First off, apologies as this title probably made some of you c++ guys throw up in your mouth a little bit. However, let me explain what I'm trying to accomplish.
I'm passing a lambda (call it funcA) as a parameter to another function (call it funcB). Is there a way for me to declare variables within funcB and access them within the scope of funcA without passing them to funcA as parameters, similar to how javascript closures work?
A simple example (which fails obviously) would be as follows:
void funcB(std::function<void()> funcA) {
int testInt = 44;
funcA();
}
int main() {
funcB([&]() {
std::cout << testInt; // undefined identifier
});
return 0;
}
Well in this particular example the idiomatic way would be to use the return value:
void funcB(std::function<int()> funcA) {
int testInt = funcA();
}
int main() {
funcB([&]() {
int testInt = 44;
std::cout >> testInt;
return testInt;
});
return 0;
}
Output parameter can be used but less usual:
void funcB(std::function<void(int&)> funcA) {
int testInt = 44;
funcA(testInt);
}
int main() {
funcB([&](int& testInt) {
std::cout >> testInt;
});
return 0;
}
It appears that I had misunderstood exactly what was going on within javascript that was allowing this behavior to occur. After digging in deeper I see now that the only time variables are available within closures are when the variable definition is within the parent scope of the closure definition as such:
var funcB = function(funcA){
funcA();
};
(function(){
var firstVar = 1;
var secondVar = 2;
funcB(function(){
alert(firstVar + secondVar);
});
})();
This same behavior could be achieved using lambda capture (as suggested by user #Marek R in question comments) in C++ as follows:
void funcB(std::function<void()> funcA) {
funcA();
}
int main() {
int firstVar = 1;
int secondVar = 2;
funcB([&]() {
std::cout << firstVar + secondVar;
});
std::cin.get();
return 0;
}
Therefore my original question stemmed from not knowing what I didn't know. To achieve what I am actually after, I will have to pass parameters to my lambda or implement a different design pattern.
Related
So I'm trying to write a recursive function that keeps track of how often it got called. Because of its recursive nature I won't be able to define an iterator inside of it (or maybe it's possible via a pointer?), since it would be redefined whenever the function gets called. So i figured I could use a param of the function itself:
int countRecursive(int cancelCondition, int counter = 0)
{
if(cancelCondition > 0)
{
return countRecursive(--cancelCondition, ++counter);
}
else
{
return counter;
}
}
Now the problem I'm facing is, that the counter would be writeable by the caller of the function, and I want to avoid that.
Then again, it wouldn't help to declare the counter as a const, right?
Is there a way to restrict the variable's manipulation to the function itself?
Or maybe my approach is deeply flawed in the first place?
The only way I can think of solving this, is to use a kind of "wrapper-function" that keeps track of how often the recursive function got called.
An example of what I want to avoid:
//inside main()
int foo {5};
int countToZero = countRecursive(foo, 10);
//countToZero would be 15 instead of 5
The user using my function should not be able to initially set the counter (in this case to 10).
You can take you function as is, and wrap it. One way I have in mind, which completely encapsulates the wrapping is by making your function a static member of a local class. To demonstrate:
int countRecursive(int cancelCondition)
{
struct hidden {
static int countRecursive(int cancelCondition, int counter = 0) {
if(cancelCondition > 0)
{
return countRecursive(--cancelCondition, ++counter);
}
else
{
return counter;
}
}
};
return hidden::countRecursive(cancelCondition);
}
Local classes are a nifty but rarely seen feature of C++. They possess some limitations, but fortunately can have static member functions. No code from outside can ever pass hidden::countRecursive an invalid counter. It's entirely under the control of the countRecursive.
