I'm still fairly rusty at c++ and I'm having trouble understanding my issue. The error message that I am receiving is "No operator '<<' matches these operands" The code I have:
for(int i = 0; i < ruleList.size(); i++)
{
cout << ruleList[i].lhs << endl;
cout << ruleList[i].rhs << endl; // Problem printing this
}
struct Rules
{
string lhs;
vector<string> rhs;
}rule;
vector<Rules> ruleList;
Would this be the appropriate way to do this? I did the lhs the same way and it works fine.
rule.rhs.push_back(token);
ruleList.push_back(rule);
There is no operator<< defined for standard containers. You will need to write a print function, something along the lines of:
void print(std::ostream& out, std::vector<std::string> const & data) {
std::copy(data.begin(), data.end(),
std::ostream_iterator<std::string>(out, " "));
}
And then use it as:
print(std::cout, ruleList[i].rhs);
std::vector does not define an operator <<. You can use a std::ostream_iterator to format a list:
std::copy( ruleList[i].rhs.begin(), ruleList[i].rhs.end(),
std::ostream_iterator< std::string >( std::cout, ", " ) );
This is a bit imperfect in that ", " is printed after the final element, but that can be worked around.
You need to write your own << operator for struct rules. It should look something like this in C++11:
struct rules {
string lhs;
std::vector<std::string> rhs;
// apparently it's a good idea to keep this out of std:: namespace
inline static std::ostream & operator << (std::ostream & out, const rules & r) {
out << r.lhs << std::endl;
//for (int i = 0; i < v.length(); i++)
for (auto & s : r.rhs) {
out << s;
}
out << std::endl;
return out;
}
}
MSDN has a write up here: http://msdn.microsoft.com/en-us/library/1z2f6c2k.aspx
Related
I'm trying to overload the << operator to print a vector that contains elements of type Position (vector).
I managed to overload the << operator for type Position, but I can't figure out how to do it for vector. Can you help?
//position.h
#ifndef POSITION_H
#define POSITION_H
#include <iostream> // using IO functions
using namespace std;
class Position {
private:
int row;
int column;
public:
friend ostream& operator<<(ostream& os, const Position& P);
friend ostream& operator<<(ostream& os, const vector<Position>& VP);
};
#endif
//position.cpp
#include"Position.h"
#include <iostream> // using IO functions
#include<vector>
ostream& operator<<(ostream& os, const Position& P)
{
os << '(' << P.row << ',' << P.column << ')' << endl;
return os;
}
ostream& operator<<(ostream& os, const vector<Position>& VP)
{
Position placeholder;
for (int i = 0; i != VP.size(); i++)
{
placeholder = VP.at(i);
cout << placeholder << endl;
}
return os;
}
int main()
{
Position C1(2, 1);
Position C2(3, 1);
Position C3(4, 1);
vector<Position> cans;
cans.push_back(C1);
cans.push_back(C2);
cans.push_back(C3);
cout << cans;
system("pause");
}
I'm sure there's a duplicate on StackOverflow somewhere, but I cannot find it. I'm a bit surprised about this question, as I don't really see a critical issue at first sight. Did you get a compiler error or segmentation fault?
Every time you assign to Placeholder, you make a copy. This is not necessary at all: you can directly access the element at the index. As you limit the index to 0 to size(), you don't have to worry about going out-of-bounds, so don't use .at(i), which throws an exception if you go out of bounds. Just use VP[i].
Also not that std::endl flushes the buffer, which is slow. A normal enter/end-line can be achieved with the '\n' character.
Finally, using namespace std; is considered bad practice.
You have multiple options for processing the elements of the vector:
index-based for loop:
for (int i = 0; i != VP.size(); ++i) {
os << VP[i] << '\n';
}
Iterator-based for loop
for (auto it = cbegin(VP); it != cend(VP); ++it) {
os << *it << '\n';
}
range-based for loop
for (auto const& el : VP) {
os << el << '\n';
}
Algorithm for(each) loop
std::for_each(cbegin(VP), cend(VP),
[](auto const& el) { os << el << '\n'; });
Copy to ostream_iterator
std::copy(cbegin(VP), cend(VP),
std::ostream_iterator<Position>(os, "\n"));
Note that here you could also write <decltype(VP)::value> instead of <Position> to keep it generic, but that might be overkill in this situation.
