How to add a list to a string? [duplicate] - c++

Question

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Printing out contents of a list from the c++ list library [duplicate]
(5 answers)
Closed 7 months ago.
I am trying to add a list to a string.
int main() {
std::cout << "Hello, welcome to Jay's Coffee!!\n";
std::string name; std::cout <<"What is your name "; std::cin >> name;
std::cout <<"Hello " << name << ", thank you so much for coming in today";
std::list <std::string> menu = {"Black Coffee" "Espresso" "Latte" "Cappucino"};
std::cout << name <<",what would you like from our menu today? Here is what we are serving.\n" << menu;
}
Returns
invalid operands to binary expression ('basic_ostream<char>' and 'std::list<std::string>' (aka 'list<basic_string<char>>'))

Answer 1

There is no operator<< for lists. You have to write a loop. For example
for (auto& item : menu)
{
std::cout << item << '\n';
}
If you think about it it's obvious why you have to do this yourself. How are you doing to separate the list items? I've chosen to put each item on a new line. You might choose to separate them with commas or spaces or some fancy format. Because there is no obvious single way to print a list there is no predefined way in the C++ library.

Answer 2

You should write code this way. In c++, you can't print a list directly.
#include <string>
#include <list>
using namespace std;
int main() {
cout << "Hello, welcome to Jay's Coffee!!\n";
string name;
cout <<"What is your name ";
cin >> name;
cout <<"Hello " << name << ", thank you so much for coming in today";
list <string> menu = {"Black Coffee", "Espresso", "Latte", "Cappucino"};
cout << name <<",what would you like from our menu today? Here is what we are serving.\n" ;
for ( string& s : menu )
{
cout << s << '\n';
}
}

Answer 3

The error message means that the operator << that you are trying to use with your object menu of the type std::list<std::string> is not defined for the class std::list<std::string>.
Also you need to separate strings in the initializer list with commas.
std::list <std::string> menu = {"Black Coffee", "Espresso", "Latte", "Cappucino"};
Otherwise the list will contain only one string due to the concatenation of string literals.
You could define such an operator as shown in the demonstration program below.
#include <iostream>
#include <string>
#include <list>
std::ostream & operator <<( std::ostream &os, const std::list<std::string>& lst )
{
for ( const auto &s : lst )
{
os << s << '\n';
}
return os;
}
int main()
{
std::list <std::string> menu =
{
"Black Coffee", "Espresso", "Latte", "Cappucino"
};
std::cout << menu;
}
The program output is
Black Coffee
Espresso
Latte
Cappucino
Or just use the range-based for loop directly in main like
std::cout << name <<",what would you like from our menu today? Here is what we are serving.\n";
for ( const auto &s : menu )
{
std::cout << s << '\n';
}
Or place the range-based for loop in a separate function similar to the operator << shown above.

Answer 4

First thing, you didn't define a list: in the declaration of menu, initializer-list elems were not separated by comma (,).
To make it reusable, I'd do it something like this:
#include <iostream>
#include <list>
static const std::string list_sep = ", ";
template<typename C> struct FormattedContainer
{
FormattedContainer(const C& cont, const std::string& sep = list_sep)
: cont_(cont)
, sep_(sep) {}
friend std::ostream& operator<<(std::ostream& os, const FormattedContainer& fc)
{
bool first = true;
for (auto&& e : fc.cont_)
{
if (first)
{
os << e;
first = false;
}
else
{
os << fc.sep_ << e;
}
}
return os;
}
const C& cont_;
const std::string sep_;
};
template<typename C>
auto make_fc(const C& cont, const std::string& sep = list_sep)
-> FormattedContainer<C>
{
return FormattedContainer<C>(cont, sep);
}
int main() {
std::list <std::string> menu = {"Black Coffee", "Espresso", "Latte", "Cappucino"};
std::cout << "What would you like from our menu today? Here is what we are serving.\n" << make_fc(menu);
}
This way, you don't need to define operator<< for something in std namespace (which might result in ambiguous calls as others might also define it, don't need to import another namespace, just simply call a wrapper around the type. You can use it with basically any container or iterable type using this method.

Answer 5

All other answers are correct but here is the better way to do the same
#include <iostream>
#include <algorithm>
#include <list>
template <typename T>
std::ostream & operator << (std::ostream & os, const std::list<T> & vec){
std::for_each (vec.begin () , vec.end() , [&](const auto& val){
std::cout << val << " ";
});
return os;
}
int main () {
std::list <std::string> menu = {"Black Coffee", "Espresso", "Latte", "Cappucino"};
std::cout << menu << "\n";
return 0;
}