If you can use something else than a free function, I would suggest to use some kind of functor to hold the count, but in case you cant, you may try to use something like this using friendship to do the trick:
#include <memory>
class Counter;
int countRecursive(int cancelCondition, std::unique_ptr<Counter> counter = nullptr);
class Counter {
int count = 0;
private:
friend int countRecursive(int, std::unique_ptr<Counter>);
Counter() = default; // the constructor can only be call within the function
// thus nobody can provide one
};
int countRecursive(int cancelCondition, std::unique_ptr<Counter> c)
{
if (c == nullptr)
c = std::unique_ptr<Counter>(new Counter());
if(cancelCondition > 0)
{
c->count++;
return countRecursive(--cancelCondition, std::move(c));
}
else
{
return c->count;
}
}
int main() {
return countRecursive(12);
}
You can encapsulate the counter:
struct counterRecParam {
counterRecParam(int c) : cancelCondition(c),counter(0) {}
private:
int cancelCondition;
int counter;
friend int countRecursive(counterRecParam);
};
Now the caller cannot modify the counter, and you only need to modify the function slightly:
int countRecursive(counterRecParam crp)
{
if(crp.cancelCondition > 0)
{
--crp.cancelCondition;
++crp.counter;
return countRecursive(crp);
}
else
{
return crp.counter;
}
}
And the implicit conversion lets you call it with an int
counterRecursive(5);
One way to do this is to use a functor. Here's a simple example:
#include <iostream>
class counter
{
public:
unsigned operator()(unsigned m, unsigned n)
{
// increment the count on every iteration
++count;
// rest of the function
if (m == 0)
{
return n + 1;
}
if (n == 0)
{
return operator()(m - 1, 1);
}
return operator()(m - 1, operator()(m, n - 1));
}
std::size_t get_count() const
{
return count;
}
private:
// call count
std::size_t count = 0;
};
int main()
{
auto f = counter();
auto res = f(4, 0);
std::cout << "Result: " << res << "\nNumber of calls: " << f.get_count() << std::endl;
return 0;
}
Output:
Result: 13
Number of calls: 107
Since the count is stored in the object itself, the user cannot overwrite it.
Have you tried using "static" counter variable. Static variables gets initialized just once, and are best candidates to be used as counter variables.
Lambdas are an awesome way to create reusable code inside a function/method without polluting the parent class. They're a very functional replacement for C-style macros most of the time.
However, there's one bit of syntactic sugar from macros that I can't seem to replicate with a lambda, and that's the ability to exit from the containing function. For example, if I need to return while checking the range of a series of ints, I can do that easily with a macro:
const int xmin(1), xmax(5);
#define CHECK_RANGE(x) { if((x) < xmin || (x) > xmax) return false; }
bool myFunc(int myint) {
CHECK_RANGE(myint);
int anotherint = myint + 2;
CHECK_RANGE(anotherint);
return true;
}
Obviously this is an oversimplified example, but the basic premise is that I'm performing the same check over and over on different variables, and I think it's more readable to encapsulate the check and related exits. Still, I know that macros aren't very safe, especially when they get really complex. However, as far as I can tell, trying to do the equivalent lambda requires awkward additional checks like so:
const int xmin(1), xmax(5);
auto check_range = [&](int x) -> bool { return !(x < xmin || x > xmax); };
bool myFunc(int myint) {
if(!check_range(myint)) return false;
int anotherint = myint + 2;
if(!check_range(anotherint)) return false;
return true;
}
Is there a way to do this with a lambda? Or am I missing some alternative solution?
Edit: I recognize that returning from inside a macro is generally a bad idea unless significant precautions are taken. I'm just wondering if it's possible.
You are correct--there's no way to return from the caller from inside a lambda. Since a lambda can be captured and stored to be called later, from inside an arbitrary caller, doing so would result in unpredictable behavior.
class Foo
{
Foo(std::function<void(int)> const& callMeLater) : func(callMeLater) {}
void CallIt(int* arr, int count)
{
for (index = count; index--;)
func(count);
// do other stuff here.
}
std::function<void(int)> func;
};
int main()
{
auto find3 = [](int arr)
{
if (arr == 3)
return_from_caller; // making up syntax here.