And the last two have a C++20 ranges equivalent:
std::ranges::for_each(VP,
[](auto const& el) { os << el << '\n'; });
std::ranges::copy(VP,
std::ostream_iterator<Position>(os, "\n"));
Note:
friend ostream& operator<<(ostream& os, const vector<Position>& VP); doesn't have to be a friend of Position! It doesn't require access to the private members.
I'm learning C++ so my question might seem a bit stupid. I wanted to build a function that print every element in a vector. I come up with that but it seems to display the address of every element. I google it and find a nice solution but i wanted to do it this way so if anyone can explain me where i'm doing something wrong.
My code :
void display_vector(std::vector<int>& to_display);
int main()
{
std::vector<int> vector_to_sort = { 2,6,7,2,1,80,2,59,8,9 };
display_vector(vector_to_sort);
}
void display_vector(std::vector<int> &to_display)
{
for (int i = 0; i < to_display.size(); i++)
{
std::cout << to_display[i] << ', ';
}
std::cout << '\n';
}
The solution i found on internet :
#include <iterator>
void display_vector(const vector<int> &v)
{
std::copy(v.begin(), v.end(),
std::ostream_iterator<int>(std::cout, " "));
}
Output of my code :
21129661129671129621129611129680112962112965911296811296911296
You use ", " instead of ', '.
You can use any of the following print mechanism in the display() function:
void display_vector(std::vector<int> &to_display)
{
//by using Normal for loop
for (auto i = to_display.begin(); i != to_display.end(); ++i) {
cout << *i << " ";
}
cout << endl;
//by using Range based for loop
for (int & i : to_display) {
cout << i << " ";
}
std::cout << '\n';
}
In this statement
std::cout << to_display[i] << ', ';
^^^^^^
you are using a multybyte character literal that has an implementation defined value.
Substitute it for string literal ", ".
As for the function then for starters if the vector is not being changed in the function then the parameter should be a constant reference.
You can use the range-based for loop to outfput elements of the vector like for example
#include <iostream>
#include <vector>
std::ostream & display_vector( const std::vector<int> &to_display, std::ostream &os = std::cout );
int main()
{
std::vector<int> vector_to_sort = { 2,6,7,2,1,80,2,59,8,9 };
display_vector(vector_to_sort) << '\n';
}
std::ostream & display_vector( const std::vector<int> &to_display, std::ostream &os )
{
for ( const auto &item : to_display )
{
os << item << ", ";
}
return os;
}
Using such a function you can for example output the vector in a text file.
Just replace below line
std::cout << to_display[i] << ', ';
with
std::cout << to_display[i] << ", ";
Also note that if you just want to display vector in function then declare parameter as const reference as shown below
void display_vector(const std::vector<int> &to_display);
The debuggers make it easy to examine vectors but I include a simple template to print out vectors of standard types and often use it when debugging data that I wish to look at with other tools.
template<class T>
void print(const std::vector<T>& v){
for (auto x: v)
std::cout << x << std::endl;
}
Hey guys I am new to C++ and I have a problem with this operator: (Also new in stackoverflow)
This is my class TestList:
class TestList{
public:
TestList() : listItems(10), position(0){};
TestList(int k) : listItems(k), position(0){};
int listItems;
int position;
std::vector<int> arr;
};
//my current operator is: What should be changed?
ostream& operator <<(ostream&, const TestList& tlist, int input){
os << tlist.arr.push_back(input);
return os;
}
//
int main() {
testList testlist(5);
testlist << 1 << 2 << 3; //how should I overload the operator to add these number to testlist.arr ?
return 0;
}
I hope someone could help me or can give me any tips? :)
The other answers are absolutely correct, I just want to say something general on operator<<. It always has the signature T operator<<(U, V), since it is always a binary operator, so it has to have exactly two arguments. Since the chain
a << b << c;
is evaluated as
(a << b) << c;
// That calls
operator<<(operator<<(a, b), c);
the types T and U should normally be the same, or at least compatible.
