Syntax Error

[marbia]

I think a silly syntax error then; i post the code..

#include <stdio.h>
#include <stdlib.h>

main()

{ int n;              
  int step=0;

    scanf("%d",&n);
    if(n==1)return 0; 
    if (n%2==0){
          n=n/2;
	  step++;
    else 
          n=3*n+1;  
          step++;}
     
   printf("%d",step);
    return 0;
}

i get back from gcc:

' error: syntax error before "else" '

then, the error stands in 'step++', i guess.. but don't know how to tweak it.. i need the function add +1 to 'step' for both cases, even and odd inputs.

any clue? thanks to all

[Salem]

    if (n%2==0){
          n=n/2;
	  step++;
    } else {   // Note the extra braces 
          n=3*n+1;  
          step++;}

[Codeplug]

I prefer:

    if ((n % 2) == 0)
    {
        n = n / 2;
        step++;
    } 
    else 
    {
        n = 3 * n + 1;  
        step++;
    }

gg

[marbia]

thanks for the input

something strange is happening.

compile: ok

the function assigns always 1 thou, whatever is the input.

here's an example;

for the input 1 -> function returns 0 and it is right;
but, if i give 3,4,7,8,10 or whatever else, it returns me always 1.


checks seems ok, compile ok, but function don't calculate in the right way the function..

[hk_mp5kpdw]

Well... it would help if we knew what your code is trying to do with the step value. I would guess you're missing a loop somewhere.

[Salem]

Given the lack of any loop at all in your code, what do you expect?

Sure, each branch of your if/else has step++, so all that you really have is (reduced)

    scanf("%d",&n);
    if(n==1)return 0; 
    step++;
    printf("%d",step);

[marbia]

sure

it should work as in this case;

input 10.

10 is even then 10/2 = 5 (step=1)
5 is odd then 5*3+1 = 16 (step=2)
16 is even then 16/2=8 (step=3)
8 is even then 8/2=4 (step=4)
4 is even then 4/2=2 (step=5)
2 is even then 2/2=1 (step=6)

when receives 1, function stops.

then, if i give the input 10 function should return me 6.

[Salem]

Well two people told you to use a loop, the rest is up to you.

[marbia]

does it mean i have to change the structure of this code?

i'm not asking someone will do a job for me, absolutely.
i've checked the code, seems to be ok, check for the '1' is ok and the return is correct, but checks for the other input fail; this is really strange, and i don't understand at the moment, where the structure fails and doesn't give me right return back.

thanks for your help, thou.

[Salem]

How about
while ( n != 1 )

Or something
*shrug*

[marbia]

ok, man i was only asking what's going on; don't know why your 'shrug'.

i don't understand, however, why the 'if' construct doesn't work properly. he makes his job nicely only for the input=1. on the other fails.
otherwise, all checks with the 'while' are correct.

thanks.

[Salem]

    if(n==1)return 0;  // Nope, this isn't a while
    if (n%2==0){ // Nope, this isn't a while
          n=n/2;
	  step++;
    else // Nope, this isn't a while
          n=3*n+1;  
          step++;}

Where exactly is your while loop?
How about posting your latest code, if you're still having problems.

[marbia]

the function works fine, while looping.

i don't really understand why the 'if' goes all the checks, over the '1', to the hell.