list<string> l;
string s;
getline(cin, s);
l.push_back(s);
getline(cin, s);
l.push_back(s);
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Using input
123
test
Printing the list 'l' results:
123
test
Shouldn't this print:
test
test
I'm a little confused as to why the string is being passed by value.
push_back
copies the string object s
passed into it. So wahetever modification you do to s
is not affecting the pushed back value.
It shouldn't, because the string is copied when you add it to the list l
. string
argument is passed by reference to getline:
istream& getline ( istream& is, string& str );
Your title is misleading; you don't have a loop anywhere in the code that you posted.
The string is being passed by value to push_back
, which makes a copy in the list
. The results you're getting are exactly what I would expect.
Pseudocode:
void list::push_back(string& s)
{
list_.add(new string(s)); // copy string
}
You see, the string is copied in push_back method. It is passed by reference just to avoid a redundant copying.
void list::push_back(string s) // copy string
{
list_.add(new string(s)); // copy string
}
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