char conversion problem

[muser]

is string converted to its integer equivalent by minusing it by 48?
the function is suppose to check the fifth digit of struct member
using the formula contained within the function.
The function isn't doing that at the moment, hence this post and the
querie above. string must be of type char.

bool checkdigitforcu stomercode( char* string )
{

int weightingfactor = 5;
int checkdigit;
int remainder;
int check1;

for(int i =0; i < 5; ++i)
check1 += (string[i] - 48) * weightingfactor ;/* problem area */
weightingfactor --;
remainder = check1 % 11;

checkdigit = 11 - remainder;

for(i=4; i<=5; i++)
if(!string[i] == checkdigit)
return false;
if(checkdigit == 11){
checkdigit = 'X';
}
else
if(checkdigit = 10){
checkdigit = '0';
}

return true;
}

[Joe C]

"muser" <ch**********@h otmail.com> wrote in message
news:f9******** *************** ***@posting.goo gle.com...

is string converted to its integer equivalent by minusing it by 48?

If you want to strip numbers from the ASCII code, I'd think you would want
to do it as follows...first , check that the char is in the range 48-57, then
take the char and "bit_and" it with binary integer 1111. The result will be
the decimal numeric equivalent to the ascii number. There's prob many
"better" ways to do this. However, if you need to process a bunch of data,
some similar method is likely efficient.

I'd use something like:

#include<iostre am>

using namespace std;

int num_seeker(char charin){
return ((47 < charin) && (charin < 58)) ? charin & 15 : -1;
}

int main(){
char g;

for(int i = 0; i < 256; ++i){
g = i;
cout << i <<" "<< g <<" "<< num_seeker(g) << endl;
}

return 0;
}

[Victor Bazarov]

"muser" <ch**********@h otmail.com> wrote...

is string converted to its integer equivalent by minusing it by 48?
No. But to convert a decimal digit in most codes (all codes that
I know of) to its decimal "value" you need to subtract '0' (zero
digit).

If you need to convert a string (are you talking of std::string?)
into its integer "equivalent ", use 'std::istringst ream'.
the function is suppose to check the fifth digit of struct member
using the formula contained within the function.
The function isn't doing that at the moment, hence this post and the
querie above. string must be of type char.
"Isn't doing" WHAT at the moment? What "formula" should it use?
If the code doesn't do what's expected of it, how should we be able
to recommend anything?

bool checkdigitforcu stomercode( char* string )
{

int weightingfactor = 5;
int checkdigit;
int remainder;
int check1;

for(int i =0; i < 5; ++i)
check1 += (string[i] - 48) * weightingfactor ;/* problem area */
If it's only the _fifth_ digit you need to check, why are you subtracting
_all_ of them?

If what you're trying to do here is to convert the five leading characters
of the 'string' array into a decimal number _manually_, that's not the right
way. The right way would be

check1 = (check1 * 10) + string[i] - '0';

(and (a) don't forget to initialise 'check1' to 0, and (b) there is a strong
assumption that all 'string' elements contain _decimal_digits _).
weightingfactor --;
remainder = check1 % 11;
Why "% 11"? Shouldn't it be "% 10"? And if you just need the 5th digit,
why not just take string[4] and look at it?

Your programming exercises are beyond me, to be honest with you. Where
are you getting all that stuff?

Try to do this: if you need something _corrected_, _describe_caref ully_
what you think it should do. Please try to avoid using "the formula
contained within the function" description. If you didn't do it right,
there is _no_ formula.

checkdigit = 11 - remainder;

for(i=4; i<=5; i++)
if(!string[i] == checkdigit)
return false;
if(checkdigit == 11){
checkdigit = 'X';
}
else
if(checkdigit = 10){
checkdigit = '0';
}

return true;
}

[Victor Bazarov]

"Joe C" <jk*****@bellso uth.net> wrote...

"muser" <ch**********@h otmail.com> wrote in message
news:f9******** *************** ***@posting.goo gle.com...

is string converted to its integer equivalent by minusing it by 48?
If you want to strip numbers from the ASCII code, I'd think you would want
to do it as follows...first , check that the char is in the range 48-57,

then take the char and "bit_and" it with binary integer 1111. The result will be the decimal numeric equivalent to the ascii number. There's prob many
"better" ways to do this. However, if you need to process a bunch of data, some similar method is likely efficient.

