Using sprintf with std::string in C++ - c++

I am using sprintf function in C++ 11, in the following way:
std::string toString()
{
std::string output;
uint32_t strSize=512;
do
{
output.reserve(strSize);
int ret = sprintf(output.c_str(), "Type=%u Version=%u ContentType=%u contentFormatVersion=%u magic=%04x Seg=%u",
INDEX_RECORD_TYPE_SERIALIZATION_HEADER,
FORAMT_VERSION,
contentType,
contentFormatVersion,
magic,
segmentId);
strSize *= 2;
} while (ret < 0);
return output;
}
Is there a better way to do this, than to check every time if the reserved space was enough? For future possibility of adding more things.

Your construct -- writing into the buffer received from c_str() -- is undefined behaviour, even if you checked the string's capacity beforehand. (The return value is a pointer to const char, and the function itself marked const, for a reason.)
Don't mix C and C++, especially not for writing into internal object representation. (That is breaking very basic OOP.) Use C++, for type safety and not running into conversion specifier / parameter mismatches, if for nothing else.
std::ostringstream s;
s << "Type=" << INDEX_RECORD_TYPE_SERIALIZATION_HEADER
<< " Version=" << FORMAT_VERSION
// ...and so on...
;
std::string output = s.str();
Alternative:
std::string output = "Type=" + std::to_string( INDEX_RECORD_TYPE_SERIALIZATION_HEADER )
+ " Version=" + std::to_string( FORMAT_VERSION )
// ...and so on...
;

The C++ patterns shown in other answers are nicer, but for completeness, here is a correct way with sprintf:
auto format = "your %x format %d string %s";
auto size = std::snprintf(nullptr, 0, format /* Arguments go here*/);
std::string output(size + 1, '\0');
std::sprintf(&output[0], format, /* Arguments go here*/);
Pay attention to
You must resize your string. reserve does not change the size of the buffer. In my example, I construct correctly sized string directly.
c_str() returns a const char*. You may not pass it to sprintf.
std::string buffer was not guaranteed to be contiguous prior to C++11 and this relies on that guarantee. If you need to support exotic pre-C++11 conforming platforms that use rope implementation for std::string, then you're probably better off sprinting into std::vector<char> first and then copying the vector to the string.
This only works if the arguments are not modified between the size calculation and formatting; use either local copies of variables or thread synchronisation primitives for multi-threaded code.

We can mix code from here https://stackoverflow.com/a/36909699/2667451 and here https://stackoverflow.com/a/7257307 and result will be like that:
template <typename ...Args>
std::string stringWithFormat(const std::string& format, Args && ...args)
{
auto size = std::snprintf(nullptr, 0, format.c_str(), std::forward<Args>(args)...);
std::string output(size + 1, '\0');
std::sprintf(&output[0], format.c_str(), std::forward<Args>(args)...);
return output;
}

A better way is to use the {fmt} library. Ex:
std::string message = fmt::sprintf("The answer is %d", 42);
It exposes also a nicer interface than iostreams and printf. Ex:
std::string message = fmt::format("The answer is {}", 42);`
See:
https://github.com/fmtlib/fmt
http://fmtlib.net/latest/api.html#printf-formatting-functions

Your code is wrong. reserve allocates memory for the string, but does not change its size. Writing into the buffer returned by c_str does not change its size either. So the string still believes its size is 0, and you've just written something into the unused space in the string's buffer. (Probably. Technically, the code has Undefined Behaviour, because writing into c_str is undefined, so anything could happen).
What you really want to do is forget sprintf and similar C-style functions, and use the C++ way of string formatting—string streams:
std::ostringstream ss;
ss << "Type=" << INDEX_RECORD_TYPE_SERIALIZATION_HEADER
<< " Version=" << FORAMT_VERSION
<< /* ... the rest ... */;
return ss.str();

Yes, there is!
In C, the better way is to associate a file with the null device and make a dummy printf of the desired output to it, to learn how much space would it take if actually printed. Then allocate appropriate buffer and sprintf the same data to it.
In C++ you could associate the output stream with a null device, too, and test the number of charactes printed with std::ostream::tellp. However, using ostringstream is a way better solution – see the answers by DevSolar or Angew.

