CODEWITHSUNDEEP
c++initializationconcatenationstringstream
user1823664
user1823664

Reputation: 1095

Difference between initializations stringstream.str( a_value ) and stringstream << a_value

Consider:

std::string        s_a, s_b;

std::stringstream  ss_1, ss_2;

// at this stage:
//     ss_1 and ss_2 have been used and are now in some strange state
//     s_a and s_b contain non-white space words

ss_1.str( std::string() );
ss_1.clear();

ss_1 << s_a;
ss_1 << s_b;

// ss_1.str().c_str() is now the concatenation of s_a and s_b, 
//                    <strike>with</strike> without space between them

ss_2.str( s_a );
ss_2.clear();

// ss_2.str().c_str() is now s_a

ss_2 << s_b;  // line ***

// ss_2.str().c_str() the value of s_a is over-written by s_b 
// 
// Replacing line *** above with "ss_2 << ss_2.str() << " " << s_b;"
//                    results in ss_2 having the same content as ss_1.

Questions:

  1. What is the difference between stringstream.str( a_value ); and stringstream << a_value; and, specifically, why does the first not allow concatenation via << but the second does?

  2. Why did ss_1 automatically get white-space between s_a and s_b, but do we need to explicitly add white space in the line that could replace line ***: ss_2 << ss_2.str() << " " << s_b;?

Upvotes: 13

Views: 2747

Answers (2)

billz
billz

Reputation: 45470

Suggest you read stringstream reference: http://en.cppreference.com/w/cpp/io/basic_stringstream

std::stringstream::str() Replaces the contents of the underlying string

operator<< Inserts data into the stream.

Upvotes: 3

Steve Lorimer
Steve Lorimer

Reputation: 28699

The problem you're experiencing is because std::stringstream is constructed by default with ios_base::openmode mode = ios_base::in|ios_base::out which is a non-appending mode.

You're interested in the output mode here (ie: ios_base::openmode mode = ios_base::out)

std::basic_stringbuf::str(const std::basic_string<CharT, Traits, Allocator>& s) operates in two different ways, depending on the openmode:

  1. mode & ios_base::ate == false: (ie: non-appending output streams):

    str will set pptr() == pbase(), so that subsequent output will overwrite the characters copied from s

  2. mode & ios_base::ate == true: (ie: appending output streams):

    str will set pptr() == pbase() + s.size(), so that subsequent output will be appended to the last character copied from s

(Note that this appending mode is new since c++11)

More details can be found here.

If you want the appending behaviour, create your stringstream with ios_base::ate:

std::stringstream ss(std::ios_base::out | std::ios_base::ate)

Simple example app here:

#include <iostream>
#include <sstream>

void non_appending()
{
    std::stringstream ss;
    std::string s = "hello world";

    ss.str(s);
    std::cout << ss.str() << std::endl;

    ss << "how are you?";
    std::cout << ss.str() << std::endl;
}

void appending()
{
    std::stringstream ss(std::ios_base::out | std::ios_base::ate);
    std::string s = "hello world";

    ss.str(s);
    std::cout << ss.str() << std::endl;

    ss << "how are you?";
    std::cout << ss.str() << std::endl;
}

int main()
{
    non_appending();
    appending();

    exit(0);
}

This will output in the 2 different ways as explained above:

hello world
how are you?
hello world
hello worldhow are you?

Upvotes: 7

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