Related
I would like to iterate through a struct which is defined in other library whose source is not under my control. So any lib which requires to define the struct with its own macros/adaptors like previous questions is not usable here. I found the closest way is using boost::hana. However, it still requires to fill up an adaptor before I can iterate through it. I attached an example here. I wonder is there any way I can automate the BOOST_HANA_ADAPT_STRUCT then I do not need to fill up all the struct member names in there (those structs in total have more than hundred members).
#include <iostream>
#include <boost/hana.hpp>
#include <typeinfo>
namespace hana=boost::hana;
struct adapt_test
{
std::string name;
int data;
};
BOOST_HANA_ADAPT_STRUCT(
adapt_test
, name
, data
);
auto names = hana::transform(hana::accessors<adapt_test>(), hana::first);
int main() {
hana::for_each(
names,
[] (auto item)
{
std::cout << hana::to<char const *>(item) << std::endl;
}
);
adapt_test s1{"a", 2};
hana::for_each(
s1,
[] (auto pair)
{
std::cout << hana::to<char const *>(hana::first(pair)) << "=" << hana::second(pair) << std::endl;
}
);
return 0;
}
You can use Boost Flat Reflection like:
struct adapt_test
{
std::string name;
int data;
};
adapt_test s1{"a", 2};
std::cout << boost::pfr::get<0>(s1) << std::endl;
std::cout << boost::pfr::get<1>(s1) << std::endl;
boost::pfr::flat_for_each_field(s1, [] (const auto& field) { std::cout << field << std::endl; } );
P.S. Respect for #apolukhin for this library.
The basic answer to your question is no.
C++ does not treat identifiers as string literal (it could be indeed useful in some cases), and there is no bridge unfortunately between these kind of strings.
Hopefully, some standard one day will bring this ability, relieving us from having to go through macros or code generation, or maybe doing differently like this: telling "please treat my struct A { int x, y; } as a pair", where the meaning would be to match type of first and second to the members x and y and then building the types so that it works, it would be really useful for tuples as well. A kind of structured template matching.
Currently the best that can be done to my knowledge is to match structs to tuple without the names (as of C++17) because of the above limitation, such as with boost::hana or boost::fusion as you do.
I mean something like:
int main()
{
void a()
{
// code
}
a();
return 0;
}
Modern C++ - Yes with lambdas!
In current versions of c++ (C++11, C++14, and C++17), you can have functions inside functions in the form of a lambda:
int main() {
// This declares a lambda, which can be called just like a function
auto print_message = [](std::string message)
{
std::cout << message << "\n";
};
// Prints "Hello!" 10 times
for(int i = 0; i < 10; i++) {
print_message("Hello!");
}
}
Lambdas can also modify local variables through **capture-by-reference*. With capture-by-reference, the lambda has access to all local variables declared in the lambda's scope. It can modify and change them normally.
int main() {
int i = 0;
// Captures i by reference; increments it by one
auto addOne = [&] () {
i++;
};
while(i < 10) {
addOne(); //Add 1 to i
std::cout << i << "\n";
}
}
C++98 and C++03 - Not directly, but yes with static functions inside local classes
C++ doesn't support that directly.
That said, you can have local classes, and they can have functions (non-static or static), so you can get this to some extend, albeit it's a bit of a kludge:
int main() // it's int, dammit!
{
struct X { // struct's as good as class
static void a()
{
}
};
X::a();
return 0;
}
However, I'd question the praxis. Everyone knows (well, now that you do, anyway :)) C++ doesn't support local functions, so they are used to not having them. They are not used, however, to that kludge. I would spend quite a while on this code to make sure it's really only there to allow local functions. Not good.
For all intents and purposes, C++ supports this via lambdas:1
int main() {
auto f = []() { return 42; };
std::cout << "f() = " << f() << std::endl;
}
Here, f is a lambda object that acts as a local function in main. Captures can be specified to allow the function to access local objects.
Behind the scenes, f is a function object (i.e. an object of a type that provides an operator()). The function object type is created by the compiler based on the lambda.
