K&R 2 exercise 2-3

[Herrcho]

Hi~ i've studied C for a few months myself,

and i'd appreciate it if anyone could improve my coding or correct it.

the following is my solution to the K&R exercise 2-3

"Write the function htoi(s), which converts a string of hexademical digits
(including an optional 0x or 0X) into its equivalent integer value.
The allowable digits are 0 through 9, a through f, and A throught F."
//*************** *************** *************** *************** **************

#include <stdio.h>

int isxdigit2(int c)
{
if ( (c >= 'a' && c <= 'f') || (c >= 'A' && c <= 'F') || (c >= '0' && c <= '9') )
return 1;
else
return 0;
}

int tolower2(int c)
{
if (c >= 'A' && c <= 'Z')
return c+32;
else
return c;
}

int power2(int base, int num)
{
int sum;
for (sum = 1; num >0 ; num --)
sum *= base;
return sum;
}

int char_to_num(int c)
{
if (c >= '0' && c <= '9')
{
return c - 48;
}
else
{
return 10 + (tolower2(c) - 'a');
}
}

int htoi(char *c)
{
int i, k, prefix = 0;
size_t sum = 0;

if (c[0] == '0' && tolower2(c[1]) == 'x')
prefix = 1;

for (i = (prefix == 1)? 2:0 ; c[i] ;i++ )
{
if (!isxdigit2(c[i]) )
{
printf("Wrong hexa number\n");
return 0;
}
c[i] = char_to_num(c[i]);
}

for (k = (prefix == 1)? 2 : 0 ; k <= i-1 ; ++k )
{
sum += c[k] * power2(16, i-1-k);
}

return sum;
}

int main()
{
char c[] = "0xAB";
printf("%u", htoi(c));

return 0;
}

//*************** *************** *************** *************** ******

when i change char c[] to char *c in main(),
it shows error, why ??

Thanks..

[Peter Pichler]

"nrk" <ra*********@de vnull.verizon.n et> wrote:

Irrwahn Grausewitz wrote:

he*********@kor net.net (Herrcho) wrote:

static const char *uppercase = "ABCDEFGHIJKLMN OPQRSTUVWXYZ";
static const char *lowercase = "abcdefghijklmn opqrstuvwxyz";

Nitpick:
static const char const *...

Nitpick, ITYM:
static const char * const ...

Here, and in other places as well :-)

Correct, though I would use:
static const char uppercase[] = ...
static const char lowercase[] = ...

There are two advantages:
1. Saving 2*sizeof(char*) bytes, and
2. Being able to apply sizeof to 'uppercase' and 'lowercase'.

I am not aware of any disadvantages ;-)

Peter

[Irrwahn Grausewitz]

nrk <ra*********@de vnull.verizon.n et> wrote:

Irrwahn Grausewitz wrote:

he*********@kor net.net (Herrcho) wrote:
<snip>

static const char *lowercase = "abcdefghijklmn opqrstuvwxyz";

Nitpick:
static const char const *...

Nitpick, ITYM:
static const char * const ...

Here, and in other places as well :-)

Umm, yes, don't know what I was thinking.
Thanks for correction. :-)

Regards
--
Irrwahn Grausewitz (ir*******@free net.de)
welcome to clc : http://www.ungerhu.com/jxh/clc.welcome.txt
clc faq-list : http://www.faqs.org/faqs/C-faq/faq/
acllc-c++ faq : http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[David Rubin]

Martin O'Brien wrote:

/* hex2int.c: Convert hexadecimal number in any locale */
/* Source: THE STANDARD C LIBRARY by P J Plauger, p34 */
/* NOTE: This code does not check for overflow. That requires */
/* additional complexity. */

#include <ctype.h>
#include <string.h>

int hex2int(const char *s)
{
int value;
static const char xd[] =
{"0123456789abc defABCDEF"};

static const char xv[] =
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15,
10, 11, 12, 13, 14, 15};

for (value = 0; isxdigit(*s); ++s)
value = (value << 4) + xv[strchr(xd, *s) - xd];

return value;

I believe htoi() should return an unsigned int (perhaps unsigned long).
AFAIK, you cannot portably represent signed values in hex. In any case,
Plauger's code exhibits undefined behavior, so this is definitely not
the implementation you want to use.

