I have requirement as follows.
I have to generate increment negative numbers from -1 to -100 which is used a unique id for a request. Like it should be like this: -1, -2, -3, ...-100, -1, -2, and so on. How can I do this effectively? I am not supposed to use Boost. C++ STL is fine. I prefer to write simple function like int GetNextID() and it should generate ID. Request sample program on how to do this effectively?
Thanks for your time and help
int ID = -1;
auto getnext = [=] mutable {
if (ID == -100) ID = -1;
return ID--;
};
Fairly basic stuff here, really. If you have to ask somebody on the Interwebs to write this program for you, you should really consider finding some educational material in C++.
I love the functor solution:
template <int limit> class NegativeNumber
{
public:
NegativeNumber() : current(0) {};
int operator()()
{
return -(1 + (current++ % limit));
};
private:
int current;
};
Then, you can define any generator with any limit and use it:
NegativeNumber<5> five;
NegativeNumber<2> two;
for (int x = 0; x < 20; ++x)
std::cout << "limit five: " << five() << "\tlimit two: " << two() << '\n';
You can also pass the generator as parameter to another function, with each funtor with its own state:
void f5(NegativeNumber<5> &n)
{
std::cout << "limit five: " << n() << '\n';
}
void f2(NegativeNumber<2> &n)
{
std::cout << "limit two: " << n() << '\n';
}
f5(five);
f2(two);
If you don't like the template solution to declare the limit, there's also the no-template version:
class NegativeNumberNoTemplate
{
public:
NegativeNumberNoTemplate(int limit) : m_limit(limit), current(0) {};
int operator()()
{
return -(1 + (current++ % m_limit));
};
private:
const int m_limit;
int current;
};
Using as argument to a function works in the same way, and it's internal state is transfered as well:
void f(NegativeNumberNoTemplate &n)
{
std::cout << "no template: " << n() << '\n';
}
NegativeNumberNoTemplate notemplate(3);
f(notemplate);
I hope you don't want to use it with threading, they're not thread safe ;)
Here you have all the examples; hope it helps.
Something like.... (haven't compiled)
class myClass
{
int number = 0;
int GetValue ()
{
return - (number = ((number+1) % 101))
}
}
Even a simple problem like this could lead you to several approximations, both in the algorithmic solution and in the concrete usage of the programming language.
This was my first solution using C++03. I preferred to switch the sign after computing the value.
#include <iostream>
int GetNextID() {
// This variable is private to this function. Be careful of not calling it
// from multiple threads!
static int current_value = 0;
const int MAX_CYCLE_VALUE = 100;
return - (current_value++ % MAX_CYCLE_VALUE) - 1;
}
int main()
{
const int TOTAL_GETS = 500;
for (int i = 0; i < TOTAL_GETS; ++i)
std::cout << GetNextID() << std::endl;
}
A different solution taking into account that the integer modulo in C++ takes the sign of the dividend (!) as commented in the Wikipedia
#include <iostream>
int GetNextID() {
// This variable is private to this function. Be careful of not calling it
// from multiple threads!
static int current_value = 0;
const int MAX_CYCLE_VALUE = 10;
return (current_value-- % MAX_CYCLE_VALUE) - 1;
}
int main()
{
const int TOTAL_GETS = 50;
for (int i = 0; i < TOTAL_GETS; ++i)
std::cout << GetNextID() << std::endl;
}
Related
Live On Coliru
I need to implement a utility class for the purpose of lookup. The value of the internal array never changes after the initialization. My initial plan was to implement this class with the help of constexpr so that the expensive initialization can be done in compilation time. However, I am not able to come up with a good solution. So far, I can only figure out the following two methods. Both methods have drawbacks.
Method One:
The implementation looks complicate for such a simple task.
Method Two:
First, I assume the initialization is done in compilation time. Is this correct?
Second, the TwoClass::m_array really should be static instead of a member variable.
Is there a better way to implement this with c++20?
