Trying to find out which character was entered

[Corey Bullard]

I'm trying to make a reverse polish notation calculator.
One of my first steps is to figure out whether an operand (number) was entered or an operator/letter (l, p, q, +, -, *, or /).
Here's what I have for this part:

     
 char buff[MAX_BUFFER_SIZE], o;
/*  This reads a string from the stack  */        if( scanf("%s", buff) < 1 )
      {
         printf("I did not understand\n");
         return 0;
      }


      /* checks to see if the user entered a number */
      if(isdigit(buff[0]) || isdigit(buff[1]))
      {
         /*turns string into number and puts value in x*/
         sscanf( buff, "%lf", &x);
         /* place statements here to handle getting a number */


       
      } else {
         /* place statements to do operations */
		  sscanf( buff, "%lf", &o);
          if (o == 'l')  {
     printf("its l");
    }
  else if (o == 'q')  {
     printf("its q");
    }
  else  (o == 'p');  {
     printf("its p");
    }

Obviously the intention of this part is not to print "its _" but I have it set for troubleshooting.
When I compile, any operator or letter I enter will print "its p".
I had it as a switch, but gave up on that and am now trying if/else.

Any tips will be great.

NOTE: We haven't learned isdigit() or sscanf() in class (that part of the program was given to us by our instructor), so therefore we are not allowed to use it anywhere else in the program.

[dmh2000]

sscanf( buff, "%lf", &o); gives you a floating point number. using a character as the variable to be assigned will give you undefined behavior. change the "%lf" to "%c"

p.s. one way to debug this kind of input is to insert printfs after each scanf operation to see what you got.

[Corey Bullard]

Thanks for the reply.
I did what you said and put

sscanf( buff, "%c", &o);
printf("%c\n", o);

and then my if/else statements.
I'm still getting the same problem. My output looks like:
v /*I entered this*/
v
its p
_

[quzah]

[/CODE]
and then my if/else statements.
I'm still getting the same problem. My output looks like:
v /*I entered this*/
v
its p
_

Of course it is.

else  (o == 'p');  {
     printf("its p");
    }

That's not an "else if", that's "else, o == 'p' ... prompting a warning from your compiler ... ; <--- END OF STATEMENT" "new statement, 'its p'"

Now I don't know exactly what you expect it to do, but that's not it:

if( a == b )
    ... do something for when a is b
else
if( a == c )
    ... do something for when a is c
...
else
    ... do something for when all else fails
...
this executes regardless of whatever happens above, assuming your code reaches this far (you didn't return or exit)

That's how an if-else chain works.


Quzah.

[Corey Bullard]

Ah, OK, I get it now.
I put an else at the end to handle all other cases (and print invalid character), and now it seems to work well.
Thanks for the help.