Hello,
I guess this might be a very dummy question, but I couldn't find an
answer in the group's archive.
I am reading DICOM files, and to store the string read I use
std::string. Unfortunately there is case where the string can be '\0\0'
Is this a defined bahavior when you do this kind of operation :
const char *s = "\0\0";
std::string a = s; //how many character will be copied
Is there a work around other than subclassing std::string to handle
properly \0 ? Knowing that I can carry the 'real' lenght around, my main
concern is how do I fill a std::string with x number of \0 ?
Thanks
Mathieu 3 3801
"Mathieu Malaterre" <mm**********@Mnycap.rr.com> wrote in message
news:%F*********************@twister.nyroc.rr.com. .. Hello,
I guess this might be a very dummy question, but I couldn't find an answer in the group's archive.
I am reading DICOM files, and to store the string read I use std::string. Unfortunately there is case where the string can be '\0\0' Is this a defined bahavior when you do this kind of operation :
const char *s = "\0\0";
NB: This will actually point to 3 times '\0'. std::string a = s; //how many character will be copied
None.
Is there a work around other than subclassing std::string to handle properly \0 ? Knowing that I can carry the 'real' lenght around, my main concern is how do I fill a std::string with x number of \0 ?
Use the std::string constructor that takes an iterator (or pointer)
range:
const char *s = "\0\0";
std::string a(s,s+2); // will copy 2 '\0'
Note that a.size() will return 2, but if you call a.c_str() instead
of a.data(), you will be guaranteed to have a third '\0' included
at the end of the returned buffer, to match C string conventions.
hth,
Ivan
-- http://ivan.vecerina.com/contact/?subject=NG_POST <- email contact form
Brainbench MVP for C++ <> http://www.brainbench.com
Ivan Vecerina wrote: "Mathieu Malaterre" <mm**********@Mnycap.rr.com> wrote in message news:%F*********************@twister.nyroc.rr.com. ..
Hello,
I guess this might be a very dummy question, but I couldn't find an answer in the group's archive.
I am reading DICOM files, and to store the string read I use std::string. Unfortunately there is case where the string can be '\0\0' Is this a defined bahavior when you do this kind of operation :
const char *s = "\0\0";
NB: This will actually point to 3 times '\0'.
std::string a = s; //how many character will be copied
None.
Is there a work around other than subclassing std::string to handle properly \0 ? Knowing that I can carry the 'real' lenght around, my main concern is how do I fill a std::string with x number of \0 ?
Use the std::string constructor that takes an iterator (or pointer) range: const char *s = "\0\0"; std::string a(s,s+2); // will copy 2 '\0'
Note that a.size() will return 2, but if you call a.c_str() instead of a.data(), you will be guaranteed to have a third '\0' included at the end of the returned buffer, to match C string conventions.
You rocks, I was -sadly- thinking I should use a vector<char>
But your advice look a lot more easy to implement/patch in my current
project.
Thanks
Mathieu
"Mathieu Malaterre" <mm**********@Mnycap.rr.com> wrote in message
news:%F*********************@twister.nyroc.rr.com. .. Hello,
I guess this might be a very dummy question, but I couldn't find an answer in the group's archive.
I am reading DICOM files, and to store the string read I use std::string. Unfortunately there is case where the string can be '\0\0' Is this a defined bahavior when you do this kind of operation :
Yes. const char *s = "\0\0"; std::string a = s; //how many character will be copied
Zero. I.e. 'a.size()' will return zero, and 'a.empty()'
will return 'true'. Any characters following the first
zero-value character at or after the address contained by
's' are not part of the 'C string' to which it points. Is there a work around
What do you want to 'work around'?
other than subclassing std::string to handle properly \0 ?
'std::string' already handles it properly.
Knowing that I can carry the 'real' lenght around,
Then length of a 'C-style' string is the sum of the number
of characters from its beginning up to the one immediately
preceding the first with a value of zero. This length
does not include the zero-valued character.
char s1[] = "\0";
char s2[] = "\0\0";
char s3[] = "\0\0\0";
char s4[] = "\0abc";
Are all strings with length of zero. 'strlen()'
will report the same.
my main concern is how do I fill a std::string with x number of \0 ?
Use a different constructor:
std::string zeros(x, 0);
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