Furthermore, it is possible but very weird to assign the result of operator<< to something (like result = (a << b))). A good rule of thumb is "My code should not be weird". Therefore the type T should mostly be a reference (so X&) since otherwise it would only be a temporary copy that is unused. And that is pretty useless most of the time.
So in 90% of all cases, your operator<< should have the signature
T& operator<<(T&, V);
I think you mean the following
TestList & operator <<( TestList &tlist , int input )
{
tlist.arr.push_back( input );
return tlist;
}
Here is a demonstrative program
#include <iostream>
#include <vector>
class TestList{
public:
TestList() : listItems(10), position(0){};
TestList(int k) : listItems(k), position(0){};
int listItems;
int position;
std::vector<int> arr;
};
TestList & operator <<( TestList &tlist , int input )
{
tlist.arr.push_back( input );
return tlist;
}
std::ostream & operator <<( std::ostream &os, const TestList &tlist )
{
for ( const auto &item : tlist.arr )
{
std::cout << item << ' ';
}
return os;
}
int main()
{
TestList testlist(5);
testlist << 1 << 2 << 3;
std::cout << testlist << '\n';
return 0;
}
The program output is
1 2 3
You can even write instead of these two statements
testlist << 1 << 2 << 3;
std::cout << testlist << '\n';
only one statement
std::cout << ( testlist << 1 << 2 << 3 ) << '\n';
Pay attention to that there is a typo in your declaration
testList testlist(5);
There should be
TestList testlist(5);
Edit: Definition of class TF:
class TF {
std::vector<V4f> waypoints;
std::vector<int> densityWaypoints;
public:
std::size_t size() const { return waypoints.size(); }
friend std::ostream& operator<<(std::ostream& str, const TF& tf);
friend std::fstream& operator<<(std::fstream& str, const TF& tf);
// methods here
};
The question may steam from the fact that I don't understand streams, so that's probably a precondition.
Is it somehow possible to overload operator<<(std::ostream, T) so that when invoked in order to display the data structure on screen, it uses one overload, and when the data structure is written to a file, another one is used? Something like this probably:
std::ostream& operator<<(std::ostream& str, const TF& tf) {
for (std::size_t i = 0; i != tf.waypoints.size(); ++i) {
str << " { "
<< tf.densityWaypoints[i] << " : "
<< tf.waypoints[i][3] << " : "
<< tf.waypoints[i][0] << " , "
<< tf.waypoints[i][1] << " , "
<< tf.waypoints[i][2]
<< " } ";
}
str << "\n";
return str;
}
std::fstream& operator<<(std::fstream& str, const TF& tf) {
str << (int)tf.size();
for (std::size_t i = 0; i != tf.waypoints.size(); ++i) {
str << tf.densityWaypoints[i]
<< tf.waypoints[i][0]
<< tf.waypoints[i][1]
<< tf.waypoints[i][2]
<< tf.waypoints[i][3];
}
This doesn't compile with a strange error (I may be tired):
error: no match for ‘operator<<’ (operand types are ‘std::fstream {aka std::basic_fstream}’ and ‘int’)
The error occurs when I add the second operator<<() overload. The first one works fine. Tried both std::ofstream and std::fstream to the same result.
But I'm not sure if it's going to work either. Sure it's possible to define a function like int writeTF(std:fstream& str, const TF&tf), but that doesn't look C++ enough to me, not to mention the strange error that will potentially appear here, too.