I'd use something like:

#include<iostre am>

using namespace std;

int num_seeker(char charin){
return ((47 < charin) && (charin < 58)) ? charin & 15 : -1;
NEVER use straight values when working with characters, it's completely
unportable. You _ought_ to just use '0' and '9' (to form the expression

('0' <= charin && charin <= '9')

Better yet, use 'std::isdigit' function from <cctype>.
}

int main(){
char g;

for(int i = 0; i < 256; ++i){
g = i;
cout << i <<" "<< g <<" "<< num_seeker(g) << endl;
}

return 0;
}

[Karl Heinz Buchegger]

muser wrote:

is string converted to its integer equivalent by minusing it by 48?
You mean: a single character, not a string.

Yes, you can do that. But do yourself a favour and replace 48
with '0'.

int AsInt;
char AsChar = '8';

AsInt = AsChar - '0';

AsInt now contains the value 8.
the function is suppose to check the fifth digit of struct member
using the formula contained within the function.
The function isn't doing that at the moment, hence this post and the
querie above. string must be of type char.

bool checkdigitforcu stomercode( char* string )
{

int weightingfactor = 5;
int checkdigit;
int remainder;
int check1;

for(int i =0; i < 5; ++i)
check1 += (string[i] - 48) * weightingfactor ;/* problem area */
weightingfactor --;
That looks strange. What is it supposed to do?

If you want to convert "32078" into an integer, then
you simply can do:

sscanf( string, "%d", &check );

But note: a 5 digit string, eg. "78542", may overflow
an int on some systems, noteable those where the maximum
signed int is 32768. So you'd better check that.

If you want to do it the hard way (by hand, insted of
sscanf) it is done this way:

check1 = 0;
for( int i = 0; i < 5; ++i )
{
check1 = 10 * check1;
check1 += string[i] - '0';
}

Consider the string "29671"
Check1 starts with a value of 0. Now the for
loop begins:

i = 0
Check1 * 10 -> 0
'2' - '0' -> 2 + Check1 -> 2

i = 1
Check1 * 10 -> 20
'9' - '0' -> 9 + Check1 -> 29

i = 3
Check1 * 10 -> 290
'6' - '0' -> 6 + Check1 -> 296

i = 4
Check1 * 10 -> 2960
'7' - '0' -> 7 + Check1 -> 2967

i = 5
Check1 * 10 -> 29670
'1' - '0' -> 1 + Check1 -> 29671

Finshed: The string "29671" has been converted to its
numerical equivalent 29671
The other thing your code: for(int i =0; i < 5; ++i)
check1 += (string[i] - 48) * weightingfactor ;/* problem area */
weightingfactor --;

could mean, is some sort of checksum over the digits, where
each digit has a different weight.
But then I guess it has to read:

for(int i =0; i < 5; ++i)
{
check1 += (string[i] - 48) * weightingfactor ;/* problem area */
weightingfactor --;
}

Note that the decrement of the weightingfactor is now enclosed
in the loop and each digit gets a different weightingfactor .
(Without that, the whole concept of weightingfactor wouldn't make
any sense, since the weightingfactor isn't used anywhere else in
your function).

Another tip: If you think that some computation doesn't work the
way you think it should, it is always a good idea to either
* use a debugger to check the variables
* or insert some outputstatement s which shed some light on what
is going on:

for( int i = 0; i < 5; ++i )
{
printf( "Check1: %d, String[%d] %c, Weight %d -> ",
check1, i, string[i], weightingfactor );

check1 += (string[i] - 48) * weightingfactor ;/* problem area */

printf( "%d\n", check1 );
}

Then you can see on your monitor what is going on and will be able to
check the computation. Once you understand what is going on and what
is wrong with that: fix the computation, use the output to verify that
it now works as you expect it to work and remove those additional output
statements.

--
Karl Heinz Buchegger
kb******@gascad .at

[Joe C]

"Victor Bazarov" <v.********@com Acast.net> wrote in message >

NEVER use straight values when working with characters, it's completely
unportable. You _ought_ to just use '0' and '9' (to form the expression

('0' <= charin && charin <= '9')

Better yet, use 'std::isdigit' function from <cctype>.

Victor, Thanks for the kick in the tail...again ;-)

I actually misread the original post...I thought he was stripping numbers
from a serial number/barcode type thingie...where portability would be
non-important.

on my original code...if you had a large file that contained a boatload of
ascii data to sort through...you could speed up my function a bit by using:

int num_seeker(char charin){
return (charin > 57) ? -1 : ((charin < 48) ? -1 : charin & 15);
}
That way you'd get your return value after just 1 comparison for all
characters (likely to be the most frequent) performing the second comparison
only if needed.

Victor...I respect your thoughts a lot. Would you ever do something like
this, or is it simply offensive to your programming sensibilities?