1 since C++11
Local classes have already been mentioned, but here is a way to let them appear even more as local functions, using an operator() overload and an anonymous class:
int main() {
struct {
unsigned int operator() (unsigned int val) const {
return val<=1 ? 1 : val*(*this)(val-1);
}
} fac;
std::cout << fac(5) << '\n';
}
I don't advise on using this, it's just a funny trick (can do, but imho shouldn't).
2014 Update:
With the rise of C++11 a while back, you can now have local functions whose syntax is a little reminiscient of JavaScript:
auto fac = [] (unsigned int val) {
return val*42;
};
For a recursive function, compile-time type deduction is not supported:
function<int(int)> factorial{ [&](int n)
{
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
} };
You can't have local functions in C++. However, C++11 has lambdas. Lambdas are basically variables that work like functions.
A lambda has the type std::function (actually that's not quite true, but in most cases you can suppose it is). To use this type, you need to #include <functional>. std::function is a template, taking as template argument the return type and the argument types, with the syntax std::function<ReturnType(ArgumentTypes)>. For example, std::function<int(std::string, float)> is a lambda returning an int and taking two arguments, one std::string and one float. The most common one is std::function<void()>, which returns nothing and takes no arguments.
Once a lambda is declared, it is called just like a normal function, using the syntax lambda(arguments).
To define a lambda, use the syntax [captures](arguments){code} (there are other ways of doing it, but I won't mention them here). arguments is what arguments the lambda takes, and code is the code that should be run when the lambda is called. Usually you put [=] or [&] as captures. [=] means that you capture all variables in the scope in which the value is defined by value, which means that they will keep the value that they had when the lambda was declared. [&] means that you capture all variables in the scope by reference, which means that they will always have their current value, but if they are erased from memory the program will crash. Here are some examples:
#include <functional>
#include <iostream>
int main(){
int x = 1;
std::function<void()> lambda1 = [=](){
std::cout << x << std::endl;
};
std::function<void()> lambda2 = [&](){
std::cout << x << std::endl;
};
x = 2;
lambda1(); //Prints 1 since that was the value of x when it was captured and x was captured by value with [=]
lambda2(); //Prints 2 since that's the current value of x and x was captured by reference with [&]
std::function<void()> lambda3 = [](){}, lambda4 = [](){}; //I prefer to initialize these since calling an uninitialized lambda is undefined behavior.
//[](){} is the empty lambda.
{
int y = 3; //y will be deleted from the memory at the end of this scope
lambda3 = [=](){
std::cout << y << endl;
};
lambda4 = [&](){
std::cout << y << endl;
};
}
lambda3(); //Prints 3, since that's the value y had when it was captured
lambda4(); //Causes the program to crash, since y was captured by reference and y doesn't exist anymore.
//This is a bit like if you had a pointer to y which now points nowhere because y has been deleted from the memory.
//This is why you should be careful when capturing by reference.
return 0;
}
You can also capture specific variables by specifying their names. Just specifying their name will capture them by value, specifying their name with a & before will capture them by reference. For example, [=, &foo] will capture all variables by value except foo which will be captured by reference, and [&, foo] will capture all variables by reference except foo which will be captured by value. You can also capture only specific variables, for example [&foo] will capture foo by reference and will capture no other variables. You can also capture no variables at all by using []. If you try to use a variable in a lambda that you didn't capture, it won't compile. Here is an example:
#include <functional>
int main(){
int x = 4, y = 5;
std::function<void(int)> myLambda = [y](int z){
int xSquare = x * x; //Compiler error because x wasn't captured
int ySquare = y * y; //OK because y was captured
int zSquare = z * z; //OK because z is an argument of the lambda
};
return 0;
}
You can't change the value of a variable that was captured by value inside a lambda (variables captured by value have a const type inside the lambda). To do so, you need to capture the variable by reference. Here is an exampmle:
#include <functional>
int main(){
int x = 3, y = 5;
std::function<void()> myLambda = [x, &y](){
x = 2; //Compiler error because x is captured by value and so it's of type const int inside the lambda
y = 2; //OK because y is captured by reference
};
x = 2; //This is of course OK because we're not inside the lambda
return 0;
}
Also, calling uninitialized lambdas is undefined behavior and will usually cause the program to crash. For example, never do this:
std::function<void()> lambda;
lambda(); //Undefined behavior because lambda is uninitialized
Examples
Here is the code for what you wanted to do in your question using lambdas:
#include <functional> //Don't forget this, otherwise you won't be able to use the std::function type
int main(){
std::function<void()> a = [](){
// code
}
a();
return 0;
}
Here is a more advanced example of a lambda:
#include <functional> //For std::function
#include <iostream> //For std::cout
int main(){
int x = 4;
std::function<float(int)> divideByX = [x](int y){
return (float)y / (float)x; //x is a captured variable, y is an argument
}
std::cout << divideByX(3) << std::endl; //Prints 0.75
return 0;
}
No.