Thank you
#include <iostream>
#include <array>
using namespace std;
//== Method One
class OneClass
{
public:
uint32_t get(size_t index)
{
return m_array[index % 0xFF];
}
struct init_static_array
{
init_static_array()
{
// std::cout << "call init_static_array\n";
for (int i = 0; i < N; i++) {
m_array[i] = i * i + 10;
}
}
};
private:
static init_static_array m_init_static_array;
enum {N = 255};
static int m_array[N]; // this is a lookup table and NEVER will be changed!
};
int OneClass::m_array[] = {0};
OneClass::init_static_array OneClass::m_init_static_array{};
//== Method Two
class TwoClass
{
public:
constexpr TwoClass()
{
//std::cout << "call TwoClass::TwoClass()\n";
for (int i = 0; i < N; i++) {
m_array[i] = i * i + 10;
}
}
int get(size_t index) const
{
return m_array[index % 0xFF];
}
private:
enum {N = 255};
int m_array[N] {0}; // this is a lookup table and NEVER will be changed!
};
int main()
{
OneClass oneclass;
cout << "1 => " << oneclass.get(1) << std::endl;
cout << "256 => " << oneclass.get(256) << std::endl;
cout << "==================\n";
TwoClass twoclass;
cout << "1 => " << twoclass.get(1) << std::endl;
cout << "256 => " << twoclass.get(256) << std::endl;
return 0;
}
Output:
call init_static_array
1 => 11
256 => 11
==================
call TwoClass::TwoClass()
1 => 11
256 => 11
If it's C++20, don't you just want this:
namespace {
constexpr std::size_t N{255u};
consteval std::array<int, N> initArr()
{
std::array<int, N> arr{};
for (int i = 0; i < N; i++) {
arr[i] = i * i + 10;
}
return arr;
}
}
class C
{
private:
constexpr static std::array<int, N> arr{initArr()};
public:
constexpr C() = default;
int get(std::size_t n) const {return arr[n%255];}
};
Neither of your methods initialized the array at compile-time.
In the first method, the array is initialized in construction of static data member m_init_static_array, e.g. before main is called.
In the second method, constexpr in function declarations, including constructors, indicates that you want them being able to be called at compile-time, e.g. in evaluation of constant expressions. It doesn't mean all invocations are evaluated at compile-time. To call constexpr functions as compile-time, you have to call them in constant expression.
One solution is to initialize the static constexpr data member internal_array of type std::array by a lambda:
#include <array>
class Demo {
public:
int get(int index) const {
return internal_array[index % N];
}
private:
static inline constexpr int N = 255;
static inline constexpr std::array<int, N> internal_array = []() {
// Cannot call non-constexpr functions at compile-time
// std::cout << "std::cout << "call init_static_array\n";
std::array<int, N> arr;
for (int i = 0; i < N; ++i)
arr[i] = i * i + 10;
return arr;
}();
};
int main() {
Demo demo;
return demo.get(10);
}
Problem
I want to create a function inside a class which function2 will use the result generated from function1. I have a small code snippet where I tried to make it easy to understand.
#include <stdio.h>
class GreaterSmaller {
public:
int greater, smaller;
};
GreaterSmaller findGreaterSmaller(int a, int b)
{
GreaterSmaller s;
if (a > b) {
s.greater = a;
s.smaller = b;
}
else {
s.greater = b;
s.smaller = a;
}
return s;
}
GreaterSmaller print()
{
GreaterSmaller s;
std::cout << s.greater << s.smaller << std::endl;
}
int main()
{
int x = 4;
int y = 3;
GreaterSmaller result;
result = findGreaterSmaller(x, y);
result = print(); // I want it to print 4 & 3
return 0;
}
P.s Just wanted to mention I am not trying to print the result in the function2 I have created that for a demo.
