I find following boost code
for (auto entry : boost::make_iterator_range(boost::filesystem::directory_iterator(dir_path), {})) {
std::cout << entry << std::endl;
}
entry can be output with dir path, I guess c++ has feature like java Class#toString to custom output string, I try following code:
class C {
str to_string() {
return "prpr";
}
};
TEST_F(FileTest, Draft1) {
C c;
std::cout << c << std::endl;
}
I expect output "prpr", but it doesn't work, how to fix it?
C++ doesn't call any "to_string" or similar function automatically. Instead you must overload the operator<< function for your class:
class C
{
public:
friend std::ostream& operator<<(std::ostream& os, C const& c)
{
return os << "prpr";
}
};
If you want to pass a class to std::cout you need to overload the << operator for std::ostream in your class.
class C {
friend std::ostream& operator<<(std::ostream& os, const C& c);
};
std::ostream& operator<<(std::ostream& os, const C& c){
os<<"prpr";
return os;
}
Related
myclass is a C++ class written by me and when I write:
myclass x;
cout << x;
How do I output 10 or 20.2, like an integer or a float value?
Typically by overloading operator<< for your class:
struct myclass {
int i;
};
std::ostream &operator<<(std::ostream &os, myclass const &m) {
return os << m.i;
}
int main() {
myclass x(10);
std::cout << x;
return 0;
}
You need to overload the << operator,
std::ostream& operator<<(std::ostream& os, const myclass& obj)
{
os << obj.somevalue;
return os;
}
Then when you do cout << x (where x is of type myclass in your case), it would output whatever you've told it to in the method. In the case of the example above it would be the x.somevalue member.
If the type of the member can't be added directly to an ostream, then you would need to overload the << operator for that type also, using the same method as above.
it's very easy, just implement :
std::ostream & operator<<(std::ostream & os, const myclass & foo)
{
os << foo.var;
return os;
}
You need to return a reference to os in order to chain the outpout (cout << foo << 42 << endl)
Even though other answer provide correct code, it is also recommended to use a hidden friend function to implement the operator<<. Hidden friend functions has a more limited scope, therefore results in a faster compilation. Since there is less overloads cluttering the namespace scope, the compiler has less lookup to do.
struct myclass {
int i;
friend auto operator<<(std::ostream& os, myclass const& m) -> std::ostream& {
return os << m.i;
}
};
int main() {
auto const x = myclass{10};
std::cout << x;
return 0;
}
Alternative:
struct myclass {
int i;
inline operator int() const
{
return i;
}
};
myclass is a C++ class written by me and when I write:
myclass x;
cout << x;
How do I output 10 or 20.2, like an integer or a float value?
Typically by overloading operator<< for your class:
struct myclass {
int i;
};
std::ostream &operator<<(std::ostream &os, myclass const &m) {
return os << m.i;
}
int main() {
myclass x(10);
std::cout << x;
return 0;
}
You need to overload the << operator,
std::ostream& operator<<(std::ostream& os, const myclass& obj)
{
os << obj.somevalue;
return os;
}
Then when you do cout << x (where x is of type myclass in your case), it would output whatever you've told it to in the method. In the case of the example above it would be the x.somevalue member.
If the type of the member can't be added directly to an ostream, then you would need to overload the << operator for that type also, using the same method as above.
it's very easy, just implement :
std::ostream & operator<<(std::ostream & os, const myclass & foo)
{
os << foo.var;
return os;
}
You need to return a reference to os in order to chain the outpout (cout << foo << 42 << endl)
Even though other answer provide correct code, it is also recommended to use a hidden friend function to implement the operator<<. Hidden friend functions has a more limited scope, therefore results in a faster compilation. Since there is less overloads cluttering the namespace scope, the compiler has less lookup to do.
struct myclass {
int i;
friend auto operator<<(std::ostream& os, myclass const& m) -> std::ostream& {
return os << m.i;
}
};
int main() {
auto const x = myclass{10};
std::cout << x;
return 0;
}
Alternative:
struct myclass {
int i;
inline operator int() const
{
return i;
}
};
myclass is a C++ class written by me and when I write:
myclass x;
cout << x;
How do I output 10 or 20.2, like an integer or a float value?
