Negative "not taking"

[mdh]

Hi All,
I understand atof is in the library, but this is part of my learning
curve.

Given

double atof( char *s){
double d = 0.00;
double sign = 1.00;
double fractprt = 1.00;
/* check for neg number */
if ( *s == '-') {
sign = -1.00;
s++;
}

while (isadigit(*s)){
d=d*10.00 + ('0' - *s++);
}
if (*s == '.'){
s++;
while(isadigit(*s)){
d=d*10.00 + ('0' - *s++);
fractprt *=10;
}
}
return (sign * d / fractprt) ;
}

where s[] equals -36.03, the return I get is 36.03 not -36.03. Is
there a reason why
"return (sign * d / fractprt) ;"

does not seem to multiply by "-1"

Thanks

[Ben Pfaff]

mdh <md**@comcast.netwrites:

d=d*10.00 + ('0' - *s++);

I think you mean *s++ - '0'.

d=d*10.00 + ('0' - *s++);

Ditto.

there a reason why
"return (sign * d / fractprt) ;"

does not seem to multiply by "-1"

Because you get the sign of each digit wrong.
--
Ben Pfaff
http://benpfaff.org

[Lew Pitcher]

On Nov 8, 9:46 am, mdh <[e-mail]:

Hi All,
I understand atof is in the library, but this is part of my learning
curve.

Given

double atof( char *s){
double d = 0.00;
double sign = 1.00;
double fractprt = 1.00;
/* check for neg number */
if ( *s == '-') {
sign = -1.00;
s++;
}

while (isadigit(*s)){
d=d*10.00 + ('0' - *s++);

Consider this: what is the result of ('0' - '3') ?
Hint: it isn't what you think it is.

}
if (*s == '.'){
s++;
while(isadigit(*s)){
d=d*10.00 + ('0' - *s++);

Again, what is the result of ('0' - '3')?

fractprt *=10;
}
}
return (sign * d / fractprt) ;

}

where s[] equals -36.03, the return I get is 36.03 not -36.03. Is
there a reason why
"return (sign * d / fractprt) ;"

does not seem to multiply by "-1"

Think of it this way: what sort of number will give you a positive
value when multiplied by -1?
Thats the sort of number that (d/fractprt) represents.
So, how did you get that sort of number? What part of your function is
suspect? What part(s) produce signed values, where the sign can be
suspect?

Thanks

[mdh]

On Nov 8, 7:00 am, Lew Pitcher <[e-mail]:

On Nov 8, 9:46 am, mdh <[e-mail]:

Think of it this way: what sort of number will give you a positive
value when multiplied by -1?

Thanks

It was a long day!!! Sometimes I think one gets so caught up in the
"progamming" aspect..esp when learning that one overlooks something
really basic.
thank you

[Mark Bluemel]

mdh wrote:

Hi All,
I understand atof is in the library, but this is part of my learning
curve.

Given

double atof( char *s){

....

d=d*10.00 + ('0' - *s++);

....

d=d*10.00 + ('0' - *s++);

where s[] equals -36.03, the return I get is 36.03 not -36.03.

Deja Vu all over again. Your name isn't Leonard Shelby by any chance?
(http://tinyurl.com/347zrf)

[Richard]

Mark Bluemel <ma**********@pobox.comwrites:

mdh wrote:

>Hi All,
I understand atof is in the library, but this is part of my learning
curve.

Given

double atof( char *s){

...

> d=d*10.00 + ('0' - *s++);

...

> d=d*10.00 + ('0' - *s++);

>where s[] equals -36.03, the return I get is 36.03 not -36.03.

Deja Vu all over again. Your name isn't Leonard Shelby by any chance?
(http://tinyurl.com/347zrf)

99.9999% of questions asked here are deja vu. It's a help group I
thought? my advice to you is to take a break if you are scoffing at the
"usual mistakes".

[CBFalconer]

Ben Pfaff wrote:

mdh <md**@comcast.netwrites:

> d=d*10.00 + ('0' - *s++);

I think you mean *s++ - '0'.

> d=d*10.00 + ('0' - *s++);

Ditto.

How about "- ('0' - *s++)" in both expressions? :-)

--
Chuck F (cbfalconer at maineline dot net)
<http://cbfalconer.home.att.net>
Try the download section.

--
Posted via a free Usenet account from http://www.teranews.com