What are you trying to do?
workaround:
int main(void)
{
struct foo
{
void operator()() { int a = 1; }
};
foo b;
b(); // call the operator()
}
Starting with C++ 11 you can use proper lambdas. See the other answers for more details.
Old answer: You can, sort-of, but you have to cheat and use a dummy class:
void moo()
{
class dummy
{
public:
static void a() { printf("I'm in a!\n"); }
};
dummy::a();
dummy::a();
}
No, it's not allowed. Neither C nor C++ support this feature by default, however TonyK points out (in the comments) that there are extensions to the GNU C compiler that enable this behavior in C.
You cannot define a free function inside another in C++.
As others have mentioned, you can use nested functions by using the gnu language extensions in gcc. If you (or your project) sticks to the gcc toolchain, your code will be mostly portable across the different architectures targeted by the gcc compiler.
However, if there is a possible requirement that you might need to compile code with a different toolchain, then I'd stay away from such extensions.
I'd also tread with care when using nested functions. They are a beautiful solution for managing the structure of complex, yet cohesive blocks of code (the pieces of which are not meant for external/general use.) They are also very helpful in controlling namespace pollution (a very real concern with naturally complex/long classes in verbose languages.)
But like anything, they can be open to abuse.
It is sad that C/C++ does not support such features as an standard. Most pascal variants and Ada do (almost all Algol-based languages do). Same with JavaScript. Same with modern languages like Scala. Same with venerable languages like Erlang, Lisp or Python.
And just as with C/C++, unfortunately, Java (with which I earn most of my living) does not.
I mention Java here because I see several posters suggesting usage of classes and class' methods as alternatives to nested functions. And that's also the typical workaround in Java.
Short answer: No.
Doing so tend to introduce artificial, needless complexity on a class hierarchy. With all things being equal, the ideal is to have a class hierarchy (and its encompassing namespaces and scopes) representing an actual domain as simple as possible.
Nested functions help deal with "private", within-function complexity. Lacking those facilities, one should try to avoid propagating that "private" complexity out and into one's class model.
In software (and in any engineering discipline), modeling is a matter of trade-offs. Thus, in real life, there will be justified exceptions to those rules (or rather guidelines). Proceed with care, though.
All this tricks just look (more or less) as local functions, but they don't work like that. In a local function you can use local variables of it's super functions. It's kind of semi-globals. Non of these tricks can do that. The closest is the lambda trick from c++0x, but it's closure is bound in definition time, not the use time.
Let me post a solution here for C++03 that I consider the cleanest possible.*
#define DECLARE_LAMBDA(NAME, RETURN_TYPE, FUNCTION) \
struct { RETURN_TYPE operator () FUNCTION } NAME;
...
int main(){
DECLARE_LAMBDA(demoLambda, void, (){
cout<<"I'm a lambda!"<<endl;
});
demoLambda();
DECLARE_LAMBDA(plus, int, (int i, int j){
return i+j;
});
cout << "plus(1,2)=" << plus(1,2) << endl;
return 0;
}
(*) in the C++ world using macros is never considered clean.