I've seen code comparing the ostream's address to that of cout. I have mixed feelings about it, but it certainly worked:
std::ostream& operator<<(std::ostream& o, Foo const&)
{
if(&o == &std::cout) {
return o << "cout";
} else {
return o << "not_cout";
}
}
demo
Note that cout outputs to standard output, it's not the same thing as "the screen".
This question already has answers here:
How can I use cout << myclass
(5 answers)
Closed 4 months ago.
I've been reading questions here for an hour or two regarding this error I'm getting and most of them forgot to #include string (which I had already done), or to overload the << operator.
Here's the code in question:
void Student::getCoursesEnrolled(const vector<Course>& c)
{
for (int i = 0; i < c.size(); i++)
{
cout << c[i] << endl;
}
}
And the error I'm getting:
Error: No operator matches these operands
operand types are: std::ostream << const Course
All I'm trying to do is return the vector. I read about overloading the << operator but we haven't learned any of that in class so I'm assuming there is another way of doing it?
I appreciate your time!
All I'm trying to do is return the vector.
Not quite; you're trying to print it using cout. And cout has no idea how to print a Course object, unless you provide an overloaded operator<< to tell it how to do so:
std::ostream& operator<<(std::ostream& out, const Course& course)
{
out << course.getName(); // for example
return out;
}
See the operator overloading bible here on StackOverflow for more information.
The problem is that operator << is not overload for type Course objects of which you are trying to output in statement
cout << c[i] << endl;
You need to overload this operator or write your own function that will output an object of type Course in std::ostream
For example let assume that below is a definition of class Course
class Course
{
private:
std::string name;
unsigned int duration;
public:
Course() : duration( 0 ) {}
Course( const std::string &s, unsigned int n ) : name( s ), duration( n ) {}
std::ostream & out( std::ostream &os ) const
{
return ( os << "Course name = " << name << ", course duration = " << duration );
}
};
When you can write
std::vector<Course> v = { { "A", 1 }, { "B", 2 }, { "C", 3 } };
for ( const Course &c : v ) c.out( std::cout ) << std::endl;
Instead member function out you can overload operator <<. For example
class Course
{
private:
std::string name;
unsigned int duration;
public:
Course() : duration( 0 ) {}
Course( const std::string &s, unsigned int n ) : name( s ), duration( n ) {}
friend std::ostream & operator <<( std::ostream &os, const Course & c )
{
return ( os << "Course name = " << c.name << ", course duration = " << c.duration );
}
};
and use it as
std::vector<Course> v = { { "A", 1 }, { "B", 2 }, { "C", 3 } };
for ( const Course &c : v ) std::cout << c << std::endl;
The stream operator << is used to "output" some representation of that object. If you don't want to overload the operator yet just pick some property to output instead:
for (int i = 0; i < c.size(); i++)
{
cout << c[i].Name << endl; // assuming Name is a property of Course
}
When you DO overload the operator you just decide then what the proper representation of a Course is:
ostream& operator<< (ostream &out, Course &c)
{
out << c.Name "(" << c.Description << ")";
return out;
}
Your Course class needs to implement an operator:
class Course
{
public:
/*
* Your code here
*/
// Probably missing this:
friend std::ostream& operator << (std::ostream& os, const Course& course)
{
os << course.name(); // etc..
return os;
};
}; // eo class Course
Since you haven't yet learned to overload operator<<, what you can do instead is to print each member of your Course class. You haven't posted the definition of Course, but perhaps it's something like this:
class Course
{
public:
int get_number() { return _number; }
const std::string& get_name() { return _name; }
private:
int _number;
std::string _name;
};
then you can say:
void Student::getCoursesEnrolled(const vector<Course>& c)
{
for (int i = 0; i < c.size(); i++)
{
cout << c[i].get_number() << " "
<< c[i].get_name() << std::endl;
}
}
Your problem is this particular part:
cout << c[i]
In your case c[i] is an object of type Course as dvnrrs correctly pointed out. So either:
implement the overloaded << operator for your object OR
if your Course object is in someway a typedef to a primitive try explicitly casting it to a string type (or similar)