I can't compile this code because the function declaration depends on the class declaration, and the class declaration depends on the function declaration. Please help.
#include <iostream>
using namespace std;
void simulate(Policy& p);
class Policy {
public:
Policy(int);
int x;
void eval();
};
int main() {
Policy p(23);
return 0;
}
Policy::Policy(int y) { x = y; }
void Policy::eval() { simulate(this); }
void simulate(Policy& p) { cout << ++p.x << endl; }
Place the prototype of your function below the definition of your class, but above the member functions.
Related
calculation() function is not working when m making input() function outside the class...has it got something to do with inline function??
#include <iostream>
using namespace std;
class price
{
public:
int pen;
int rubber;
int scale;
void input()
{
cout<<"enter the variables\n";
cin>>pen>>rubber>>scale;
cout<<"\n"<<pen<<" "<<rubber<<" "<<scale;
}
};
void calculate(price p)
{
int rate[2],total;
rate[0]=p.pen*5;
rate[1]=p.rubber*3;
rate[2]=p.scale*4;
total=rate[0]+rate[1]+rate[2];
cout<<"\n"<<total;
}
int main()
{
price a,b,c;
a.input();
calculate(a);
return 0;
}
No we don't. inline has no effect at all on the semantics of a C++ function. It's only effect is on how that function is treated by the linker.
I want to call a class member function that's nested within a namespace from a different file, but I don't know how.
For example:
How to call a class member function someFunc() that's located in code.h and nested within namespace "program" from main.cpp.
//code.h
#include <iostream>
namespace program
class test {
private:
int x;
public:
test()
{
test::x = 10;
};
someFunc()
{
cout << x << " ";
};
};
There are some problems in your code
#include <iostream>
namespace program { // <-- braces missing
class test
{
private:
int x;
public:
test()
{
test::x = 10; // <-- test:: is unnecessary but no error
};
void someFunc() // <-- return type missing
{
std::cout << x << " "; // <-- namespace std missing
};
};
} // <-- braces missing
int main() {
program::test t;
t.someFunc();
}
I try to compile the following code:
#include <cppunit/extensions/HelperMacros.h>
#include "tested.h"
class TestTested : public CppUnit::TestFixture
{
CPPUNIT_TEST_SUITE(TestTested);
CPPUNIT_TEST(check_value);
CPPUNIT_TEST_SUITE_END();
public:
void check_value();
};
CPPUNIT_TEST_SUITE_REGISTRATION(TestTested);
void TestTested::check_value() {
tested t(3);
int expected_val = t.getValue(); // <----- Line 18.
CPPUNIT_ASSERT_EQUAL(7, expected_val);
}
As a result I get:
testing.cpp:18:32: Error: void-value is not ignored where it should be
EDDIT
To make the example complete I post the code of the tested.h and tested.cpp:
tested.h
#include <iostream>
using namespace std;
class tested {
private:
int x;
public:
tested(int int_x);
void getValue();
};
tested.cpp
#include <iostream>
using namespace std;
tested::tested(int x_inp) {
x = x_inp;
}
int tested::getValue() {
return x;
}
you declare void getValue(); in the class tested.. change to int getValue();.
A void function cannot return a value.
You are getting a value of int from the API getValue(), hence it should return an int.
Your class definition doesn't match the implementation:
In your header you've declared it in the following way (as an aside, you might want to look into some naming conventions).
class tested {
private:
int x;
public:
tested(int int_x);
void getValue();
};
You've declared getValue() as void, i.e no return. Doesn't make much sense for a getter to return nothing, does it?
However, in the .cpp file you've implemented getValue() like so:
int tested::getValue() {
return x;
}
You need to update the getValue() method signature in the header type so that its return type matches the implementation (int).