You can use an implementation of sprintf() into a std::string I wrote that uses vsnprintf() under the hood.
It splits the format string into sections of plain text which are just copied to the destination std::string and sections of format fields (such as %5.2lf) which are first vsnprintf()ed into a buffer and then appended to the destination.
https://gitlab.com/eltomito/bodacious-sprintf

Related

Passing a .war file using a http POST request not working [duplicate]

If I want to construct a std::string with a line like:
std::string my_string("a\0b");
Where i want to have three characters in the resulting string (a, null, b), I only get one. What is the proper syntax?
Since C++14
we have been able to create literal std::string
#include <iostream>
#include <string>
int main()
{
using namespace std::string_literals;
std::string s = "pl-\0-op"s; // <- Notice the "s" at the end
// This is a std::string literal not
// a C-String literal.
std::cout << s << "\n";
}
Before C++14
The problem is the std::string constructor that takes a const char* assumes the input is a C-string. C-strings are \0 terminated and thus parsing stops when it reaches the \0 character.
To compensate for this, you need to use the constructor that builds the string from a char array (not a C-String). This takes two parameters - a pointer to the array and a length:
std::string x("pq\0rs"); // Two characters because input assumed to be C-String
std::string x("pq\0rs",5); // 5 Characters as the input is now a char array with 5 characters.
Note: C++ std::string is NOT \0-terminated (as suggested in other posts). However, you can extract a pointer to an internal buffer that contains a C-String with the method c_str().
Also check out Doug T's answer below about using a vector<char>.
Also check out RiaD for a C++14 solution.
If you are doing manipulation like you would with a c-style string (array of chars) consider using
std::vector<char>
You have more freedom to treat it like an array in the same manner you would treat a c-string. You can use copy() to copy into a string:
std::vector<char> vec(100)
strncpy(&vec[0], "blah blah blah", 100);
std::string vecAsStr( vec.begin(), vec.end());
and you can use it in many of the same places you can use c-strings
printf("%s" &vec[0])
vec[10] = '\0';
vec[11] = 'b';
Naturally, however, you suffer from the same problems as c-strings. You may forget your null terminal or write past the allocated space.
I have no idea why you'd want to do such a thing, but try this:
std::string my_string("a\0b", 3);
What new capabilities do user-defined literals add to C++? presents an elegant answer: Define
std::string operator "" _s(const char* str, size_t n)
{
return std::string(str, n);
}
then you can create your string this way:
std::string my_string("a\0b"_s);
or even so:
auto my_string = "a\0b"_s;
There's an "old style" way:
#define S(s) s, sizeof s - 1 // trailing NUL does not belong to the string
then you can define
std::string my_string(S("a\0b"));
The following will work...
std::string s;
s.push_back('a');
s.push_back('\0');
s.push_back('b');
You'll have to be careful with this. If you replace 'b' with any numeric character, you will silently create the wrong string using most methods. See: Rules for C++ string literals escape character.
For example, I dropped this innocent looking snippet in the middle of a program
// Create '\0' followed by '0' 40 times ;)
std::string str("\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00", 80);
std::cerr << "Entering loop.\n";
for (char & c : str) {
std::cerr << c;
// 'Q' is way cooler than '\0' or '0'
c = 'Q';
}
std::cerr << "\n";
for (char & c : str) {
std::cerr << c;
}
std::cerr << "\n";
Here is what this program output for me:
Entering loop.
Entering loop.
vector::_M_emplace_ba
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ
That was my first print statement twice, several non-printing characters, followed by a newline, followed by something in internal memory, which I just overwrote (and then printed, showing that it has been overwritten). Worst of all, even compiling this with thorough and verbose gcc warnings gave me no indication of something being wrong, and running the program through valgrind didn't complain about any improper memory access patterns. In other words, it's completely undetectable by modern tools.
You can get this same problem with the much simpler std::string("0", 100);, but the example above is a little trickier, and thus harder to see what's wrong.
Fortunately, C++11 gives us a good solution to the problem using initializer list syntax. This saves you from having to specify the number of characters (which, as I showed above, you can do incorrectly), and avoids combining escaped numbers. std::string str({'a', '\0', 'b'}) is safe for any string content, unlike versions that take an array of char and a size.
In C++14 you now may use literals
using namespace std::literals::string_literals;
std::string s = "a\0b"s;
std::cout << s.size(); // 3
Better to use std::vector<char> if this question isn't just for educational purposes.
anonym's answer is excellent, but there's a non-macro solution in C++98 as well:
template <size_t N>
std::string RawString(const char (&ch)[N])
{
return std::string(ch, N-1); // Again, exclude trailing `null`
}
With this function, RawString(/* literal */) will produce the same string as S(/* literal */):
std::string my_string_t(RawString("a\0b"));
std::string my_string_m(S("a\0b"));
std::cout << "Using template: " << my_string_t << std::endl;
std::cout << "Using macro: " << my_string_m << std::endl;
Additionally, there's an issue with the macro: the expression is not actually a std::string as written, and therefore can't be used e.g. for simple assignment-initialization:
std::string s = S("a\0b"); // ERROR!
...so it might be preferable to use:
#define std::string(s, sizeof s - 1)
Obviously you should only use one or the other solution in your project and call it whatever you think is appropriate.
I know it is a long time this question has been asked. But for anyone who is having a similar problem might be interested in the following code.
CComBSTR(20,"mystring1\0mystring2\0")
Almost all implementations of std::strings are null-terminated, so you probably shouldn't do this. Note that "a\0b" is actually four characters long because of the automatic null terminator (a, null, b, null). If you really want to do this and break std::string's contract, you can do:
std::string s("aab");
s.at(1) = '\0';
but if you do, all your friends will laugh at you, you will never find true happiness.

