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..

[RoRsOaMrPiEo]

On Thu, 19 Feb 2004 21:12:59 -0000, "Martin"
<martin.o_brien @[no-spam]which.net> wrote:

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()).
ok, ok
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.
I assume ascii or ebcdic and a two's complement pc
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.
Do you show some numeric examples?
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.

[RoRsOaMrPiEo]

On Fri, 20 Feb 2004 07:05:54 GMT, RoRsOaMrPiEo <[e-mail]> wrote:

I assume that CHAR_MAX < 512

#include <stdio.h>
#include <assert.h>
#include <limits.h>

int htoi1(char s[])
{unsigned i=0, n=0, d, h;
char c;

assert(s!=NULL) ;
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;
}
char* volte1(void)
{static char xd[512] = {16};
xd['0']= 0; xd['1']= 1; xd['2']= 2; xd['3']= 3;
xd['4']= 4; xd['5']= 5; xd['6']= 6; xd['7']= 7;
xd['8']= 8; xd['9']= 9;
xd['a']=10; xd['b']=11; xd['c']=12; xd['d']=13;
xd['e']=14; xd['f']=15;
xd['A']=10; xd['B']=11; xd['C']=12; xd['D']=13;
xd['E']=14; xd['F']=15;
return xd;
}
int htoi2(char* s)
{unsigned i=0, n=0, d, h;
static char *p=0;
char c;

assert(s!=NULL) ;
if(p==0) p=volte1();

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= c>0 ? p[c]: 16;
if(d==16)
break;
if( n>(UINT_MAX/16) || ( h=(n*16) )> UINT_MAX - d )
{n= ((int)n)>0 ? INT_MAX : INT_MIN; break;}
n= h + d;
c=s[++i];
}
return (int) n;
}
int main(void)
{char b[30]={0};
int c, ch, j;

assert(CHAR_MAX <=512);
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=htoi2(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;
}

Do you like it?

[RoRsOaMrPiEo]

On Fri, 20 Feb 2004 08:44:50 GMT, RoRsOaMrPiEo <[e-mail]> wrote:

On Fri, 20 Feb 2004 07:05:54 GMT, RoRsOaMrPiEo <[e-mail]> wrote:

I assume that CHAR_MAX < 512

I assume '0'...'9','a'.. .'f','A'...'F' < 512
#include <stdio.h>
#include <assert.h>
#include <limits.h>

/* It is wrong ...*/
int htoi1(char s[])
{unsigned i=0, n=0, d, h;
char c;

assert(s!=NULL) ;
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;
}
char* volte1(void)
{static char xd[512] = {16};
xd['0']= 0; xd['1']= 1; xd['2']= 2; xd['3']= 3; xd['4']= 4;
xd['5']= 5;
xd['6']= 6; xd['7']= 7; xd['8']= 8; xd['9']= 9;
xd['a']=10; xd['b']=11; xd['c']=12; xd['d']=13; xd['e']=14;
xd['f']=15;
xd['A']=10; xd['B']=11; xd['C']=12; xd['D']=13; xd['E']=14;
xd['F']=15;
return xd;
}

int ci_siamo_o_no(v oid)
{if('9'>= 512 ||
'a'>= 512 || 'b'>= 512 || 'c'>= 512 || 'd'>= 512 || 'e'>= 512 ||
'f'>= 512 ||
'A'>= 512 || 'B'>= 512 || 'C'>= 512 || 'D'>= 512 || 'E'>= 512 ||
'F'>= 512
) return 0;
return 1;
}

int htoi2(char* s)
{unsigned i=0, n=0, d, h;
static char *p=0;
char c;

assert(s!=NULL) ;
if(p==0) p=volte1();