};
Foo foo(find3);
};
Is there a way to do this with a lambda?
Not exactly like the macro but your lambda, instead of returning a bool, can throw a special exception (of type bool, by example)
auto check_range
= [](int x) { if ( (x < xmin) || (x > xmax) ) throw bool{false}; };
and the function myFunc() can intercept this special type
bool myFunc (int myint)
{
try
{
check_range(myint);
int anotherint = myint + 2;
check_range(anotherint);
return true;
}
catch ( bool e )
{ return e; }
}
For a single check_range() call, this is (I suppose) a bad idea; if you have a lot of calls, I suppose can be interesting.
The following is a full working example
#include <iostream>
constexpr int xmin{1}, xmax{5};
auto check_range
= [](int x) { if ( (x < xmin) || (x > xmax) ) throw bool{false}; };
bool myFunc (int myint)
{
try
{
check_range(myint);
int anotherint = myint + 2;
check_range(anotherint);
return true;
}
catch ( bool e )
{ return e; }
}
int main ()
{
std::cout << myFunc(0) << std::endl; // print 0
std::cout << myFunc(3) << std::endl; // print 1
std::cout << myFunc(7) << std::endl; // print 0
}
No better way to do this than just to use the return value of the lambda and then return from the calling function. Macros are ew for this.
As it stands in C++, that is the idiomatic way to exit from a function that uses another condition to determine whether or not to exit.
Not C++11, but people have hacked C++2a coroutines to basically do this.
It would look a bit like:
co_await check_range(foo);
where the co_await keyword indicates that in some cases, this coroutine could return early with an incomplete result. In your cases, this incomplete result would be non-resumabable error.
The playing around I saw was with optionals, and required using a shared ptr, but things may improve before it is standardized.
Let define a structure parser :
struct parser {
int (*buffer_push_strategy)();
int (*escape_buffer_push_strategy)();
int (*do_callback_strategy)();
};
I have an initialization function :
int parser_init() {
if (some_condition) {
parser->buffer_push_strategy = buffer_push_strategy1;
parser->escape_buffer_push_strategy = escape_buffer_push_strategy1;
parser->do_callback_strategy = do_callback_strategy1;
}
else {
parser->buffer_push_strategy = buffer_push_strategy2;
parser->escape_buffer_push_strategy = escape_buffer_push_strategy2;
parser->do_callback_strategy = do_callback_strategy2;
}
return 0;
}
where the strategy functions are defined somewhere.
Ok, so my interest is to determine which strategy has been used when I write the unit tests. Any idea how to accomplish that?
I saw something on internet about is_pointer function from C++ 11, but I don`t think this would help me.
parser is a variable:
struct parserT {
int (*buffer_push_strategy)();
int (*escape_buffer_push_strategy)();
int (*do_callback_strategy)();
} parser;
If you want to know which the strategy is, you could use:
int strategy= (parser->buffer_push_strategy == buffer_push_strategy1) ? 1 : 2;
Perhaps, you prefer to store the strategy number:
int parser_init() {
if (some_condition) {
parser->buffer_push_strategy = buffer_push_strategy1;
parser->escape_buffer_push_strategy = escape_buffer_push_strategy1;
parser->do_callback_strategy = do_callback_strategy1;
return 1;
}
else {
parser->buffer_push_strategy = buffer_push_strategy2;
parser->escape_buffer_push_strategy = escape_buffer_push_strategy2;
parser->do_callback_strategy = do_callback_strategy2;
return 2;
}
}
Then, you could init the parser as:
const int STRATEGY= parser_init();
You can compare function pointers
if(p.buffer_push_strategy == buffer_push_strategy1)
See https://ideone.com/QQzL1c
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Declaring and initializing a variable in a Conditional or Control statement in C++
Instead of this...
int value = get_value();
if ( value > 100 )
{
// Do something with value.