Creating binary (custom length) string in C++ [duplicate]

If I want to construct a std::string with a line like:
std::string my_string("a\0b");
Where i want to have three characters in the resulting string (a, null, b), I only get one. What is the proper syntax?
Since C++14
we have been able to create literal std::string
#include <iostream>
#include <string>
int main()
{
using namespace std::string_literals;
std::string s = "pl-\0-op"s; // <- Notice the "s" at the end
// This is a std::string literal not
// a C-String literal.
std::cout << s << "\n";
}
Before C++14
The problem is the std::string constructor that takes a const char* assumes the input is a C-string. C-strings are \0 terminated and thus parsing stops when it reaches the \0 character.
To compensate for this, you need to use the constructor that builds the string from a char array (not a C-String). This takes two parameters - a pointer to the array and a length:
std::string x("pq\0rs"); // Two characters because input assumed to be C-String
std::string x("pq\0rs",5); // 5 Characters as the input is now a char array with 5 characters.
Note: C++ std::string is NOT \0-terminated (as suggested in other posts). However, you can extract a pointer to an internal buffer that contains a C-String with the method c_str().
Also check out Doug T's answer below about using a vector<char>.
Also check out RiaD for a C++14 solution.
If you are doing manipulation like you would with a c-style string (array of chars) consider using
std::vector<char>
You have more freedom to treat it like an array in the same manner you would treat a c-string. You can use copy() to copy into a string:
std::vector<char> vec(100)
strncpy(&vec[0], "blah blah blah", 100);
std::string vecAsStr( vec.begin(), vec.end());
and you can use it in many of the same places you can use c-strings
printf("%s" &vec[0])
vec[10] = '\0';
vec[11] = 'b';
Naturally, however, you suffer from the same problems as c-strings. You may forget your null terminal or write past the allocated space.
I have no idea why you'd want to do such a thing, but try this:
std::string my_string("a\0b", 3);
What new capabilities do user-defined literals add to C++? presents an elegant answer: Define
std::string operator "" _s(const char* str, size_t n)
{
return std::string(str, n);
}
then you can create your string this way:
std::string my_string("a\0b"_s);
or even so:
auto my_string = "a\0b"_s;
There's an "old style" way:
#define S(s) s, sizeof s - 1 // trailing NUL does not belong to the string
then you can define
std::string my_string(S("a\0b"));
The following will work...
std::string s;
s.push_back('a');
s.push_back('\0');
s.push_back('b');
You'll have to be careful with this. If you replace 'b' with any numeric character, you will silently create the wrong string using most methods. See: Rules for C++ string literals escape character.
For example, I dropped this innocent looking snippet in the middle of a program
// Create '\0' followed by '0' 40 times ;)
std::string str("\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00", 80);
std::cerr << "Entering loop.\n";
for (char & c : str) {
std::cerr << c;
// 'Q' is way cooler than '\0' or '0'
c = 'Q';
}
std::cerr << "\n";
for (char & c : str) {
std::cerr << c;
}
std::cerr << "\n";
Here is what this program output for me:
Entering loop.
Entering loop.
vector::_M_emplace_ba
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ
That was my first print statement twice, several non-printing characters, followed by a newline, followed by something in internal memory, which I just overwrote (and then printed, showing that it has been overwritten). Worst of all, even compiling this with thorough and verbose gcc warnings gave me no indication of something being wrong, and running the program through valgrind didn't complain about any improper memory access patterns. In other words, it's completely undetectable by modern tools.