But we can declare a function inside main():
int main()
{
void a();
}
Although the syntax is correct, sometimes it can lead to the "Most vexing parse":
#include <iostream>
struct U
{
U() : val(0) {}
U(int val) : val(val) {}
int val;
};
struct V
{
V(U a, U b)
{
std::cout << "V(" << a.val << ", " << b.val << ");\n";
}
~V()
{
std::cout << "~V();\n";
}
};
int main()
{
int five = 5;
V v(U(five), U());
}
=> no program output.
(Only Clang warning after compilation).
C++'s most vexing parse again
Yes, and you can do things with them that even C++20 Lambdas don't support. Namely, pure recursive calls to themselves & related functions.
For example, the Collatz Conjecture is that a certain simple recursive function will ultimately produce "1" for ANY positive integer N. Using an explicit local struct and functions, I can write a single self-contained function to run the test for any "N".
constexpr std::optional<int> testCollatzConjecture(int N) {
struct CollatzCallbacks {
constexpr static int onEven(int n) {
return recurse(n >> 1); // AKA "n/2"
}
constexpr static int onOdd(int n) {
if(n==1) return 1; // Break recursion. n==1 is only possible when n is odd.
return recurse(3 * n + 1);
}
constexpr static int recurse(int n) {
return (n%2) ? onOdd(n) : onEven(n); // (n%2) == 1 when n is odd
}
};
// Error check
if(N < 0) return {};
// Recursive call.
return CollatzCallbacks::recurse(N);
}
Notice some things that even c++20 lambdas couldn't do here:
I didn't need std::function<> glue OR lambda captures ("[&]") just to enable my local recursive functions call themselves, or each other. I needed 3 plain-old-functions with names, and that's all I had to write.
My code is more readable and (due to (1)) will also run much faster.
I cleanly separate the recursive logic in "CollatzCallbacks" from the rest of "testCollatzConjecture". It all runs in an isolated sandbox.
I was able to make everything "constexpr" and state-less, so it can all run at compile time for any constant value. AFAIK I'd need c++23 just to achieve the recursion part with state-less lambdas.
Remember: Lambda functions are really just compiler-generated local structs like "CollatzCallbacks", only they're unnamed and only have a single "operator()" member function. You can always write more complex local structs and functions directly, especially in cases like this where you really need them.
EDIT: This question was originally titled "Using std::bind to create inline function," but that's not really what I'm after: I just want a simple way to alias functions.
I would like to expose std::chrono::high_resolution_clock::now as a standalone function. That is, I would like to do the following:
auto current_time = std::bind(std::chrono::high_resolution_clock::now);
Unfortunately, since this is in a header file, it results in multiple definitions of current_time at link-time. Is there a way to return an inline function from std::bind?
Here's what I do if I want to create a simple function alias
constexpr auto &&now = std::chrono::high_resolution_clock::now;
and if I want to create a full wrapper alias that will be inlined
template<typename ... Args>
inline constexpr auto now(Args &&... args) -> decltype(std::chrono::high_resolution_clock::now(std::forward<Args>(args)...)){
return std::chrono::high_resolution_clock::now(std::forward<Args>(args)...);
}
The reason why I use a universal reference auto&& in the alias definition is because of the possibility of addressof(now) == addressof(std::chrono::high_resolution_clock::now).
On my system with G++ 4.9.2 running this:
constexpr auto &&now_ref = std::chrono::high_resolution_clock::now;
constexpr auto now_var = std::chrono::high_resolution_clock::now;
template<typename ... Args>
inline constexpr auto now_wrapper(Args &&... args)
-> decltype(std::chrono::high_resolution_clock::now(std::forward<Args>(args)...)){
return std::chrono::high_resolution_clock::now(std::forward<Args>(args)...);
}
int main(int argc, char *argv[]){
std::cout << std::hex << std::showbase;
std::cout << (uintptr_t)std::addressof(std::chrono::high_resolution_clock::now) << '\n';
std::cout << (uintptr_t)std::addressof(now_wrapper<>) << '\n';
std::cout << (uintptr_t)std::addressof(now_var) << '\n';
std::cout << (uintptr_t)std::addressof(now_ref) << '\n';
}
I get the following results:
0x4007c0
0x400a50
0x400ae8
0x4007c0
Showing that only the auto&& is actually a direct alias of the function, whereas all other methods have some level of indirection. (although, after compilation they may be replaced by inlined function calls. maybe.)
