Each undeclared identifier is reported only once
prog2.c:6: error: for each function it appears in.)
prog2.c:7: error: `y' undeclared (first use in this function)

And the program is

Quote:

#include<stdio.h>
int swap(int *,int *);
int swap(*x,*y)
{
int t;
t=*x;
*x=*y;
*y=t;
}
int main(void)
{
int a=10,b=20;
swap(&a,&b);
printf("a=%d",a);
printf("b=%d",b);
return 0;
}

Thanks!

Nylex · 01-05-2011, 03:35 PM

I've reported this to be moved to Programming, since it's more suitable there.

First of all, the prototype

int swap(int *,int *);

is unnecessary, since you're defining the function before main().

Secondly, the line

int swap(*x,*y)

is incorrect, as you've not given types for the parameters x and y, other than specifying that they're pointers. Edit: also, surely this function should return void, as it doesn't return anything.

Edit: also, in future, put your code between CODE tags, rather than QUOTE tags, as the former preserve indentation.

johnsfine · 01-05-2011, 04:09 PM

Quote:

Originally Posted by Nylex

First of all, the prototype

int swap(int *,int *);

is unnecessary, since you're defining the function before main().

I think that is a matter of style, not an error, so I hope your opinion on that didn't distract Ajit Gunge from understanding the important info you gave next.

Quote:

the line

int swap(*x,*y)

is incorrect, as you've not given types for the parameters x and y, other than specifying that they're pointers.

That is the major error.

Quote:

surely this function should return void, as it doesn't return anything.

Correct, but I think most compilers give just a warning, not an error for that mistake.

Quote:

also, in future, put your code between CODE tags

I'll repeat that as well.

In addition:

Code:

   printf("a=%d",a);
   printf("b=%d",b);

I expect the OP will understand the minor error in those lines better by seeing the output of the program before correcting that error, than if I pointed it out now.

Quote:

Originally Posted by Ajit Gunge

Thanks!

I think it is more polite to save that for the post you put at the end of the thread telling us that you understood our answers and your program is now working.

Obviously, feel free to first post follow up questions if you didn't understand the answers or your program still doesn't work. Though unfortunately common, it is impolite to skip posting a final status (it worked or you gave up).

Nylex · 01-05-2011, 04:13 PM

Quote:

Originally Posted by johnsfine

I think that is a matter of style, not an error, so I hope your opinion on that didn't distract Ajit Gunge from understanding the important info you gave next.

Fair enough.

Quote:

Correct, but I think most compilers give just a warning, not an error for that mistake.

True. I'm used to compiling with -Wall with gcc.

Ajit Gunge · 02-09-2011, 06:36 AM

Thanks for your comments.The program has worked now.Sorry for the delay though.

MTK358 · 02-09-2011, 08:38 AM

Mark the thread as solved.

ta0kira · 02-09-2011, 01:01 PM

Quote:

Originally Posted by Ajit Gunge

Code:

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

The pre-standard style of argument definition would have you do it like this:

Code:

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

(Don't copy that! C hasn't been like that since the 80s.) Did you, by chance, learn from an old reference?
Kevin Barry