do-while loop for checking users input using c

Question

I'm using this piece of code to read users input and check if it is a number or not.But sincerly it just works for numbers and letters. I want it to work with every char. For example "!?%". I have already tried to change the "isalnum" by "isascii" but that does not work.

#include <stdio.h>
#include <ctype.h>

int main ()
  {

    int a;
    int b = 1;
    char c ;

     do
     { 
       printf("Please type in a number: ");

       if (scanf("%d", &a) == 0)
       {
         printf("Your input is not correct\n");
         do 
         {
           c = getchar();
         }
         while (isalnum(c));  
         ungetc(c, stdin);    
       }
       else
       { 
         printf("Thank you! ");
         b--;
       }

     }
     while(b != 0); 

     getchar();
     getchar();

     return 0;
  }

I don't understand your question. What characters do you want the test to reject? — Oliver Charlesworth
– Oliver Charlesworth, Commented Nov 13, 2010 at 12:00
Are these numbers? -34, 4.5, 00000000000000000000000000003, 9.87654321, -23E-34, 0x42, 0b101010101, ... — pmg
– pmg, Commented Nov 13, 2010 at 12:08
No just simple numbers like 1, 2, 3, 4 and so on. The problem is that characters like !?% are not correctly rejected like letters. — Ordo
– Ordo, Commented Nov 14, 2010 at 20:20

pmg · Accepted Answer · 2010-11-13 12:19:50Z

2

Unless you have specific requirements, you should use fgets and sscanf

while (1) {
    char buf[1000];
    printf("Please input a number: ");
    fflush(stdout);
    if (!fgets(buf, sizeof buf, stdin)) assert(0 && "error in fgets. shouldn't have hapenned ..."):
    /* if enter pending, remove all pending input characters */
    if (buf[strlen(buf) - 1] != '\n') {
        char tmpbuf[1000];
        do {
            if (!fgets(tmpbuf, sizeof tmpbuf, stdin)) assert(0 && "error in fgets. shouldn't have hapenned ...");
        } while (buf[strlen(tmpbuf) - 1] != '\n');
    }
    if (sscanf(buf, "%d", &a) == 1) break; /* for sufficiently limited definition of  "numbers" */
    printf("That was not a number. Try again\n");
}

edited Nov 13, 2010 at 12:19

answered Nov 13, 2010 at 12:13

pmg

109k14 gold badges132 silver badges203 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

user411313 · Accepted Answer · 2010-11-13 13:17:09Z

2

A correct way in strictly C89 with clearing input buffer, checking overflow looks like:

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

int readLong(long *l)
{
  char *e,in[20];
  fgets( in,20,stdin );
  if(!strchr(in,'\n')) while( getchar()!='\n' );
  else *strchr(in,'\n')=0;
  errno=0;
  *l=strtol(in,&e,10);
  return *in&&!*e&&!errno;
}

int main()
{
  long l;
  if( readLong(&l) )
    printf("long-input was OK, long = %ld",l);
  else
    puts("error on long-input");
  return 0;
}

answered Nov 13, 2010 at 13:17

user411313

4,00021 silver badges16 bronze badges

3 Comments

pmg Over a year ago

+1 but you should increase the size of in. It is possible that long is 64 bits and -9223372036854775808 (20 chars including '-') is valid but not accepted by your function.

Ordo Over a year ago

Sorry but this i too complicated. I just want a simple solution. Is there anything i could use instead of isalnum(), i tried isascii() but this doesn't work for me.

user411313 Over a year ago

Its too difficult for you to call a function? Its not to difficult for you to handle inputbuffer yourself everytime?

Collectives™ on Stack Overflow

do-while loop for checking users input using c

2 Answers 2

Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related