/david

--
Andre, a simple peasant, had only one thing on his mind as he crept
along the East wall: 'Andre, creep... Andre, creep... Andre, creep.'
-- unknown

[Martin O'Brien]

David wrote:

I believe htoi() should return an
unsigned int (perhaps unsigned long). AFAIK,
you cannot portably represent signed values in hex.
In any case, Plauger's code exhibits undefined
behavior, so this is definitely not the
implementation you want to use.

The mistake is mine. I wrapped Plauger's code in a function definition
and wrongly put the return type as int.

Plauger says this:

---
To convert a hexadecimal number in any locale, write:

#include <ctype.h>
#include <string.h>
....
static const char xd[] =
{"0123456789abc defABCDEF"};

static const char xv[] =
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15,
10, 11, 12, 13, 14, 15};

for (value = 0; isxdigit(*s); ++s)
value = (value << 4) + xv[strchr(xd, *s) - xd];

Note that this code does not check for overflow. That requires
additional complexity.
---

Martin
http://martinobrien.co.uk/

[CBFalconer]

Martin O'Brien wrote:

David wrote:

I believe htoi() should return an
unsigned int (perhaps unsigned long). AFAIK,
you cannot portably represent signed values in hex.
In any case, Plauger's code exhibits undefined
behavior, so this is definitely not the
implementation you want to use.

The mistake is mine. I wrapped Plauger's code in a function
definition and wrongly put the return type as int.

Plauger says this:

---
To convert a hexadecimal number in any locale, write:

#include <ctype.h>
#include <string.h>
...
static const char xd[] =
{"0123456789abc defABCDEF"};

static const char xv[] =
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15,
10, 11, 12, 13, 14, 15};

for (value = 0; isxdigit(*s); ++s)
value = (value << 4) + xv[strchr(xd, *s) - xd];

Note that this code does not check for overflow. That requires
additional complexity.

It still exhibits undefined behaviour. You need to define value.
Also s, although we can make some assumptions about that.

--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!

[Martin]

"CBFalconer " wrote:

It still exhibits undefined behaviour. You need to define value.
Also s, although we can make some assumptions about that.

I'm puzzled. How can it exhibit anything if it doesn't compile?

Martin

[RoRsOaMrPiEo]

On 19 Feb 2004 00:33:08 -0800, ma************@ which.net (Martin
O'Brien) wrote:

David wrote:

I believe htoi() should return an
unsigned int (perhaps unsigned long). AFAIK,
you cannot portably represent signed values in hex.
In any case, Plauger's code exhibits undefined
behavior, so this is definitely not the
implementation you want to use.

The mistake is mine. I wrapped Plauger's code in a function definition
and wrongly put the return type as int.

Plauger says this:

---
To convert a hexadecimal number in any locale, write:

#include <ctype.h>
#include <string.h>
...
static const char xd[] =
{"0123456789abc defABCDEF"};

static const char xv[] =
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15,
10, 11, 12, 13, 14, 15};

for (value = 0; isxdigit(*s); ++s)
value = (value << 4) + xv[strchr(xd, *s) - xd];

Note that this code does not check for overflow. That requires
additional complexity.
---

Martin
http://martinobrien.co.uk/

htoi1 "check for overflow" ? Is it right?
How checking for overflow?