Define the second method as taking an argument of the first type, and pass it when you call it, as such:
void print(GreaterSmaller &s)
{
std::cout << s.greater << s.smaller << std::endl;
}
print(result); // I want it to print 4 & 3
I'm creating a node system (similar to eg. UE4 or Blender's Cycles) in which i can create nodes of different types and use them later. At the moment I have 2 classes of nodes with output functions like these:
class InputInt
{
public:
int output()
{
int x;
std::cin>>x;
return x;
}
};
class RandomInt
{
public:
int rand10()
{
int x;
x = rand()%10;
return x;
}
int rand100()
{
int x;
x = rand()%100;
return x;
}
};
I don't pass anything to these nodes. Now I want to create a node which takes and output function from and object of one of above classes. Here is how I implemented it to use InputInt node only:
class MultiplyBy2
{
typedef int (InputInt::*func)();
func input_func;
InputInt *obj;
public:
MultiplyBy2(InputInt *object, func i): obj(object), input_func(i) {}
int output()
{
return (obj->*input_func)()*2;
}
};
Having this I can create and use object of MultiplyBy2 in main() and it works perfectly.
int main()
{
InputInt input;
MultiplyBy2 multi(&input, input.output);
std::cout<<multi.output()<<std::endl;
}
It doesn't obviously work for object of RandomInt as I have to pass *InputInt object to MultiplyBy2 object. Is there a way to make MultiplyBy2 take any kind of an object with its output function eg. like this?
int main()
{
RandomInt random;
MultiplyBy2 multi2(&random, random.rand10);
std::cout<<multi2.output()<<std::endl;
}
An alternative approach, using a common base class with virtual methods:
#include <iostream>
struct IntOp {
virtual int get() = 0;
};
struct ConstInt: IntOp {
int n;
explicit ConstInt(int n): n(n) { }
virtual int get() override { return n; }
};
struct MultiplyIntInt: IntOp {
IntOp *pArg1, *pArg2;
MultiplyIntInt(IntOp *pArg1, IntOp *pArg2): pArg1(pArg1), pArg2(pArg2) { }
virtual int get() override { return pArg1->get() * pArg2->get(); }
};
int main()
{
ConstInt i3(3), i4(4);
MultiplyIntInt i3muli4(&i3, &i4);
std::cout << i3.get() << " * " << i4.get() << " = " << i3muli4.get() << '\n';
return 0;
}
Output:
3 * 4 = 12
Live Demo on coliru
As I mentioned std::function in post-answer conversation with OP, I fiddled a bit with this idea and got this:
#include <iostream>
#include <functional>
struct MultiplyIntInt {
std::function<int()> op1, op2;
MultiplyIntInt(std::function<int()> op1, std::function<int()> op2): op1(op1), op2(op2) { }
int get() { return op1() * op2(); }
};
int main()
{
auto const3 = []() -> int { return 3; };
auto const4 = []() -> int { return 4; };
auto rand100 = []() -> int { return rand() % 100; };
MultiplyIntInt i3muli4(const3, const4);
MultiplyIntInt i3muli4mulRnd(
[&]() -> int { return i3muli4.get(); }, rand100);
for (int i = 1; i <= 10; ++i) {
std::cout << i << ".: 3 * 4 * rand() = "
<< i3muli4mulRnd.get() << '\n';
}
return 0;
}
Output:
1.: 3 * 4 * rand() = 996
2.: 3 * 4 * rand() = 1032
3.: 3 * 4 * rand() = 924
4.: 3 * 4 * rand() = 180
5.: 3 * 4 * rand() = 1116
6.: 3 * 4 * rand() = 420
7.: 3 * 4 * rand() = 1032
8.: 3 * 4 * rand() = 1104
9.: 3 * 4 * rand() = 588
10.: 3 * 4 * rand() = 252
Live Demo on coliru
With std::function<>, class methods, free-standing functions, and even lambdas can be used in combination. So, there is no base class anymore needed for nodes. Actually, even nodes are not anymore needed (explicitly) (if a free-standing function or lambda is not counted as "node").
I must admit that graphical dataflow programming was subject of my final work in University (though this is a long time ago). I remembered that I distinguished
demand-driven execution vs.
data-driven execution.
Both examples above are demand-driven execution. (The result is requested and "pulls" the arguments.)
