stringstream string and char conversion confusion

Question

My question can be boiled down to  where does the string returned from stringstream str   c str   live in memory  and why can t it be assigned to a const char    This code example will explain it better than I can   include  lt string gt   include  lt sstream gt   include  lt iostream gt   using namespace std   int main         stringstream ss  this is a string n         string str ss str          const char  cstr1   str c str         const char  cstr2   ss str   c str         cout  lt  lt  cstr1      Prints correctly          lt  lt  cstr2       ERROR  prints out garbage      system  PAUSE         return 0      The assumption that stringstream str   c str   could be assigned to a const char  led to a bug that took me a while to track down   For bonus points  can anyone explain why replacing the cout statement with  cout  lt  lt  cstr               Prints correctly      lt  lt  ss str   c str      Prints correctly      lt  lt  cstr2               Prints correctly         prints the strings correctly   I m compiling in Visual Studio 2008

User · Answer

What you re doing is creating a temporary  That temporary exists in a scope determined by the compiler  such that it s long enough to satisfy the requirements of where it s going   As soon as the statement const char  cstr2   ss str   c str    is complete  the compiler sees no reason to keep the temporary string around  and it s destroyed  and thus your const char   is pointing to free d memory   Your statement string str ss str     means that the temporary is used in the constructor for the string variable str that you ve put on the local stack  and that stays around as long as you d expect  until the end of the block  or function you ve written  Therefore the const char   within is still good memory when you try the cout

User · Answer

stringstream str   returns a temporary string object that s destroyed at the end of the full expression  If you get a pointer to a C string from that  stringstream str   c str     it will point to a string which is deleted where the statement ends  That s why your code prints garbage       You could copy that temporary string object to some other string object and take the C string from that one   const std  string tmp   stringstream str    const char  cstr   tmp c str      Note that I made the temporary string const  because any changes to it might cause it to re-allocate and thus render cstr invalid  It is therefor safer to not to store the result of the call to str   at all and use cstr only until the end of the full expression   use c str  stringstream str   c str        Of course  the latter might not be easy and copying might be too expensive  What you can do instead is to bind the temporary to a const reference  This will extend its lifetime to the lifetime of the reference       const std  string amp  tmp   stringstream str         const char  cstr   tmp c str        IMO that s the best solution  Unfortunately it s not very well known

User · Answer

In this line   const char  cstr2   ss str   c str      ss str   will make a copy of the contents of the stringstream  When you call c str   on the same line  you ll be referencing legitimate data  but after that line the string will be destroyed  leaving your char  to point to unowned memory

User · Answer

The std  string object returned by ss str   is a temporary object that will have a life time limited to the expression  So you cannot assign a pointer to a temporary object without getting trash   Now  there is one exception  if you use a const reference to get the temporary object  it is legal to use it for a wider life time  For example you should do    include  lt string gt   include  lt sstream gt   include  lt iostream gt   using namespace std   int main         stringstream ss  this is a string n         string str ss str          const char  cstr1   str c str         const std  string amp  resultstr   ss str        const char  cstr2   resultstr c str         cout  lt  lt  cstr1          Prints correctly          lt  lt  cstr2           No more error   cstr2 points to resultstr memory that is still alive as we used the const reference to keep it for a time       system  PAUSE         return 0      That way you get the string for a longer time   Now  you have to know that there is a kind of optimisation called RVO that say that if the compiler see an initialization via a function call and that function return a temporary  it will not do the copy but just make the assigned value be the temporary  That way you don t need to actually use a reference  it s only if you want to be sure that it will not copy that it s necessary  So doing    std  string resultstr   ss str     const char  cstr2   resultstr c str      would be better and simpler

User · Answer

The ss str   temporary is destroyed after initialization of cstr2 is complete  So when you print it with cout  the c-string that was associated with that std  string temporary has long been destoryed  and thus you will be lucky if it crashes and asserts  and not lucky if it prints garbage or does appear to work    const char  cstr2   ss str   c str      The C-string where cstr1 points to  however  is associated with a string that still exists at the time you do the cout - so it correctly prints the result    In the following code  the first cstr is correct  i assume it is cstr1 in the real code    The second prints the c-string associated with the temporary string object ss str    The object is destroyed at the end of evaluating the full-expression in which it appears  The full-expression is the entire cout  lt  lt      expression - so while the c-string is output  the associated string object still exists  For cstr2 - it is pure badness that it succeeds  It most possibly internally chooses the same storage location for the new temporary which it already chose for the temporary used to initialize cstr2  It could aswell crash    cout  lt  lt  cstr               Prints correctly      lt  lt  ss str   c str      Prints correctly      lt  lt  cstr2               Prints correctly         The return of c str   will usually just point to the internal string buffer - but that s not a requirement  The string could make up a buffer if its internal implementation is not contiguous for example  that s well possible - but in the next C   Standard  strings need to be contiguously stored     In GCC  strings use reference counting and copy-on-write  Thus  you will find that the following holds true  it does  at least on my GCC version   string a    hello   string b a   assert a c str      b c str       The two strings share the same buffer here  At the time you change one of them  the buffer will be copied and each will hold its separate copy  Other string implementations do things different  though

[c++] stringstream, string, and char* conversion confusion

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