formating an stdin input

[momotaro]

the idea that am trying to impliment is to read an ip adress intred by the user in the format 'xxx.yyy.zzz.mm m' , detect all possible errors then cut it into xxx, yyy, zzz, mmm, each one must be put in a member of a structure.
am confused i could play with getc, getchar...
but plz if u have a better idea and may be easier plz tell me

[JosAH]

the idea that am trying to impliment is to read an ip adress intred by the user in the format 'xxx.yyy.zzz.mm m' , detect all possible errors then cut it into xxx, yyy, zzz, mmm, each one must be put in a member of a structure.
am confused i could play with getc, getchar...
but plz if u have a better idea and may be easier plz tell me

For a bit of thorough checking read your string (a potential IP/4 address) and:

1) check if the string length == 15
2) at positions 3, 7 and 11 should be dots.
3) at all other positions should be digits.
4) the three numbers should be in the range [0,255]

Making an int number out of three consecutive character positions containing
the ASCII representation of digits is easy.

scanf("%d" ...) is a nono here because it accepts leading spaces which you
don't want. The function 'isdigit()' in file ctype.h is your friend.

kind regards,

Jos

[momotaro]

but the string is not alway 15!
for example: 34.54.234.1 is a valide IP adress yet it's not a 15 string lenght!

[momotaro]

am thinking about a function that keps reading my string findind the positions of the dot's then returning them what do you think.?

[JosAH]

but the string is not alway 15!
for example: 34.54.234.1 is a valide IP adress yet it's not a 15 string lenght!

Ah, ok, the rules have to be a bit more relaxed then; are you allowed to use
regular expression? as in \d+\.\d+\.\d+\. \d+
That'd be the Java way of doing it; the Boost library has a regular expression
parser and matcher too.

If you can't use those follow that pattern 'manually':

1) read one or more digits
2) read a dot.
3) read one or more digits
4) repeat steps 2) and 3) three times.

If any of the steps fail or if the number read wasn't in the [0,255] interval,
the pattern wasn't a valid IP/4 address.

kind regards,

Jos

[momotaro]

no we can't but the dot idea solve 90% of the problem still left with how to turn the red digits into integers!

[momotaro]

could I use this function:
[CODE=c]
int atoi(const char *s);
[/CODE]

[JosAH]

could I use this function:
[CODE=c]
int atoi(const char *s);
[/CODE]

I think you could if you take a few precautions:

1) the atoi function stops when it read a non-digit; you have to find where that non-digit is;
2) atoi() doesn't care about the range of the int; you have to take care of that yourself.

kind regards,

Jos

[momotaro]

the have been modified(*)

pre: - array pos[3] containing the dots' position in the IP adress
- array IP[15] containing the IP string
post:- return 4 arrays xx[3], yy[3], zz[3], tt[3] containing the IP/4 digits

problem(remaini ng from 1st version): - unexpected result while checking using printf
example: - instead having "xx[3] = 123" am having 123:)
- third part of the IP 123 and some bizard char instead of 123
questions: - please help(if there is any other comment...)

CODE:
[CODE=c]
for(i = 0; i < 4; i++)
{
if(i == 0)
{
for(j = i; j < pos[i]; j++)
{
if(isdigit(IP[j]))
xx[j] = IP[j];
}
printf("xxx part\n");
printf("%s\n", xx);
}
else if(i == 1)
{
for(j = pos[i-1]+1; j < pos[i]; j++)
{
if(isdigit(IP[j]))
{
yy[y] = IP[j];
y++; //****
}
}
printf("yyy part\n");
printf("%s\n", yy);
}
else if(i == 2)
{
for(j = pos[i-1]; j < pos[i]; j++)
{
if(isdigit(IP[j]))
{
zz[z] = IP[j];
z++; //****
}
}
printf("zzz part\n");
printf("%s\n", zz);
}
else if(i == 3)
{
for(j = pos[i-1]+1; j < 15; j++)
{
if(isdigit(IP[j]))
{
mm[m] = IP[j];
m++; // ******
}
}
printf("mmm part\n");
printf("%s\n", mm);
}
}
[/CODE]