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Determing Input

Posted on 2004-09-10
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Last Modified: 2010-04-15
I'm trying to make a Roman to Int and vice versa program with C. The main question is here, how can I determine the input so if the user enters a 1 then it would convert it to roman and if he entered a letter then it would call the roman to int function!


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Question by:fluked
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Expert Comment

by:brettmjohnson
ID: 12029693
Try the system library macros isalpha() and isdigit(), defined in <ctype.h>

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Accepted Solution

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lesniles earned 100 total points
ID: 12029739
The "right" way would be to use something like isdigit():

#include <ctype.h>
#include <stdlib.h>
...

   int i;
   i = 0;
   while ( isblank(s[i]) ) i++; /* Skip leading whitespace */
   if ( isdigit(s[i]) ) {
      to_roman(atoi(s));
   } else {
      to_int(&(s[i]));
   }

A more compact version uses the fact that atoi() returns 0 for
strings that do not represent decimal numbers, and the fact that
there is no roman numeral zero:

#include <stdlib.h>
...
   int n;
   n = atoi(s);
   if ( n != 0 ) {
      to_roman(n);
   } else {
      to_int(s);
   }

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Author Comment

by:fluked
ID: 12050780
ok check this out your above answer is okay but im trying to do this

if ( argument 1 == isaletter ) && ( argument 2 == is a number )
 do

end

how would that work tho ?
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Author Comment

by:fluked
ID: 12050913
how can i also convert a char* to an integer
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LVL 23

Expert Comment

by:brettmjohnson
ID: 12050976
Check out the atoi() and strtol() library routines.
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Author Comment

by:fluked
ID: 12050988
im looking for an example, thats why i increased the points to 200 :)
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LVL 23

Assisted Solution

by:brettmjohnson
brettmjohnson earned 100 total points
ID: 12051092
int i = atoi("20");
long l = strtol("123456789", 10);
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LVL 23

Expert Comment

by:brettmjohnson
ID: 12051099
If you actually read the documentation for atoi() or strtol(), you would have
found that the usage is so trivial that you don't really benefit by example.
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