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= c>0 && (unsigned) c < 512u ? p[c]: 16;
if(d==16)
break;
if( n>(UINT_MAX/16) || ( h=(n*16) )> UINT_MAX - d )
{n= ((int)n)>0 ? INT_MAX : INT_MIN; break;}
n= h + d;
c=s[++i];
}
return (int) n;
}
int main(void)
{char b[30]={0};
int c, ch, j;

c=ci_siamo_o_no ();
assert(c==1);

do{printf("\nLi sta di numeri esadecinmali(1 termina)>");
do{scanf("%29s" , &b);
c=htoi1(b);
printf("\nParte nza=%s\n",b);
printf("Arrivo =%d\n", c);
c=htoi2(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;
}

[RoRsOaMrPiEo]

On Fri, 20 Feb 2004 01:41:33 GMT, CBFalconer <cb********@yah oo.com>
wrote:

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.

Do you say that you can make a "htoi()" better than my "htoi2()" ?

[Richard Bos]

RoRsOaMrPiEo <[e-mail]> wrote:

On Fri, 20 Feb 2004 01:41:33 GMT, CBFalconer <cb********@yah oo.com>
wrote:

RoRsOaMrPiEo wrote:

#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.

Do you say that you can make a "htoi()" better than my "htoi2()" ?

Easily. Why do work which the library will do for you?

#include <limits.h>
#include <stdlib.h>

unsigned int htoi(char *txt)
{
unsigned long tmp=strtoul(txt , 0, 16);
if (tmp>UINT_MAX) return UINT_MAX;
return tmp;
}

Shorter, sweeter, doesn't waste value space on negative numbers which
you don't parse anyway, doesn't invoke undefined behaviour on overflow,
and is portable no matter what your character set it. Feel free.

Richard

[RoRsOaMrPiEo]

On Fri, 20 Feb 2004 10:13:10 GMT, rl*@hoekstra-uitgeverij.nl (Richard
Bos) wrote:

RoRsOaMrPiEo <[e-mail]> wrote:

On Fri, 20 Feb 2004 01:41:33 GMT, CBFalconer <cb********@yah oo.com>
wrote:

>RoRsOaMrPiEo wrote:
>>
>> #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.
Do you say that you can make a "htoi()" better than my "htoi2()" ?

Easily. Why do work which the library will do for you?

#include <limits.h>
#include <stdlib.h>

unsigned int htoi(char *txt)
{
unsigned long tmp=strtoul(txt , 0, 16);
if (tmp>UINT_MAX) return UINT_MAX;
return tmp;
}

but here htoi(8eeeeeeeee eeeeeeeeeeeeee) ==-1

this seems htoi(8eeeeeeeee eeeeeeeeeeeeee) == INT_MAX

unsigned int htoi(char *txt)
{unsigned long tmp=strtoul(txt , 0, 16);
if (tmp>=UINT_MAX)
return INT_MAX;
return tmp;
}
Shorter, sweeter, doesn't waste value space on negative numbers which
you don't parse anyway, doesn't invoke undefined behaviour on overflow,
and is portable no matter what your character set it. Feel free.

Richard

but my htoi2 seems 2*faster than library htoi()

C:>2_7_c
test is ok
Differenza libreria =6.000000 s
Differenza mia =2.000000 s

C:>2_7_c
test is ok
Differenza libreria =6.000000 s
Differenza mia =3.000000 s

C:>2_7_c
test is ok
Differenza libreria =6.000000 s
Differenza mia =3.000000 s

C:>2_7_c
test is ok
Differenza libreria =6.000000 s
Differenza mia =3.000000 s

C:>2_7_c
test is ok
Differenza libreria =6.000000 s
Differenza mia =3.000000 s
#include <stdio.h>
#include <assert.h>
#include <limits.h>
#include <stdlib.h>
#include <time.h>

char* riempi_hex(char * a, int lunghezza)
{static char ary[]="0123456789abc defABCDEF";
int i, j, lungh;

assert( a!=NULL && lunghezza>13 );
lungh = rand() % 12;
for( j=0; j<lungh; ++j )
a[j]=ary[ rand() % 21 ];