So, my last sample is dedicated to show a simplified data-driven execution (in principle):
#include <iostream>
#include <vector>
#include <functional>
struct ConstInt {
int n;
std::vector<std::function<void(int)>> out;
ConstInt(int n): n(n) { eval(); }
void link(std::function<void(int)> in)
{
out.push_back(in); eval();
}
void eval()
{
for (std::function<void(int)> &f : out) f(n);
}
};
struct MultiplyIntInt {
int n1, n2; bool received1, received2;
std::vector<std::function<void(int)>> out;
void set1(int n) { n1 = n; received1 = true; eval(); }
void set2(int n) { n2 = n; received2 = true; eval(); }
void link(std::function<void(int)> in)
{
out.push_back(in); eval();
}
void eval()
{
if (received1 && received2) {
int prod = n1 * n2;
for (std::function<void(int)> &f : out) f(prod);
}
}
};
struct Print {
const char *text;
explicit Print(const char *text): text(text) { }
void set(int n)
{
std::cout << text << n << '\n';
}
};
int main()
{
// setup data flow
Print print("Result: ");
MultiplyIntInt mul;
ConstInt const3(3), const4(4);
// link nodes
const3.link([&mul](int n) { mul.set1(n); });
const4.link([&mul](int n) { mul.set2(n); });
mul.link([&print](int n) { print.set(n); });
// done
return 0;
}
With the dataflow graph image (provided by koman900 – the OP) in mind, the out vectors represent outputs of nodes, where the methods set()/set1()/set2() represent inputs.
Output:
Result: 12
Live Demo on coliru
After connection of graph, the source nodes (const3 and const4) may push new results to their output which may or may not cause following operations to recompute.
For a graphical presentation, the operator classes should provide additionally some kind of infrastructure (e.g. to retrieve a name/type and the available inputs and outputs, and, may be, signals for notification about state changes).
Surely, it is possible to combine both approaches (data-driven and demand-driven execution). (A node in the middle may change its state and requests new input to push new output afterwards.)
You can use templates.
template <typename UnderlyingClass>
class MultiplyBy2
{
typedef int (UnderlyingClass::*func)();
func input_func;
UnderlyingClass *obj;
public:
MultiplyBy2(UnderlyingClass *object, func i) : obj(object), input_func(i) {}
int output()
{
return (obj->*input_func)() * 2;
}
};
int main()
{
// test
InputInt ii;
MultiplyBy2<InputInt> mii{ &ii, &InputInt::output };
RandomInt ri;
MultiplyBy2<RandomInt> mri{ &ri, &RandomInt::rand10 };
}
This is a bit convoluted. However I think you should be making an interface or class that returns a value and the objects should inherit from this. Then the operator class can take any class that inherits from the base/interface. Eg Make an BaseInt class that stores an int and has the output method/ RandomInt and InputInt should inherit from BaseInt
The first time the function is called it returns 1. Every other time it's called it returns 1+the number returned the last time it was called.
int counter()
{
return counter()++;
}
Is this valid?
Sounds like you want a static variable:
int counter()
{
static int count = 0; // initialized only once across all calls
return ++count;
}
You can use static.
int counter()
{
static int counter = 0;
return ++counter;
}
Static (in the function body context) is just a nice keyword that makes a variable essentially global, but not semantic-wise. So you cannot use it outside of the function, but it gets initialized with all the globals (it means once near the start point) and it is stored where other global variables are.
Just for the record, since c++ label. You could write a functor class, with a private count variable and overload operator(). Or even probably a template that would be able to wrap any function and count its calls.
Also note that in your try, you call counter recursively with no exit point with a fixed value. It will never finish (unless you count stack overflowing as end).
Apart from defining a static local variable in the function as it was already shown you can define a static class member function that will increment a static data member of the same class. For example
#include <iostream>
class Counter
{
public:
static int Count() { return ( ++cnt ); }
static int Reset() { return ( cnt = 0 ); }
private:
static int cnt;
};
int Counter::cnt = 0;
int main()
{
for ( int i = 0; i < 10; i++ ) std::cout << Counter::Count() << ' ';
std::cout << std::endl;
std::cout << "\nResetting the counter to " << Counter::Reset() << std::endl;
for ( int i = 0; i < 10; i++ ) std::cout << Counter::Count() << ' ';
std::cout << std::endl;
}
If using C++11, use a std::function and a mutable lambda expression:
#include <iostream>
#include <functional>
using namespace std;
int main()
{
function<int()> func = [&]()
{
static int i = 1;
return i++;
};
for(int i = 0; i < 10; i++)
cout << func() << endl;
return 0;
}
This question probably only makes sense for people with knowledge on programming languages supporting closures. If you don't, please do not comment "why would you like to do this?": there are tons of legitimate reasons to do that.