[Karl Heinz Buchegger]

Joe C wrote:

"Victor Bazarov" <v.********@com Acast.net> wrote in message >

NEVER use straight values when working with characters, it's completely
unportable. You _ought_ to just use '0' and '9' (to form the expression

('0' <= charin && charin <= '9')

Better yet, use 'std::isdigit' function from <cctype>.

Victor, Thanks for the kick in the tail...again ;-)

I actually misread the original post...I thought he was stripping numbers
from a serial number/barcode type thingie...where portability would be
non-important.

on my original code...if you had a large file that contained a boatload of
ascii data to sort through...you could speed up my function a bit by using:

int num_seeker(char charin){
return (charin > 57) ? -1 : ((charin < 48) ? -1 : charin & 15);
}
That way you'd get your return value after just 1 comparison for all
characters (likely to be the most frequent) performing the second comparison
only if needed.

Victor...I respect your thoughts a lot. Would you ever do something like
this, or is it simply offensive to your programming sensibilities?

Can't talk for Victor.
But even then I would do:

return (charin > '9') ? -1 : ((charin < '0') ? -1 : charin - '0');

Much simpler to not have to remember the code values for the digits.
(On most CPU's I think its reasonable to assume that a subtraction
runs in the same time as an and masking).

--
Karl Heinz Buchegger
kb******@gascad .at

[Victor Bazarov]

"Karl Heinz Buchegger" <kb******@gasca d.at> wrote...

Joe C wrote:

"Victor Bazarov" <v.********@com Acast.net> wrote in message >

NEVER use straight values when working with characters, it's completely unportable. You _ought_ to just use '0' and '9' (to form the expression
('0' <= charin && charin <= '9')

Better yet, use 'std::isdigit' function from <cctype>.

Victor, Thanks for the kick in the tail...again ;-)

I actually misread the original post...I thought he was stripping numbers from a serial number/barcode type thingie...where portability would be
non-important.

on my original code...if you had a large file that contained a boatload of ascii data to sort through...you could speed up my function a bit by using:
int num_seeker(char charin){
return (charin > 57) ? -1 : ((charin < 48) ? -1 : charin & 15);
}
That way you'd get your return value after just 1 comparison for all
characters (likely to be the most frequent) performing the second comparison only if needed.

Victor...I respect your thoughts a lot. Would you ever do something like this, or is it simply offensive to your programming sensibilities?

Can't talk for Victor.
But even then I would do:

return (charin > '9') ? -1 : ((charin < '0') ? -1 : charin - '0');

Much simpler to not have to remember the code values for the digits.
(On most CPU's I think its reasonable to assume that a subtraction
runs in the same time as an and masking).

Karl, you read my mind. Couple more of these and I'll let you talk
for me. :-)))

Victor

[Gene Wirchenko]

On Wed, 17 Dec 2003 20:26:06 +0100, Karl Heinz Buchegger
<kb******@gasca d.at> wrote:

[snip]

Can't talk for Victor.
But even then I would do:

return (charin > '9') ? -1 : ((charin < '0') ? -1 : charin - '0');

Much simpler to not have to remember the code values for the digits.
(On most CPU's I think its reasonable to assume that a subtraction
runs in the same time as an and masking).

It also has the advantage that it does not depend on the
character set being ASCII.

Sincerely,

Gene Wirchenko

[Jerry Coffin]

In article <f9************ **************@ posting.google. com>,
ch**********@ho tmail.com says...

is string converted to its integer equivalent by minusing it by 48?

No. A single character might be, depending on the character set in use,
but a character is not a string.

If I'm reading your code correctly, I'd use something like this:

for (i=0; i<5;i++)
if ( isdigit(string[i]))
check1 += (string[i]-'0') * weightingfactor--;

The part where you have 'if (checkdigit = 10)' is almost certainly an
error -- you're doing an assignment instead of a comparison. In any
case, I'd probably cheat, and make this table-driven:

char digits[] = "0123456789 0X";

// loop from above.

checkdigit = digits[11 - (check1 % 11)];

I'm not quite sure I follow your 'if (!string[i] == checkdigit)'.
Perhaps you wanted: 'if (string[i] != checkdigit)'? comparison ('==')
has lower precedence than logical negation ('!'), so as-is, you're doing
logical negation on 'string[i]', and then comparing the result of that
to checkdigit. Logical negation is _normally_ done on a logical value,
and it looks like string[i] will be a character rather than boolean.

--
Later,
Jerry.

The universe is a figment of its own imagination.