std::basic_stringbuf::str
From cppreference.com
< cpp | io | basic stringbuf
std::basic_string<CharT, Traits, Allocator> str() const;
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void str( const std::basic_string<CharT, Traits, Allocator>& s);
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Gets and sets the underlying string.
1) Creates and returns a std::basic_string object containing a copy of this
std::basic_stringbuf
's underlying character sequence. For input-only streams, the returned string contains the characters from the range [eback(), egptr()). For input/output or output-only streams, contains the characters from pbase() to the last character in the sequence regardless of egptr() and epptr().
The member character sequence in a buffer open for writing can be over-allocated for efficiency purposes. In that case, only the initialized characters are returned: these characters are the ones that were obtained from the string argument of the constructor, the string argument of the most recent call to the second overload of str() , or from an write operation. Typical implementation that uses over-allocation maintains a high-watermark pointer to track the end of the initialized part of the buffer and this overload returns the characters from pbase() to the high-watermark pointer |
(since C++11) |
2) Deletes the entire underlying character sequence of this
std::basic_stringbuf
and then configures a new underlying character sequence containing a copy of the contents of s
. The pointers of std::basic_streambuf are initialized as follows:
- For input streams (mode & ios_base::in == true), eback() points at the first character, gptr() == eback(), and egptr() == eback() + s.size(): the subsequent input will read the first character copied from
s
. - For output streams (mode & ios_base::out == true), pbase() points at the first character and epptr() >= pbase() + s.size() (epptr is allowed to point farther so that the following
sputc()
wouldn't immediately calloverflow()
)- For append streams (mode & ios_base::ate == true), pptr() == pbase() + s.size(), so that subsequent output will be appended to the last character copied from
s
(since C++11) - For no-appending output streams, pptr() == pbase(), so that subsequent output will overwrite the characters copied from
s
.
- For append streams (mode & ios_base::ate == true), pptr() == pbase() + s.size(), so that subsequent output will be appended to the last character copied from
Contents |
[edit] Parameters
s | - | a string object holding the replacement character sequence |
[edit] Return value
1) A string object holding a copy of this buffer's underlying character sequence.
2) (none)
[edit] Notes
This function is typically accessed through std::basic_stringstream::str().
[edit] Example
Run this code
#include <sstream> #include <iostream> int main() { int n; std::istringstream in; // could also use in("1 2") in.rdbuf()->str("1 2"); // set the get area in >> n; std::cout << "after reading the first int from \"1 2\", the int is " << n << ", str() = \"" << in.rdbuf()->str() << "\"\n"; // or in.str() std::ostringstream out("1 2"); out << 3; std::cout << "after writing the int '3' to output stream \"1 2\"" << ", str() = \"" << out.str() << "\"\n"; std::ostringstream ate("1 2", std::ios_base::ate); // C++11 ate << 3; std::cout << "after writing the int '3' to append stream \"1 2\"" << ", str() = \"" << ate.str() << "\"\n"; }
Output:
after reading the first int from "1 2", the int is 1, str() = "1 2" after writing the int '3' to output stream "1 2", str() = "3 2" after writing the int '3' to append stream "1 2", str() = "1 23"
[edit] See also
gets or sets the contents of underlying string device object (public member function of std::basic_stringstream )
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