I don't think there is anyway to do this as bind is not constexpr.
Also lambdas are not constexpr-able.
Edit: there is this trick to make a constexpr-like lambda http://pfultz2.com/blog/2014/09/02/static-lambda/
Adding another answer 'cause it takes a very different tack to what you want.
std::bind isn't necessary in this case, because no 'binding' is happening.
However I feel this could lead to some confusing problems down the line, since current_time isn't really an alias in the same way that using delcarations are.
#include <iostream>
#include <chrono>
using namespace std;
auto constexpr current_time = std::chrono::high_resolution_clock::now;
int main() {
auto now = current_time();
cout << std::chrono::system_clock::to_time_t(now) << endl;
return 0;
}
Using GCC it is possible to create a "function alias", but only for functions which are defined in the same translation unit and for which you know the mangled name, so it's not possible to do reliably for std::chrono::high_resolution_clock::now()
See the alias attribute at https://gcc.gnu.org/onlinedocs/gcc/Function-Attributes.html
Keep it simple.
const auto current_time = std::chrono::high_resolution_clock::now;
I am trying to write a C++ class that allows me to access certain matrix elements by a string lookup. I wanted to create a 'static' class that can do this, such as:
#include <unordered_map>
namespace Mine {
static double AA[3][4] = {
{5.04964676394959,-0.693207030363152,0.0422140829479668,-0.000968959310672217},
{2.6044054979329,0.288475262243944,-0.0208805589126506,0.000380899394040856},
{-4.32707864788065,1.07090008760872,-0.0777874445746693,0.00165150952598117}
};
static unordered_map<std::string, double[3][4]> Mine::parameter_store = { {"AA", AA}};
With the idea being that I would have several matrices, and could look them up based on a key. However, this seems to totally and utterly fail with the following error:
error: object expression of non-scalar type 'double [3][4]' cannot be used in a pseudo-destructor expression
Is it possible to build a lookup table this way in C++?
#include <unordered_map>
#include <vector>
namespace Mine{
template<class T>
using Matrix = std::vector<std::vector<T>>;
Matrix<double> AA = {
{5.04964676394959,-0.693207030363152,0.0422140829479668,-0.000968959310672217},
{2.6044054979329,0.288475262243944,-0.0208805589126506,0.000380899394040856},
{-4.32707864788065,1.07090008760872,-0.0777874445746693,0.00165150952598117}
};
static std::unordered_map<std::string, Matrix<double>* > parameter_store = { {"AA", &AA}};
}
#include <iostream>
int main()
{
std::cout << (*Mine::parameter_store["AA"])[0][0] << std::endl;
std::cout << (*Mine::parameter_store["AA"])[0][1] << std::endl;
std::cout << (*Mine::parameter_store["AA"])[1][2] << std::endl;
}
output
5.04965
-0.693207
-0.0208806
The Matrix<> template used here causes each row to store its length even though that's redundant. You can avoid this by used a std::array (but then you're locked into each matrix having equal dimensions since that's part of the type information) or using some library like Boost that provides a multidimensional array. That's an extremely small inefficiency though and unless you know you need to it might be best to not worry about that.
You can try wrapping double[3][4] in a structure/class
structure myMatrix {
double arr[3][4];
//if you want to initialize it
myMatrix(double[3][4] p){
//copy matrix here
}
};
I mean something like:
int main()
{
void a()
{
// code
}
a();
return 0;
}
Modern C++ - Yes with lambdas!