It is common in functional languages to define local functions that capture the already defined local variables. In C++, that would look like (but of course is illegal):
#include <iostream>
using namespace std;
int main()
{
int x = 0;
int f() { return x + 1; }
cout << f() << endl; // would print 1
x = 2;
cout << f() << endl; // would print 3
}
To allow this, C++11 introduces lambda functions, so it is actually possible to do it in a rather nice way (though, not as nice as it generally is in functional languages ;-) ):
#include <iostream>
using namespace std;
int main()
{
int x = 0;
auto f = [&] () { return x + 1; };
cout << f() << endl; // actually compiles and prints 1
x = 2;
cout << f() << endl; // actually compiles and prints 3
}
My question is: now that it is possible to automatically capture free variables by reference for functions, wouldn't that be nice to be possible to do for locally defined structs? Ideally, I would love to be able to write:
int main()
{
int x = 0;
struct A
{
int y;
A(int y) : y(y) {}
int f() { return x + y; };
};
A a1(1);
A a2(2);
cout << a1.f() << endl; // would print 1
cout << a2.f() << endl; // would print 2
x = 2;
cout << a1.f() << endl; // would print 3
cout << a2.f() << endl; // would print 4
}
The only workaround I've found is to manually pass as argument to the constructor all the non-local (free) variables, which is a bit of a pain when there are plenty of them:
#include <iostream>
using namespace std;
int main()
{
int x = 0;
struct A
{
// meaningful members
int y;
int f() { return x + y; };
// free variables
int & x;
// Constructor
A(
// meaningful arguments
int y,
// capturing free variables
int & x
) : y(y), x(x) {}
};
A a1(1, x);
A a2(2, x);
cout << a1.f() << endl; // prints 1
cout << a2.f() << endl; // prints 2
x = 2;
cout << a1.f() << endl; // prints 3
cout << a2.f() << endl; // prints 4
}
Do you know of any other workaround that would avoid manually passing as argument all free variables, or do you know if these kind of "environment-aware" locally-defined structs are considered for future extensions of C++? (i.e., C++1y?)
What you ask for is not available, but you can get similar results by combining functions with a combination of lambdas and binders:
auto lambda = [](int i) { return x+i; };
auto a1 = std::bind(lambda,1);
auto a2 = std::bind(lambda,2);
Depending on the amount and shape of changes, you could invert the solution and have a struct that takes the lambda with the capture and then adds it's own logic.
I don't find this particularly beautiful, and I'm not entirely sure it's compliant, but neither g++ nor clang++ complains about this:
#include <iostream>
int main()
{
int x = 1;
auto l = [&](int p){
auto ll0 = [&, p]{ return p + x + 5; };
auto ll1 = [&, p]{ return p + x * 2; };
struct
{
decltype(ll0) l0;
decltype(ll1) l1;
} ret{ll0, ll1};
return ret;
};
std::cout << l(42).l0() << '\n';
auto lo = l(21);
std::cout << lo.l1() << '\n';
}
I think the creation of the unnamed struct could possibly be automated by a macro.
The lambda expressions of C++ are the capturing mechanism and inline object literals of a sort. Depending on your exact purpose, they may be more convenient than a local struct definition.
As a motivating example, consider the following:
// environment
int offset = 42;
struct local_type {
// capture environment 'by-hand'
int offset;
// purpose of the local type is to expose two overloads
int operator()(int x) const
{ return x + offset; }
double operator()(double x) const
{ return x + offset; }
} f { offset };
You can turn this on its head by doing:
int offset = 42;
auto f = make_overload([=](int x) { return offset + x; },
[=](double x) { return offset + x; });
The lambda expressions take care of capturing, the make_overload combinator takes care of building the desired object -- here, one that has an overloaded operator(). (It would be implemented best by making use of inheritance.)
This approach makes sense if you know that you'll (re)use make_overload from various places. For one-time, special uses there's no avoiding writing a specialty type, whether local or not.