Vlearns
asked on
extracting a sub section of the string
i have a const char* p whose value is printed as below:
(gdb) p p
$5 = 0x1394510 "(HEADER MESSAGE-ID <8d8vhsk5qqpfl7lyetqsbdsb.
where
p is defined and initialized as
const char* p = str.c_str();
i need to extract 8d8vhsk5qqpfl7lyetqsbdsb.1
how do i do that in the best efficient way?
(gdb) p p
$5 = 0x1394510 "(HEADER MESSAGE-ID <8d8vhsk5qqpfl7lyetqsbdsb.
where
p is defined and initialized as
const char* p = str.c_str();
i need to extract 8d8vhsk5qqpfl7lyetqsbdsb.1
how do i do that in the best efficient way?
SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
Looking at that, I would rather copy str into a new string to avoid any const literal string issues. I'll revise this.
I am not sure why the above version works on cygwin. I would have thought that the constant literal string would not allow modification by strtok_r. So, I copied the original string into a new area.
string str = "(HEADER MESSAGE-ID <8d8vhsk5qqpfl7lyetqsbdsb.1264020499132@email.android.com>)";
char* p = (char *)malloc( str.size() + 1 );
string result;
char * ptr;
strncpy( p, str.c_str(), str.size() + 1 );
p = strtok_r(p, "<", &ptr);
result = strtok_r(NULL, ">", &ptr);
free(p);
cout << result << endl;
When using strtok_r, you should check to see if the returned value is NULL. If NULL, then the delimiter was not found, and you stop processing the string.
If you do not define p as you said, then I think this is a cleaner version:
string str = "(HEADER MESSAGE-ID <8d8vhsk5qqpfl7lyetqsbdsb.1264020499132@email.android.com>)";
string result;
int bpos, epos;
if( ( (bpos = str.find("<") )!= string::npos ) &&
( (epos = str.find(">") )!= string::npos ) &&
(bpos < epos) ) {
result = str.substr( bpos+1, epos - bpos -1 );
}
else {
cout << "<> pair not found" << endl;
}
cout << "result = " << result << endl;
phoffric,
You can simplify your last code sample by starting the second find from bpos, meaning you won't need the bpos < epos check.
-Rx.
You can simplify your last code sample by starting the second find from bpos, meaning you won't need the bpos < epos check.
-Rx.
good point :)
ASKER CERTIFIED SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.