Typically by overloading operator<< for your class:
struct myclass {
int i;
};
std::ostream &operator<<(std::ostream &os, myclass const &m) {
return os << m.i;
}
int main() {
myclass x(10);
std::cout << x;
return 0;
}
You need to overload the << operator,
std::ostream& operator<<(std::ostream& os, const myclass& obj)
{
os << obj.somevalue;
return os;
}
Then when you do cout << x (where x is of type myclass in your case), it would output whatever you've told it to in the method. In the case of the example above it would be the x.somevalue member.
If the type of the member can't be added directly to an ostream, then you would need to overload the << operator for that type also, using the same method as above.
it's very easy, just implement :
std::ostream & operator<<(std::ostream & os, const myclass & foo)
{
os << foo.var;
return os;
}
You need to return a reference to os in order to chain the outpout (cout << foo << 42 << endl)
Even though other answer provide correct code, it is also recommended to use a hidden friend function to implement the operator<<. Hidden friend functions has a more limited scope, therefore results in a faster compilation. Since there is less overloads cluttering the namespace scope, the compiler has less lookup to do.
struct myclass {
int i;
friend auto operator<<(std::ostream& os, myclass const& m) -> std::ostream& {
return os << m.i;
}
};
int main() {
auto const x = myclass{10};
std::cout << x;
return 0;
}
Alternative:
struct myclass {
int i;
inline operator int() const
{
return i;
}
};
Like in this code :
#include <iostream>
enum class A {
a,
b
};
std::ostream& operator<<(std::ostream& os, A val)
{
return os << val;
}
int main() {
auto a = A::a;
std::cout << a;
return 0;
}
When I did not provide std::ostream& operator<<(std::ostream& os, A val) the program didn't compile because A::a didn't have any function to go with <<. But now when I've already provided it, it produces garbage in my terminal and on ideone, it produces a runtime error (time limit exceeded).
std::ostream& operator<<(std::ostream& os, A val) {
return os << val;
}
This causes infinite recursion. Remember that os << val is really seen to the compiler operator<<(os,val) in this instance. What you want to do is print the underlying value of the enum. Fortunately, there is a type_trait that allows you to expose the underlying type of the enum and then you can cast the parameter to that type and print it.
#include <iostream>
#include <type_traits>
enum class A {
a, b
};
std::ostream& operator<<(std::ostream& os, A val) {
return os << static_cast<std::underlying_type<A>::type>(val);
}
int main() {
auto a = A::a;
std::cout << a;
}
std::ostream& operator<<(std::ostream& os, A val)
{
return os << val; // Calls the function again.
// Results in infinite recursion.
}
Try
std::ostream& operator<<(std::ostream& os, A val)
{
return os << static_cast<int>(val);
}
Which operator do I have to overload if I want to use sth like this?
MyClass C;
cout<< C;
The output of my class would be string.
if you've to overload operator<< as:
std::ostream& operator<<(std::ostream& out, const MyClass & obj)
{
//use out to print members of obj, or whatever you want to print
return out;
}
If this function needs to access private members of MyClass, then you've to make it friend of MyClass, or alternatively, you can delegate the work to some public function of the class.
For example, suppose you've a point class defined as:
struct point
{
double x;
double y;
double z;
};
Then you can overload operator<< as:
std::ostream& operator<<(std::ostream& out, const point & pt)
{
out << "{" << pt.x <<"," << pt.y <<"," << pt.z << "}";
return out;
}
And you can use it as:
point p1 = {10,20,30};
std::cout << p1 << std::endl;
Output:
{10,20,30}
Online demo : http://ideone.com/zjcYd
Hope that helps.
The stream operator: <<
You should declare it as a friend of your class:
class MyClass
{
//class declaration
//....
friend std::ostream& operator<<(std::ostream& out, const MyClass& mc);
}
std::ostream& operator<<(std::ostream& out, const MyClass& mc)
{
//logic here
}
You should implement operator<< as a free function.