You can get this same problem with the much simpler std::string("0", 100);, but the example above is a little trickier, and thus harder to see what's wrong.
Fortunately, C++11 gives us a good solution to the problem using initializer list syntax. This saves you from having to specify the number of characters (which, as I showed above, you can do incorrectly), and avoids combining escaped numbers. std::string str({'a', '\0', 'b'}) is safe for any string content, unlike versions that take an array of char and a size.
In C++14 you now may use literals
using namespace std::literals::string_literals;
std::string s = "a\0b"s;
std::cout << s.size(); // 3
Better to use std::vector<char> if this question isn't just for educational purposes.
anonym's answer is excellent, but there's a non-macro solution in C++98 as well:
template <size_t N>
std::string RawString(const char (&ch)[N])
{
return std::string(ch, N-1); // Again, exclude trailing `null`
}
With this function, RawString(/* literal */) will produce the same string as S(/* literal */):
std::string my_string_t(RawString("a\0b"));
std::string my_string_m(S("a\0b"));
std::cout << "Using template: " << my_string_t << std::endl;
std::cout << "Using macro: " << my_string_m << std::endl;
Additionally, there's an issue with the macro: the expression is not actually a std::string as written, and therefore can't be used e.g. for simple assignment-initialization:
std::string s = S("a\0b"); // ERROR!
...so it might be preferable to use:
#define std::string(s, sizeof s - 1)
Obviously you should only use one or the other solution in your project and call it whatever you think is appropriate.
I know it is a long time this question has been asked. But for anyone who is having a similar problem might be interested in the following code.
CComBSTR(20,"mystring1\0mystring2\0")
Almost all implementations of std::strings are null-terminated, so you probably shouldn't do this. Note that "a\0b" is actually four characters long because of the automatic null terminator (a, null, b, null). If you really want to do this and break std::string's contract, you can do:
std::string s("aab");
s.at(1) = '\0';
but if you do, all your friends will laugh at you, you will never find true happiness.

C++ Better Practice to Use of Streams and Buffers

I know a way to read from a stream and use it like below:
strstream s; // It can be another standard stream type
// ...
while (!s.eof())
{
char buf[MAX];
s.read(buf, sizeof (buf));
int count = s.gcount();
THIRD_PARTY_FUNCTION(buf, count);
// ...
}
but this code has an abusing point, It first copies data from the stream to buf and then passes buf to THIRD_PARTY_FUNCTION.
Is there any way to reform the code to something like below(I mean below code avoids an extra copy) ?
strstream s; // It can be another standard stream type
// ...
while (!s.eof())
{
char *buf = A_POINTER_TO_DATA_OF_STREAM(s);
int count = AVAIABLE_DATA_SIZE_OF_STREAM(s);
// Maybe it needs s.seekg(...) here
THIRD_PARTY_FUNCTION(buf, count);
// ...
}
Something like this might work for you.
char buffer[2000];
std::istream& s = getStreamReference();
s.rdbuf()->pubsetbuf(buffer, 2000);
while(s)
{
THIRD_PARTY_FUNCTION(buffer, s.rdbuf()->in_avail());
s.ignore(s.rdbuf()->in_avail());
// Not sure this may go into an infinite loop.
// Its late here so I have not tested it.
}
Note sure I care about the cost of copying a 2K buffer.
The profiling would have to show that this is a real hotspot that is causing a significant degrade in performance before I would look at making this kind of optimization. Readability is going to be my most important factor here 99% of the time.
You can convert a std::stringstream to a c-style string by first calling its member method str to get an std::string and then call the member function c_str of that to convert it to a c-style null-terminated char[].