In current versions of c++ (C++11, C++14, and C++17), you can have functions inside functions in the form of a lambda:
int main() {
// This declares a lambda, which can be called just like a function
auto print_message = [](std::string message)
{
std::cout << message << "\n";
};
// Prints "Hello!" 10 times
for(int i = 0; i < 10; i++) {
print_message("Hello!");
}
}
Lambdas can also modify local variables through **capture-by-reference*. With capture-by-reference, the lambda has access to all local variables declared in the lambda's scope. It can modify and change them normally.
int main() {
int i = 0;
// Captures i by reference; increments it by one
auto addOne = [&] () {
i++;
};
while(i < 10) {
addOne(); //Add 1 to i
std::cout << i << "\n";
}
}
C++98 and C++03 - Not directly, but yes with static functions inside local classes
C++ doesn't support that directly.
That said, you can have local classes, and they can have functions (non-static or static), so you can get this to some extend, albeit it's a bit of a kludge:
int main() // it's int, dammit!
{
struct X { // struct's as good as class
static void a()
{
}
};
X::a();
return 0;
}
However, I'd question the praxis. Everyone knows (well, now that you do, anyway :)) C++ doesn't support local functions, so they are used to not having them. They are not used, however, to that kludge. I would spend quite a while on this code to make sure it's really only there to allow local functions. Not good.
For all intents and purposes, C++ supports this via lambdas:1
int main() {
auto f = []() { return 42; };
std::cout << "f() = " << f() << std::endl;
}
Here, f is a lambda object that acts as a local function in main. Captures can be specified to allow the function to access local objects.
Behind the scenes, f is a function object (i.e. an object of a type that provides an operator()). The function object type is created by the compiler based on the lambda.
1 since C++11
Local classes have already been mentioned, but here is a way to let them appear even more as local functions, using an operator() overload and an anonymous class:
int main() {
struct {
unsigned int operator() (unsigned int val) const {
return val<=1 ? 1 : val*(*this)(val-1);
}
} fac;
std::cout << fac(5) << '\n';
}
I don't advise on using this, it's just a funny trick (can do, but imho shouldn't).
2014 Update:
With the rise of C++11 a while back, you can now have local functions whose syntax is a little reminiscient of JavaScript:
auto fac = [] (unsigned int val) {
return val*42;
};
For a recursive function, compile-time type deduction is not supported:
function<int(int)> factorial{ [&](int n)
{
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
} };
You can't have local functions in C++. However, C++11 has lambdas. Lambdas are basically variables that work like functions.
A lambda has the type std::function (actually that's not quite true, but in most cases you can suppose it is). To use this type, you need to #include <functional>. std::function is a template, taking as template argument the return type and the argument types, with the syntax std::function<ReturnType(ArgumentTypes)>. For example, std::function<int(std::string, float)> is a lambda returning an int and taking two arguments, one std::string and one float. The most common one is std::function<void()>, which returns nothing and takes no arguments.
Once a lambda is declared, it is called just like a normal function, using the syntax lambda(arguments).
To define a lambda, use the syntax [captures](arguments){code} (there are other ways of doing it, but I won't mention them here). arguments is what arguments the lambda takes, and code is the code that should be run when the lambda is called. Usually you put [=] or [&] as captures. [=] means that you capture all variables in the scope in which the value is defined by value, which means that they will keep the value that they had when the lambda was declared. [&] means that you capture all variables in the scope by reference, which means that they will always have their current value, but if they are erased from memory the program will crash. Here are some examples:
#include <functional>
#include <iostream>
int main(){
int x = 1;
std::function<void()> lambda1 = [=](){
std::cout << x << std::endl;
};
std::function<void()> lambda2 = [&](){
std::cout << x << std::endl;
};
x = 2;
lambda1(); //Prints 1 since that was the value of x when it was captured and x was captured by value with [=]
lambda2(); //Prints 2 since that's the current value of x and x was captured by reference with [&]
std::function<void()> lambda3 = [](){}, lambda4 = [](){}; //I prefer to initialize these since calling an uninitialized lambda is undefined behavior.