a[j]=0;
return a;
}

unsigned int htoi(char *txt)
{unsigned long tmp=strtoul(txt , 0, 16);
if (tmp>=UINT_MAX)
return INT_MAX;
return tmp;
}

char* volte1(void)
{static char xd[512] = {16};
xd['0']= 0; xd['1']= 1; xd['2']= 2; xd['3']= 3; xd['4']= 4; xd['5']=
5;
xd['6']= 6; xd['7']= 7; xd['8']= 8; xd['9']= 9;
xd['a']=10; xd['b']=11; xd['c']=12; xd['d']=13; xd['e']=14;
xd['f']=15;
xd['A']=10; xd['B']=11; xd['C']=12; xd['D']=13; xd['E']=14;
xd['F']=15;
return xd;
}

int ci_siamo_o_no(v oid)
{if('9'>= 512 ||
'a'>= 512 || 'b'>= 512 || 'c'>= 512 || 'd'>= 512 || 'e'>= 512 ||
'f'>= 512 ||
'A'>= 512 || 'B'>= 512 || 'C'>= 512 || 'D'>= 512 || 'E'>= 512 ||
'F'>= 512
) return 0;
return 1;
}

int htoi2(char* s)
{unsigned i=0, n=0, d, h;
static char *p=0;
char c;

assert(s!=NULL) ;
if(p==0) p=volte1();

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= c>0 && (unsigned) c < 512u ? p[c]: 16;
if(d==16)
break;
if( n>(UINT_MAX/16) || ( h=(n*16) )> UINT_MAX - d )
{n= ((int)n)>0 ? INT_MAX : INT_MIN; break;}
n= h + d;
c=s[++i];
}
return (int) n;
}
int main(void)
{char b[30]={0};
int c, ch, j;
time_t x, y;

c=ci_siamo_o_no ();
assert(c==1);
srand(time(0));

for(j=0; j<700000; ++j)
{riempi_hex(b, 30);
c=htoi(b); ch=htoi2(b);
if(c!=ch && c-ch!=-1)
{printf("\nPart enza=%s\n",b);
printf("Arrivo =%d\n", c );
printf("Arrivo1 =%d\nTest fails \n", ch);
}
}

printf("test is ok \n");
x=time(0);
for(j=0; j<7000000; ++j)
{riempi_hex(b, 30);
c=htoi(b);
if(c==9999999)
{printf("\nPart enza=%s\n",b);
printf("Arrivo =%d\n", c );
}
}

y=time(0);
printf("Differe nza libreria =%f s \n", difftime(y,x) );

x=time(0);
for(j=0; j<7000000; ++j)
{riempi_hex(b, 30);
c=htoi2(b);
if(c==9999999)
{printf("\nPart enza=%s\n",b);
printf("Arrivo =%d\n", c );
}
}
y=time(0);
printf("Differe nza mia =%f s \n", difftime(y,x) );

return 0;
}

[Richard Bos]

RoRsOaMrPiEo <[e-mail]> wrote:

On Fri, 20 Feb 2004 10:13:10 GMT, rl*@hoekstra-uitgeverij.nl (Richard
Bos) wrote:

RoRsOaMrPiEo <[e-mail]> wrote:

Do you say that you can make a "htoi()" better than my "htoi2()" ?

Easily. Why do work which the library will do for you?

#include <limits.h>
#include <stdlib.h>

unsigned int htoi(char *txt)
{
unsigned long tmp=strtoul(txt , 0, 16);
if (tmp>UINT_MAX) return UINT_MAX;
return tmp;
}

but here htoi(8eeeeeeeee eeeeeeeeeeeeee) ==-1

this seems htoi(8eeeeeeeee eeeeeeeeeeeeee) == INT_MAX

Imprimis, htoi(8eeeeeeeee eeeeeeeeee) is a syntax error.
Secundis, htoi("8eeeeeeee eeeeeeeeeeee") invokes undefined behaviour for
your implementation (and thus has a license to return anything,
including garbage, and even crash the machine), and returns UINT_MAX
(not INT_MAX) for my implementation, without being undefined.