Is it possible to use an std::string for read()?

Is it possible to use an std::string for read() ?
Example :
std::string data;
read(fd, data, 42);
Normaly, we have to use char* but is it possible to directly use a std::string ? (I prefer don't create a char* for store the result)
Thank's
Well, you'll need to create a char* somehow, since that's what the
function requires. (BTW: you are talking about the Posix function
read, aren't you, and not std::istream::read?) The problem isn't
the char*, it's what the char* points to (which I suspect is what
you actually meant).
The simplest and usual solution here would be to use a local array:
char buffer[43];
int len = read(fd, buffer, 42);
if ( len < 0 ) {
// read error...
} else if ( len == 0 ) {
// eof...
} else {
std::string data(buffer, len);
}
If you want to capture directly into an std::string, however, this is
possible (although not necessarily a good idea):
std::string data;
data.resize( 42 );
int len = read( fd, &data[0], data.size() );
// error handling as above...
data.resize( len ); // If no error...
This avoids the copy, but quite frankly... The copy is insignificant
compared to the time necessary for the actual read and for the
allocation of the memory in the string. This also has the (probably
negligible) disadvantage of the resulting string having an actual buffer
of 42 bytes (rounded up to whatever), rather than just the minimum
necessary for the characters actually read.
(And since people sometimes raise the issue, with regards to the
contiguity of the memory in std:;string: this was an issue ten or more
years ago. The original specifications for std::string were designed
expressedly to allow non-contiguous implementations, along the lines of
the then popular rope class. In practice, no implementor found this
to be useful, and people did start assuming contiguity. At which point,
the standards committee decided to align the standard with existing
practice, and require contiguity. So... no implementation has ever not
been contiguous, and no future implementation will forego contiguity,
given the requirements in C++11.)
No, you cannot and you should not. Usually, std::string implementations internally store other information such as the size of the allocated memory and the length of the actual string. C++ documentation explicitly states that modifying values returned by c_str() or data() results in undefined behaviour.
If the read function requires a char *, then no. You could use the address of the first element of a std::vector of char as long as it's been resized first. I don't think old (pre C++11) strings are guarenteed to have contiguous memory otherwise you could do something similar with the string.
No, but
std::string data;
cin >> data;
works just fine. If you really want the behaviour of read(2), then you need to allocate and manage your own buffer of chars.
Because read() is intended for raw data input, std::string is actually a bad choice, because std::string handles text. std::vector seems like the right choice to handle raw data.
Using std::getline from the strings library - see cplusplus.com - can read from an stream and write directly into a string object. Example (again ripped from cplusplus.com - 1st hit on google for getline):
int main () {
string str;
cout << "Please enter full name: ";
getline (cin,str);
cout << "Thank you, " << str << ".\n";
}
So will work when reading from stdin (cin) and from a file (ifstream).

How do you construct a std::string with an embedded null?

