Allan Ashton <allanasht...@mywitsend99.freeserve.co.uk> wrote in message

> Would anyone please tell me how I can do this. i.e.

>     char * string;
>     int integer1;

>     // input string for numeric values only using scanf....
>          ..........

>     // set integer1 somehow to the value of string..

Dann Corbit wrote:
> Allan Ashton <allanasht...@mywitsend99.freeserve.co.uk> wrote in message
> news:381e300c.14153455@news.freeserve.co.uk...
> > Would anyone please tell me how I can do this. i.e.

> >     char * string;
> >     int integer1;

> >     // input string for numeric values only using scanf....
> >          ..........

> >     // set integer1 somehow to the value of string..

> Your spec is rather incomplete.
> The scanf() function has a conversion specifier for integers.
> Else, you can go char by char for isdigit() and multiply by 10 each time.
> Don't forget to allocate memory for string.

Al Bowers (abow...@gate.net) wrote:

> Here is a function, asctoint() that will do this. However, it is dangerous to
> use on user input because the string may not convert to a valid int. You have
> the hazard of int underflow or overflow. The string may not be a valid
> number. etc.
> #include <stdio.h>
> int asctoint(const char *string) {
>    const char *s1,*s2,*num = "0123456789";
>    int sum = 0,negative = 0;
>    if(*string == '-'){
>       negative++;

>       string++;
>       }
>    for(s1 = string;*s1 != '\0';s1++) {
>       for(s2 = num;*s2 != '\0' ; s2++)
>   if(*s2 == *s1) break;
>       if(*s2 == '\0') return sum;

>       sum *= 10;
>       sum += s2 - num;
>       }

>    return negative?sum*-1:sum;

>    }

Steven Huang wrote:

> Al Bowers (abow...@gate.net) wrote:
> [...]
> > Here is a function, asctoint() that will do this. However, it is dangerous to
> > use on user input because the string may not convert to a valid int. You have
> > the hazard of int underflow or overflow. The string may not be a valid
> > number. etc.

> > #include <stdio.h>

> > int asctoint(const char *string) {
> >    const char *s1,*s2,*num = "0123456789";
> >    int sum = 0,negative = 0;

> >    if(*string == '-'){
> >       negative++;

> This is stylistic, of course, but I personally don't like to set a
> flag with an arithmetic operation.  I prefer:

>   #define POSITIVE 1
>   #define NEGATIVE 2

>   int sign = POSITIVE;

>   sign = NEGATIVE;

> >       string++;
> >       }
> >    for(s1 = string;*s1 != '\0';s1++) {
> >       for(s2 = num;*s2 != '\0' ; s2++)
> >   if(*s2 == *s1) break;
> >       if(*s2 == '\0') return sum;

> There's a bug here.  If I entered "-1a", the function will return 1,
> not -1.

> >       sum *= 10;
> >       sum += s2 - num;
> >       }

> Why not just:

>   for (s1 = string; *s1; s1++)
>     if (*s1 >= '0' && *s1 <= '9')
>     {
>       sum *= 10;
>       sum += *s1 - '0';
>     }
>     else
>     {
>       return (sign == NEGATIVE) ? -sum : sum;
>     }

Kien Ha (Kien...@Mitel.COM) wrote:

> There is a bug here too :-).

> For strings such as "-1", "-12345" or
> "12345", the else clause will never be reached and the for loop ends
> when *s1 == '\0';
> So pull the return statement out of the for loop.

>Kien Ha (Kien...@Mitel.COM) wrote:
>[...]
>> There is a bug here too :-).

>Nice try, but no dice.  :)

>> For strings such as "-1", "-12345" or
>> "12345", the else clause will never be reached and the for loop ends
>> when *s1 == '\0';
>> So pull the return statement out of the for loop.

>You should've read further down, where I wrote:

>   >    return negative?sum*-1:sum;

>   Or simpler yet,

>      ... ? -sum : sum;

>The return wasn't missing... from my mind, at least.  ;)