}
... is it possible to reduce the scope of value to only where it is needed:
if ( int value = get_value() > 100 )
{
// Obviously this doesn't work. get_value() > 100 returns true,
// which is implicitly converted to 1 and assigned to value.
}
If you want specific scope for value, you can introduce a scope block.
#include <iostream>
int get_value() {
return 101;
}
int main() {
{
int value = get_value();
if(value > 100)
std::cout << "Hey!";
} //value out of scope
}
Can you declare a variable and compare it within the if() statement? No.
Can you declare a variable and compare it in such a way that the scope is tightly-bound to the if() block? Yes!
You can either declare a variable:
if (int x = 5) {
// lol!
}
or you can do things with one:
int x = foo();
if (x == 5) {
// wahey!
}
You can't do both!
You can cheat a little where the only thing you need to do is compare with true, because the declaration itself evaluates to the value of the new object.
So, if you have:
int foo()
{
return 0;
}
Then this:
if (int x = foo()) {
// never reached
}
is equivalent to:
{
int x = foo();
if (x) {
// never reached
}
}
This final syntax, using a standalone scope block, is also your golden bullet for more complex expressions:
{
int x = foo();
if (x > bar()) {
// wahooza!
}
}
Put it in a function:
void goodName(int value) {
if(value > 100) {
// Do something with value.
}
}
//...
goodName(get_value());
How about using for instead?
for (int value = get_value(); value > 100; value = 0) {
//...
}
If you want to go C++11 on it, you can use a lambda:
[](int value = get_value()) {
if (value > 100) {
//...
std::cout << "value:" << value;
}
}();
Or you could just add an extra set of braces for a nested scope, although it's not exactly pretty:
{
int value = get_value();
if ( value > 100 )
{
// Do something with value.
}
}
//now value is out of scope
You can write a small function which can do the comparison for you and return the value the if comparison returns true, else return 0 to avoid executing the if block:
int greater_than(int right, int left)
{
return left > right ? left : 0;
}
Then use it as:
if ( int value = greater_than(100, get_value()))
{
//wow!
}
Or you can use for as other answer said. Or manually put braces to reduce the scope of the variable.
At any rate, I would not write such code in production code.
Don't write code for machines. Write code for humans. Machines will understand anything as long as you follow their grammar; humans understand what is readable to them. So readability should be your priority over unnecessary scoping.
In this particular case, you can bodge it:
if (int value = (get_value() > 100 ? get_value() : 0)) {
...
}
I don't really recommend it, though. It doesn't work for all possible tests that you might want to perform, and it calls get_value() twice.
I have the following Python snippet that I would like to reproduce using C++:
from itertools import count, imap
source = count(1)
pipe1 = imap(lambda x: 2 * x, source)
pipe2 = imap(lambda x: x + 1, pipe1)
sink = imap(lambda x: 3 * x, pipe2)
for i in sink:
print i
I've heard of Boost Phoenix, but I couldn't find an example of a lazy transform behaving in the same way as Python's imap.
Edit: to clarify my question, the idea is not only to apply functions in sequence using a for, but rather to be able to use algorithms like std::transform on infinite generators. The way the functions are composed (in a more functional language like dialect) is also important, as the next step is function composition.
Update: thanks bradgonesurfing, David Brown, and Xeo for the amazing answers! I chose Xeo's because it's the most concise and it gets me right where I wanted to be, but David's was very important into getting the concepts through. Also, bradgonesurfing's tipped Boost::Range :).
Employing Boost.Range:
int main(){
auto map = boost::adaptors::transformed; // shorten the name
auto sink = generate(1) | map([](int x){ return 2*x; })
| map([](int x){ return x+1; })
| map([](int x){ return 3*x; });
for(auto i : sink)
std::cout << i << "\n";
}
Live example including the generate function.