//[](){} is the empty lambda.
{
int y = 3; //y will be deleted from the memory at the end of this scope
lambda3 = [=](){
std::cout << y << endl;
};
lambda4 = [&](){
std::cout << y << endl;
};
}
lambda3(); //Prints 3, since that's the value y had when it was captured
lambda4(); //Causes the program to crash, since y was captured by reference and y doesn't exist anymore.
//This is a bit like if you had a pointer to y which now points nowhere because y has been deleted from the memory.
//This is why you should be careful when capturing by reference.
return 0;
}
You can also capture specific variables by specifying their names. Just specifying their name will capture them by value, specifying their name with a & before will capture them by reference. For example, [=, &foo] will capture all variables by value except foo which will be captured by reference, and [&, foo] will capture all variables by reference except foo which will be captured by value. You can also capture only specific variables, for example [&foo] will capture foo by reference and will capture no other variables. You can also capture no variables at all by using []. If you try to use a variable in a lambda that you didn't capture, it won't compile. Here is an example:
#include <functional>
int main(){
int x = 4, y = 5;
std::function<void(int)> myLambda = [y](int z){
int xSquare = x * x; //Compiler error because x wasn't captured
int ySquare = y * y; //OK because y was captured
int zSquare = z * z; //OK because z is an argument of the lambda
};
return 0;
}
You can't change the value of a variable that was captured by value inside a lambda (variables captured by value have a const type inside the lambda). To do so, you need to capture the variable by reference. Here is an exampmle:
#include <functional>
int main(){
int x = 3, y = 5;
std::function<void()> myLambda = [x, &y](){
x = 2; //Compiler error because x is captured by value and so it's of type const int inside the lambda
y = 2; //OK because y is captured by reference
};
x = 2; //This is of course OK because we're not inside the lambda
return 0;
}
Also, calling uninitialized lambdas is undefined behavior and will usually cause the program to crash. For example, never do this:
std::function<void()> lambda;
lambda(); //Undefined behavior because lambda is uninitialized
Examples
Here is the code for what you wanted to do in your question using lambdas:
#include <functional> //Don't forget this, otherwise you won't be able to use the std::function type
int main(){
std::function<void()> a = [](){
// code
}
a();
return 0;
}
Here is a more advanced example of a lambda:
#include <functional> //For std::function
#include <iostream> //For std::cout
int main(){
int x = 4;
std::function<float(int)> divideByX = [x](int y){
return (float)y / (float)x; //x is a captured variable, y is an argument
}
std::cout << divideByX(3) << std::endl; //Prints 0.75
return 0;
}
No.
What are you trying to do?
workaround:
int main(void)
{
struct foo
{
void operator()() { int a = 1; }
};
foo b;
b(); // call the operator()
}
Starting with C++ 11 you can use proper lambdas. See the other answers for more details.
Old answer: You can, sort-of, but you have to cheat and use a dummy class:
void moo()
{
class dummy
{
public:
static void a() { printf("I'm in a!\n"); }
};
dummy::a();
dummy::a();
}
No, it's not allowed. Neither C nor C++ support this feature by default, however TonyK points out (in the comments) that there are extensions to the GNU C compiler that enable this behavior in C.
You cannot define a free function inside another in C++.
As others have mentioned, you can use nested functions by using the gnu language extensions in gcc. If you (or your project) sticks to the gcc toolchain, your code will be mostly portable across the different architectures targeted by the gcc compiler.
However, if there is a possible requirement that you might need to compile code with a different toolchain, then I'd stay away from such extensions.
I'd also tread with care when using nested functions. They are a beautiful solution for managing the structure of complex, yet cohesive blocks of code (the pieces of which are not meant for external/general use.) They are also very helpful in controlling namespace pollution (a very real concern with naturally complex/long classes in verbose languages.)
But like anything, they can be open to abuse.
It is sad that C/C++ does not support such features as an standard. Most pascal variants and Ada do (almost all Algol-based languages do). Same with JavaScript. Same with modern languages like Scala. Same with venerable languages like Erlang, Lisp or Python.
