CODEWITHSUNDEEP
catoi
user3250373
user3250373

Reputation: 193

Usage of atoi in c program

I have this code but here atoi does not work at all If i gave base address then also it causes prob

int main()
{
    char *token;
    int value;
    char data = "23+100";
    int i = 0;
    while(data[i] != '+' )
    {
        value = atoi(data[i]);
        push(value,&top1);
        i++;
    }
}

Upvotes: 0

Views: 375

Answers (3)

Jabberwocky
Jabberwocky

Reputation: 50776

Have a look at the strtol function. Following sample shows how to use it.

int main()
{
  char *token;
  int value;
  char *data = "23+100-2*7";
  int i = 0;
  char *endp ;
  token = data ;

  do
  {
    value = strtol(token, &endp, 10) ;
    token = endp + 1 ;
    printf ("value: %d\n", value) ;

    if (*endp != 0)
      printf ("operator: %c\n", *endp) ;

  } while (*endp != 0) ;
}

Upvotes: 0

unwind
unwind

Reputation: 399793

This:

char data = "23+100";

really shouldn't have compiled.

It should be:

const char *data = "23+100";

Which will make atoi(data[i]); even less correct than it was before.

Also, you're not supposed to use atoi() on sub-strings like that. Perhaps you should look at sscanf() or strtok(), or (perhaps best if I understand what you're after correctly) strtol().

The latter allows you to parse a long integer until a non-digit character is found, and also lets you figure out which character that was after the call.

Upvotes: 1

KARTHIK BHAT
KARTHIK BHAT

Reputation: 1420

array initialization is wrong it could be either

char data[] = "23+100"

or as unwind told

const char *data = "23+100"

and one more thing your logic for going through the array might be wrong.

As soon as + is encountered the loop is finished and 100 won't be converted and push'd onto the stack(hoping your trying to achieve that)

Upvotes: 0

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