#include <iostream>
using namespace std;
class B
{
public:
int getMsg(int i)
{
return i + 1;
}
};
class A
{
B b;
public:
void run()
{
taunt(b.getMsg);
}
void taunt(int (*msg)(int))
{
cout << (*msg)(1) << endl;
}
};
int main()
{
A a;
a.run();
}
The above code has a class B inside a class A, and class A has a method taunt that takes a function as an argument. class B's getMsg is passed into taunt...The above code generated the following error message: "error: no matching function for call to 'A::taunt()'"
What's causing the error message in the above code? Am I missing something?
Update:
#include <iostream>
using namespace std;
class B
{
public:
int getMsg(int i)
{
return i + 1;
}
};
class A
{
B b;
public:
void run()
{
taunt(b.getMsg);
}
void taunt(int (B::*msg)(int))
{
cout << (*msg)(1) << endl;
}
};
int main()
{
A a;
a.run();
}
t.cpp: In member function 'void A::run()':
Line 19: error: no matching function for call to 'A::taunt()'
compilation terminated due to -Wfatal-errors.
I'm still getting the same error after changing (*msg)(int) to (B::*msg)(int)
b.getMsg is not the correct way to form a pointer to member, you need &B::getMsg.
(*msg)(1) is not the correct way to call a function through a pointer to member you need to specify an object to call the function on, e.g. (using a temporary) (B().*msg)(1).
The right way to do such things in OOP is to use interfaces so all you need to do is to define an interface and implement it in B class after that pass the pointer of instance which implements this interface to your method in class A.
class IB{
public:
virtual void doSomething()=0;
};
class B: public IB{
public:
virtual void doSomething(){...}
};
class A{
public:
void doSomethingWithB(IB* b){b->doSomething();}
};
This works in VS 2010. The output is the same on all lines:
#include <iostream>
#include <memory>
#include <functional>
using namespace std;
using namespace std::placeholders;
class A
{
public:
int foo(int a, float b)
{
return int(a*b);
}
};
int main(int argc, char* argv[])
{
A temp;
int x = 5;
float y = 3.5;
auto a = std::mem_fn(&A::foo);
cout << a(&temp, x, y) << endl;
auto b = std::bind(a, &temp, x, y);
cout << b() << endl;
auto c = std::bind(std::mem_fn(&A::foo), &temp, _1, y);
cout << c(5) << endl;
}
Basically, you use std::mem_fn to get your callable object for the member function, and then std::bind if you want to bind additional parameters, including the object pointer itself. I'm pretty sure there's a way to use std::ref to encapsulate a reference to the object too if you'd prefer that. I also included the _1 forwarding marker just for another way to specify some parameters in the bind, but not others. You could even specify everything BUT the class instance if you wanted the same parameters to everything but have it work on different objects. Up to you.
If you'd rather use boost::bind it recognizes member functions and you can just put it all on one line a bit to be a bit shorter: auto e = boost::bind(&A::foo, &temp, x, y) but obviously it's not much more to use completely std C++11 calls either.
I am trying to figure out if there's any known pattern/idiom in c++ for what I am trying to do here. Class A must be composed of an object that has a function whose argument must also be of type A. The following code doesn't compile since typeid may not be used in a constant expression. Any suggestions?
#include <iostream>
#include <typeinfo>
using namespace std;
template <typename T>
struct B {
int f(T& i) { cout << "Hello\n"; }
};
class A {
B<typeid(A)> b;
};
int main()
{
A k;
}
Your stated requirements don't need templates at all, just a forward declaration:
#include <iostream>
class A; // forward declare A
struct B {
int f(A &i); // declaration only, definition needs the complete type of A
};
class A {
B b;
};
int B::f(A &i) { std::cout << "Hello\n"; } // define f()
int main()
{
A k;
}
You are looking for B<A> b; The following program compiles without error or warning on g++ 4.4.3.
#include <iostream>
#include <typeinfo>
using namespace std;
template <typename T>
struct B {
int f(T& i) { cout << "Hello\n"; return 0; }
};
class A {
public:
B<A> b;
};
int main()
{
A k;
return k.b.f(k);
}
Note: If you are using templates only to avoid forward declaration, my solution is wrong. But, I'll leave it here in case you are using templates for some other legitimate reason.