And just as with C/C++, unfortunately, Java (with which I earn most of my living) does not.
I mention Java here because I see several posters suggesting usage of classes and class' methods as alternatives to nested functions. And that's also the typical workaround in Java.
Short answer: No.
Doing so tend to introduce artificial, needless complexity on a class hierarchy. With all things being equal, the ideal is to have a class hierarchy (and its encompassing namespaces and scopes) representing an actual domain as simple as possible.
Nested functions help deal with "private", within-function complexity. Lacking those facilities, one should try to avoid propagating that "private" complexity out and into one's class model.
In software (and in any engineering discipline), modeling is a matter of trade-offs. Thus, in real life, there will be justified exceptions to those rules (or rather guidelines). Proceed with care, though.
All this tricks just look (more or less) as local functions, but they don't work like that. In a local function you can use local variables of it's super functions. It's kind of semi-globals. Non of these tricks can do that. The closest is the lambda trick from c++0x, but it's closure is bound in definition time, not the use time.
Let me post a solution here for C++03 that I consider the cleanest possible.*
#define DECLARE_LAMBDA(NAME, RETURN_TYPE, FUNCTION) \
struct { RETURN_TYPE operator () FUNCTION } NAME;
...
int main(){
DECLARE_LAMBDA(demoLambda, void, (){
cout<<"I'm a lambda!"<<endl;
});
demoLambda();
DECLARE_LAMBDA(plus, int, (int i, int j){
return i+j;
});
cout << "plus(1,2)=" << plus(1,2) << endl;
return 0;
}
(*) in the C++ world using macros is never considered clean.
But we can declare a function inside main():
int main()
{
void a();
}
Although the syntax is correct, sometimes it can lead to the "Most vexing parse":
#include <iostream>
struct U
{
U() : val(0) {}
U(int val) : val(val) {}
int val;
};
struct V
{
V(U a, U b)
{
std::cout << "V(" << a.val << ", " << b.val << ");\n";
}
~V()
{
std::cout << "~V();\n";
}
};
int main()
{
int five = 5;
V v(U(five), U());
}
=> no program output.
(Only Clang warning after compilation).
C++'s most vexing parse again
Yes, and you can do things with them that even C++20 Lambdas don't support. Namely, pure recursive calls to themselves & related functions.
For example, the Collatz Conjecture is that a certain simple recursive function will ultimately produce "1" for ANY positive integer N. Using an explicit local struct and functions, I can write a single self-contained function to run the test for any "N".
constexpr std::optional<int> testCollatzConjecture(int N) {
struct CollatzCallbacks {
constexpr static int onEven(int n) {
return recurse(n >> 1); // AKA "n/2"
}
constexpr static int onOdd(int n) {
if(n==1) return 1; // Break recursion. n==1 is only possible when n is odd.
return recurse(3 * n + 1);
}
constexpr static int recurse(int n) {
return (n%2) ? onOdd(n) : onEven(n); // (n%2) == 1 when n is odd
}
};
// Error check
if(N < 0) return {};
// Recursive call.
return CollatzCallbacks::recurse(N);
}
Notice some things that even c++20 lambdas couldn't do here:
I didn't need std::function<> glue OR lambda captures ("[&]") just to enable my local recursive functions call themselves, or each other. I needed 3 plain-old-functions with names, and that's all I had to write.
My code is more readable and (due to (1)) will also run much faster.
I cleanly separate the recursive logic in "CollatzCallbacks" from the rest of "testCollatzConjecture". It all runs in an isolated sandbox.
I was able to make everything "constexpr" and state-less, so it can all run at compile time for any constant value. AFAIK I'd need c++23 just to achieve the recursion part with state-less lambdas.
Remember: Lambda functions are really just compiler-generated local structs like "CollatzCallbacks", only they're unnamed and only have a single "operator()" member function. You can always write more complex local structs and functions directly, especially in cases like this where you really need them.