Shorter, sweeter, doesn't waste value space on negative numbers which
you don't parse anyway, doesn't invoke undefined behaviour on overflow,
and is portable no matter what your character set it. Feel free.

but my htoi2 seems 2*faster than library htoi()

There is no library htoi() (not in Standard C, at least), and some
slight slowness, while not ideal, is at least to be preferred above a
license to crash your machine at any inconvenient moment.

Richard

[RoRsOaMrPiEo]

On Fri, 20 Feb 2004 10:18:29 GMT, RoRsOaMrPiEo <[e-mail]> wrote:

On Fri, 20 Feb 2004 01:41:33 GMT, CBFalconer <cb********@yah oo.com>
wrote:

RoRsOaMrPiE o 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.

Do you say that you can make a "htoi()" better than my "htoi2()" ?

Excuse me if I gave offence to you
It would be easy if we use pointer instead of index
but

int htoi3(char* s)
{unsigned n=0, d, h;
static char *p=0;
char c;

assert(s!=NULL) ;
if(p==0) p=volte1();

while( (c=*s)==' ' || c=='\t' || c=='\n' )
s++;
if( c=='0' && ( s[1]=='x' || s[1]=='X') )
{s += 2; c = *s;}
while(1)
{d= c>0 && (unsigned) c < 512u ? p[c]: 16;
if(d==16)
break;
if( n>(UINT_MAX/16) || ( h=(n*16) )> UINT_MAX - d )
{n= ((int)n)>0 ? INT_MAX : INT_MIN; break;}
n= h + d;
c=*++s;
}
return (int) n;
}

seems slower of htoi2 in my sys.

[RoRsOaMrPiEo]

On Fri, 20 Feb 2004 12:32:52 GMT, rl*@hoekstra-uitgeverij.nl (Richard
Bos) wrote:

> #include <limits.h>
> #include <stdlib.h>
>
> unsigned int htoi(char *txt)
> {
> unsigned long tmp=strtoul(txt , 0, 16);
> if (tmp>UINT_MAX) return UINT_MAX;
> return tmp;
> }
but here htoi(8eeeeeeeee eeeeeeeeeeeeee) ==-1

this seems htoi(8eeeeeeeee eeeeeeeeeeeeee) == INT_MAX

Imprimis, htoi(8eeeeeeeee eeeeeeeeee) is a syntax error.
Secundis, htoi("8eeeeeeee eeeeeeeeeeee") invokes undefined behaviour for
your implementation (and thus has a license to return anything,
including garbage, and even crash the machine), and returns UINT_MAX
(not INT_MAX) for my implementation, without being undefined.

so your is htou and not htoi

>Shorter, sweeter, doesn't waste value space on negative numbers which
>you don't parse anyway, doesn't invoke undefined behaviour on overflow,
>and is portable no matter what your character set it. Feel free.

but my htoi2 seems 2*faster than library htoi()

There is no library htoi() (not in Standard C, at least),

ok, htoi that use library (strtoul(txt, 0, 16))
and some
slight slowness, while not ideal, is at least to be preferred above a
license to crash your machine at any inconvenient moment.
crash where in code?
Richard

[Martin]

"Richard Bos" wrote:

Why do work which the library will do for you?

#include <limits.h>
#include <stdlib.h>

unsigned int htoi(char *txt)
{
unsigned long tmp=strtoul(txt , 0, 16);
if (tmp>UINT_MAX) return UINT_MAX;
return tmp;
}

Shorter, sweeter, doesn't waste value space on
negative numbers which you don't parse anyway,
doesn't invoke undefined behaviour on overflow,
and is portable no matter what your character set it.
Feel free.

Isn't there a possibility that UINT_MAX == ULONG_MAX?

Martin
http://martinobrien.co.uk/