I think the most idiomatic way to do this in C++ is with iterators. Here is a basic iterator class that takes an iterator and applies a function to its result:
template<class Iterator, class Function>
class LazyIterMap
{
private:
Iterator i;
Function f;
public:
LazyIterMap(Iterator i, Function f) : i(i), f(f) {}
decltype(f(*i)) operator* () { return f(*i); }
void operator++ () { ++i; }
};
template<class Iterator, class Function>
LazyIterMap<Iterator, Function> makeLazyIterMap(Iterator i, Function f)
{
return LazyIterMap<Iterator, Function>(i, f);
}
This is just a basic example and is still incomplete as it has no way to check if you've reached the end of the iterable sequence.
Here's a recreation of your example python code (also defining a simple infinite counter class).
#include <iostream>
class Counter
{
public:
Counter (int start) : value(start) {}
int operator* () { return value; }
void operator++ () { ++value; }
private:
int value;
};
int main(int argc, char const *argv[])
{
Counter source(0);
auto pipe1 = makeLazyIterMap(source, [](int n) { return 2 * n; });
auto pipe2 = makeLazyIterMap(pipe1, [](int n) { return n + 1; });
auto sink = makeLazyIterMap(pipe2, [](int n) { return 3 * n; });
for (int i = 0; i < 10; ++i, ++sink)
{
std::cout << *sink << std::endl;
}
}
Apart from the class definitions (which are just reproducing what the python library functions do), the code is about as long as the python version.
I think the boost::rangex library is what you are looking for. It should work nicely with the new c++lambda syntax.
int pipe1(int val) {
return 2*val;
}
int pipe2(int val) {
return val+1;
}
int sink(int val) {
return val*3;
}
for(int i=0; i < SOME_MAX; ++i)
{
cout << sink(pipe2(pipe1(i))) << endl;
}
I know, it's not quite what you were expecting, but it certainly evaluates at the time you want it to, although not with an iterator iterface. A very related article is this:
Component programming in D
Edit 6/Nov/12:
An alternative, still sticking to bare C++, is to use function pointers and construct your own piping for the above functions (vector of function pointers from SO q: How can I store function pointer in vector?):
typedef std::vector<int (*)(int)> funcVec;
int runPipe(funcVec funcs, int sinkVal) {
int running = sinkVal;
for(funcVec::iterator it = funcs.begin(); it != funcs.end(); ++it) {
running = (*(*it))(running); // not sure of the braces and asterisks here
}
return running;
}
This is intended to run through all the functions in a vector of such and return the resulting value. Then you can:
funcVec funcs;
funcs.pushback(&pipe1);
funcs.pushback(&pipe2);
funcs.pushback(&sink);
for(int i=0; i < SOME_MAX; ++i)
{
cout << runPipe(funcs, i) << endl;
}
Of course you could also construct a wrapper for that via a struct (I would use a closure if C++ did them...):
struct pipeWork {
funcVec funcs;
int run(int i);
};
int pipeWork::run(int i) {
//... guts as runPipe, or keep it separate and call:
return runPipe(funcs, i);
}
// later...
pipeWork kitchen;
kitchen.funcs = someFuncs;
int (*foo) = &kitchen.run();
cout << foo(5) << endl;
Or something like that. Caveat: No idea what this will do if the pointers are passed between threads.
Extra caveat: If you want to do this with varying function interfaces, you will end up having to have a load of void *(void *)(void *) functions so that they can take whatever and emit whatever, or lots of templating to fix the kind of pipe you have. I suppose ideally you'd construct different kinds of pipe for different interfaces between functions, so that a | b | c works even when they are passing different types between them. But I'm going to guess that that's largely what the Boost stuff is doing.
Depending on the simplicity of the functions :
#define pipe1(x) 2*x
#define pipe2(x) pipe1(x)+1
#define sink(x) pipe2(x)*3
int j = 1
while( ++j > 0 )
{
std::cout << sink(j) << std::endl;
}