If I want to construct a std::string with a line like:
std::string my_string("a\0b");
Where i want to have three characters in the resulting string (a, null, b), I only get one. What is the proper syntax?
Since C++14
we have been able to create literal std::string
#include <iostream>
#include <string>
int main()
{
using namespace std::string_literals;
std::string s = "pl-\0-op"s; // <- Notice the "s" at the end
// This is a std::string literal not
// a C-String literal.
std::cout << s << "\n";
}
Before C++14
The problem is the std::string constructor that takes a const char* assumes the input is a C-string. C-strings are \0 terminated and thus parsing stops when it reaches the \0 character.
To compensate for this, you need to use the constructor that builds the string from a char array (not a C-String). This takes two parameters - a pointer to the array and a length:
std::string x("pq\0rs"); // Two characters because input assumed to be C-String
std::string x("pq\0rs",5); // 5 Characters as the input is now a char array with 5 characters.
Note: C++ std::string is NOT \0-terminated (as suggested in other posts). However, you can extract a pointer to an internal buffer that contains a C-String with the method c_str().
Also check out Doug T's answer below about using a vector<char>.
Also check out RiaD for a C++14 solution.
If you are doing manipulation like you would with a c-style string (array of chars) consider using
std::vector<char>
You have more freedom to treat it like an array in the same manner you would treat a c-string. You can use copy() to copy into a string:
std::vector<char> vec(100)
strncpy(&vec[0], "blah blah blah", 100);
std::string vecAsStr( vec.begin(), vec.end());
and you can use it in many of the same places you can use c-strings
printf("%s" &vec[0])
vec[10] = '\0';
vec[11] = 'b';
Naturally, however, you suffer from the same problems as c-strings. You may forget your null terminal or write past the allocated space.
I have no idea why you'd want to do such a thing, but try this:
std::string my_string("a\0b", 3);
What new capabilities do user-defined literals add to C++? presents an elegant answer: Define
std::string operator "" _s(const char* str, size_t n)
{
return std::string(str, n);
}
then you can create your string this way:
std::string my_string("a\0b"_s);
or even so:
auto my_string = "a\0b"_s;
There's an "old style" way:
#define S(s) s, sizeof s - 1 // trailing NUL does not belong to the string
then you can define
std::string my_string(S("a\0b"));
The following will work...
std::string s;
s.push_back('a');
s.push_back('\0');
s.push_back('b');
You'll have to be careful with this. If you replace 'b' with any numeric character, you will silently create the wrong string using most methods. See: Rules for C++ string literals escape character.
For example, I dropped this innocent looking snippet in the middle of a program
// Create '\0' followed by '0' 40 times ;)
std::string str("\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00", 80);
std::cerr << "Entering loop.\n";
for (char & c : str) {
std::cerr << c;
// 'Q' is way cooler than '\0' or '0'
c = 'Q';
}
std::cerr << "\n";
for (char & c : str) {
std::cerr << c;
}
std::cerr << "\n";
Here is what this program output for me:
Entering loop.
Entering loop.
vector::_M_emplace_ba
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ
That was my first print statement twice, several non-printing characters, followed by a newline, followed by something in internal memory, which I just overwrote (and then printed, showing that it has been overwritten). Worst of all, even compiling this with thorough and verbose gcc warnings gave me no indication of something being wrong, and running the program through valgrind didn't complain about any improper memory access patterns. In other words, it's completely undetectable by modern tools.
You can get this same problem with the much simpler std::string("0", 100);, but the example above is a little trickier, and thus harder to see what's wrong.
Fortunately, C++11 gives us a good solution to the problem using initializer list syntax. This saves you from having to specify the number of characters (which, as I showed above, you can do incorrectly), and avoids combining escaped numbers. std::string str({'a', '\0', 'b'}) is safe for any string content, unlike versions that take an array of char and a size.
In C++14 you now may use literals
using namespace std::literals::string_literals;
std::string s = "a\0b"s;
std::cout << s.size(); // 3
Better to use std::vector<char> if this question isn't just for educational purposes.
anonym's answer is excellent, but there's a non-macro solution in C++98 as well:
template <size_t N>
std::string RawString(const char (&ch)[N])
{
return std::string(ch, N-1); // Again, exclude trailing `null`
}
With this function, RawString(/* literal */) will produce the same string as S(/* literal */):
std::string my_string_t(RawString("a\0b"));
std::string my_string_m(S("a\0b"));
std::cout << "Using template: " << my_string_t << std::endl;
std::cout << "Using macro: " << my_string_m << std::endl;
Additionally, there's an issue with the macro: the expression is not actually a std::string as written, and therefore can't be used e.g. for simple assignment-initialization:
std::string s = S("a\0b"); // ERROR!
...so it might be preferable to use:
#define std::string(s, sizeof s - 1)
Obviously you should only use one or the other solution in your project and call it whatever you think is appropriate.
I know it is a long time this question has been asked. But for anyone who is having a similar problem might be interested in the following code.
CComBSTR(20,"mystring1\0mystring2\0")
Almost all implementations of std::strings are null-terminated, so you probably shouldn't do this. Note that "a\0b" is actually four characters long because of the automatic null terminator (a, null, b, null). If you really want to do this and break std::string's contract, you can do:
std::string s("aab");
s.at(1) = '\0';
but if you do, all your friends will laugh at you, you will never find true happiness.