#include <stdio.h>
int lower(int c)
{if(c>='A' && c<='Z') return c+'a'-'A';
else return c;
}

int isdigit(int c)
{if(c>='0' && c<='9') return 1; else return 0;}

int ispace(int c)
{ if(c==' ' || c=='\t' || c=='\n') return 1; else return 0;}
int htoi1(char s[])
{unsigned i=0, n=0, d, h;
char c;

while( (c=s[i])==' ' || c=='\t' || c=='\n' )
i++;
if( c=='0' && ( s[i+1]=='x' || s[i+1]=='X') )
{i+=2; c=s[i];}
while(1)
{d= ('0'<=c && c<='9' ) ? c - '0' :
( c>='A' && c<='F' ) ? c - 'A' + 10:
( c>='a' && c<='f' ) ? c - 'a' + 10: 16;
if(d==16)
break;
if( (h=16 * n) < n || (n=h+d)<h || n<d )
{n= ((int)n)>0 ? ((~0u)>>1) : ((~0u)>>1)+1; break;}
c=s[++i];
}
return (int) n;
}
int htoi(char s[])
{int i=0,n=0;
while(ispace(s[i])) i++;
if(s[i]=='0'&& lower(s[i+1])=='x') i+=2;
for( ; ;++i)
if( isdigit(s[i]) ) n=16*n + (s[i]-'0');
else if((lower(s[i])>='a' && lower(s[i])<='f'))
n=16*n+lower(s[i])-'a'+ 10;
else break;
return n;
}

main()
{char b[30];
int c, ch, j;
do{printf("\nLi sta di numeri esadecinmali(1 termina)>");
do{scanf("%s", &b);
c=htoi1(b);
printf("\nParte nza=%s\n",b);
printf("Arrivo =%d\n", c);
c=htoi(b);
printf("\nParte nza=%s\n",b);
printf("Arrivo =%d\n", c);
if((ch=fgetc(st din) )=='\n') break;
else ungetc(ch,stdin );
} while(ch!=EOF );
}while(c!=1);
return 0;
}

[Martin]

Hello RoRsOaMrPiEo.

Your code is very busy. Why are you writing functions lower(), isdigit() and
isspace() when the Standard Library has them? (in the C Library lower() is
called tolower()).

And this thread should have convinced you that although you can assume the
characters '0' to '9' have consecutive values, you can not make that
assumption about upper and lower case letters. Which is why Plauger uses a
table of values xv, to define the values of the characters in xd.

Also, Chuck has already said that my function wrapper around Plauger's
suggestion was faulty because it returned int, which was an oversight on my
part, and your function now does the same, and even casts the results of its
calculations (in an unsigned int) to int before returning it.

I think what Plauger means about additional complexity for overflow is to
flag a warning if overflow is indicated and stop the conversion. It's what
I'd do anyway.

--
Martin
http://martinobrien.co.uk/
"RoRsOaMrPi Eo" wrote:

htoi1 "check for overflow" ? Is it right?
How checking for overflow?

#include <stdio.h> [...]
} while(ch!=EOF ); }while(c!=1);
return 0;
}

[Martin Dickopp]

"Martin" <martin.o_brien @[no-spam]which.net> writes:

"CBFalconer " wrote:

It still exhibits undefined behaviour. You need to define value.
Also s, although we can make some assumptions about that.

I'm puzzled. How can it exhibit anything if it doesn't compile?

"Undefined behavior" means that the standard poses no requirements on
how a conforming implementation behaves. If a compiler is part of the
implementation, not compiling the code at all is therefore just as
allowed as generating executable code.

(FWIW, there are cases of undefined behavior which only occurs at
run-time for some input data. An obvious example is any program which
calls the `gets' function. In such cases, the compiler must compile the
program.)
--
,--. Martin Dickopp, Dresden, Germany ,= ,-_-. =.
/ ,- ) http://www.zero-based.org/ ((_/)o o(\_))
\ `-' `-'(. .)`-'
`-. Debian, a variant of the GNU operating system. \_/

[CBFalconer]

RoRsOaMrPiEo wrote:

.... snip ...
#include <stdio.h>

int lower(int c)
{if(c>='A' && c<='Z') return c+'a'-'A';
else return c;
}

Already non-portable code, with foul formatting. I see at least
four blanks you could have saved.

--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!