Problem reading and analyzing 3 variables

I'm following a C tutorial with exercisess, and in one of them you have to make a program that when given 3 variables tells the user the biggest and smallest number... This is what I have, but I get a tonne of errors!

Code:

#include <stdio.h>
void main()
{
    int x,y,z;
    printf("Escribe tres numeros\n");
    scanf("%d %d %d", &x, &y, &z);
    if x>y && x>z && y>z {
        printf("Big numer is %d Small number is d%\n", &x,&z);
    }else if x>y && x>z && y<z{
        printf("Big numer is %d Small number is d%\n", &x,&y);
    }else if y>x && y>z && x>z{
        printf("Big numer is %d Small number is d%\n", &y,&z);
    }else if y>x && y>z && x<z{
        printf("Big numer is %d Small number is d%\n", &y,&x);
    }else if z>x && z>y && y>x{
        printf("Big numer is %d Small number is d%\n", &z,&x);
    }else
        printf("Big numer is %d Small number is d%\n", &z,&y);
    }
}

Errors I get include: parse error in line 7 and "return type of 'main' is not 'int' "
Thanks

ps I should stop using scanf...

This:

Code:

if x>y && x>z && y>z {

Needs to be this:

Code:

if( x>y && x>z && y>z ){

And likewise for your other else if statements.

... and you're still doing that void main thing others have told you about. It may work fine but it is really frowned upon and you should get in the habit of saying int main instead.

[edit]Almost forgot... you don't put & in front of the variables when you print them.[/edit]

you need parens around your if statement, I dont know what tutorial you're following, but main returns an INT (as is pretty clearly specified by the error your getting) so change it to int main(void) and throw a return statement in there at the end.

example:

Code:

int main(void)
{
  if (dead_horse)
    beat;
  return 0;
}

Looking at your code, I can already see a half-dozen errors, and these are only related to syntax. Anyways, I re-written your stuff, as it should be corrected. I suggest you look at the comments I have left

Code:

#include <stdio.h>
int main()   /*** to avoid warning, put return type of main to int  ***/
{
    int x,y,z;
    printf("Escribe tres numeros\n");
    scanf("%d %d %d", &x, &y, &z);
    if (x>y && x>z && y>z)  /** parentheses  **/
    {
        printf("Big numer is %d Small number is %d\n", x,z);  /*** print values, not addresses ***/
    }
    else
    { 
       if (x>y && x>z && y<z)/** parentheses  **/
       {
          printf("Big numer is %d Small number is %d\n", x,y);/*** print values, not addresses ***/
       }
       else
       {
          if (y>x && y>z && x>z)/** parentheses  **/
          {
             printf("Big numer is %d Small number is %d\n", y,z);/*** print values, not addresses ***/
          }
          else
          { 
             if (y>x && y>z && x<z)/** parentheses  **/
             {
                printf("Big numer is %d Small number is %d\n",y,x);/*** print values, not addresses ***/
             }
             else
             { 
                if (z>x && z>y && y>x)/** parentheses  **/
                {
                   printf("Big numer is %d Small number is %d\n", z,x);/*** print values, not addresses ***/
                }
                else
                {
                   printf("Big numer is %d Small number is %d\n", z,y);/*** print values, not addresses ***/
                }
             }  /** forgottten brace  ***/
          }  /*  forgotten brace  ***/
       }   /* forgtotten brace ***/
   }
   
   return 0;   /**** to return int  ***/
}

Also, 'd%' in the middle of a printf statement will print 'd%'. I think what you wanted was '%d'. I took the liberty of correcting that too.

Also, 'd%' in the middle of a printf statement will print 'd%'.

Actually, it will print 'd' followed by whatever special escape character translation printf() makes depending on the character immediately following the %. In this program's case it tries to convert %\ which it doesn't understand and will give you some error message related to that. To print '%' using printf() you need to tell printf() to print "%%".

the "d%" I could have fixed myself, but thanks for the harder stuff guys...
Is "void main()" ever useful?

Well sure! If you're looking to write illegal C code then void main() is a great way to accomplish it. Everything is useful at some time to someone.

void main() is not necessarily "illegal". If the implementation documents it as being okay, it's then an implementation defined program, it's just no longer strictly conforming.

If you want to write a strictly conforming C program (one that relies on no implementation defined, undefined, or unspecified behaviour), then you need to have main return an int.

See http://homepages.tesco.net/~J.deBoyn...void-main.html for why blanket statements like "void main is illegal" are wrong.

Originally Posted by cwr

See http://homepages.tesco.net/~J.deBoyn...void-main.html for why blanket statements like "void main is illegal" are wrong.

Q - How many [...] engineers does it take to change a light bulb?
A - None. [...] will just redefine Darkness (TM) to be the new industry standard.

C90 was the standard, once.

Originally Posted by Dave_Sinkula

C90 was the standard, once.

That page applies also to C99, the C99 standard still states that main can have some implementation defined format for its return.