I'm writing some code that would greatly benefit from the concise syntax of lambdas, which were introduced with C++ 11. Is this supported by the compiler?
How do I specify the compiler flags when compiling using Energia or embedXcode?
As of February 2018, up to C++14 is supported with some limitations:
http://processors.wiki.ti.com/index.php/C%2B%2B_Support_in_TI_Compilers
There isn't much about this topic on the TI site, or, at least, I don't know enough C++ to give you a detailed and precise response.
The implementation of the embedded ABI is described in this document that is mainly a derivation of the Itanium C++ ABI. It explains nothing about the implementation of lambdas nor the auto, keyword (or probably I'm not able to derive this information from the documentation).
Thus I decided to directly test in Energia. Apparently the g++ version is 4.6.3, thus it should support both.
And in fact (from a compilation point of view, I don't have my MSP here to test the code) it can compile something like:
// In template.hpp
#ifndef TEMPLATE_HPP_
#define TEMPLATE_HPP_
template<class T>
T func(T a) {
auto c = [&](int n) { return n + a; };
return c(0);
}
#endif /* TEMPLATE_HPP_ */
// in the sketch main
#include "template.hpp"
void setup() { int b = func<int>(0); }
void loop() { }
(the template works only if in an header, in the main sketch raises an error). To compile this sketch I had to modify one internal file of the editor. The maximum supported standard seems to be -std=c++0x, and the compilation flags are in the file:
$ENERGIA_ROOT/hardware/energia/msp430/platform.txt
in my setup the root is in /opt/energia. Inside that file I modified line 32 (compiler.cpp.flags) and added the option. Notice that -std=c++11 is not supported (raises an error).
compiler.cpp.flags=-std=c++0x -c -g -O2 {compiler.mlarge_flag} {compiler.warning_flags} -fno-exceptions -ffunction-sections -fdata-sections -fno-threadsafe-statics -MMD
Unfortunately I have zero experience with embedXcode :\
Mimic std::function
std::function is not provided, thus you have to write some sort of class that mimics it. Something like:
// callback.hpp
#ifndef CALLBACK_HPP_
#define CALLBACK_HPP_
template <class RET, class ARG>
class Callback {
RET (*_f)(ARG);
public:
Callback() : _f(0) { };
Callback(RET (*f)(ARG)) : _f(f) { };
bool is_set() const { return (_f) ? true : false; }
RET operator()(ARG a) const { return is_set() ? _f(a) : 0; }
};
#endif /* CALLBACK_HPP_ */
// sketch
#include "callback.hpp"
// | !! empty capture!
void setup() { // V
auto clb = Callback<int, char>([](char c) { return (int)c; });
if (clb.is_set())
auto b = clb('a');
}
void loop() {}
may do the work, and it uses a simple trick:
The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. [C++11 standard 5.1.2]
As soon as you leave the capture empty, you are assured to have a "conversion" to a function pointer, thus you can store it without issues. The code I have written:
requires a first template RET that is the returned type
requires a second template ARG that is one argument for the callback. In the majority of the case you may consider to use void* as common argument (cast a struct pointer in a void pointer and use it as argument, to counter-cast in the function, the operation costs nothing)
implements two constructors: the empty constructor initialize the function pointer to NULL, while the second directly assigns the callback. Notice that the copy constructor is missing, you need to implement it.
implements a method to call the function (overloading the operator ()) and to check if the callback actually exists.
Again: this stuff compiles with no warnings, but I don't know if it works on the MSP430, since I cannot test it (it works on a common amd64 linux system).
Related
I'm the beginner with C++.
Could some one explain the output after compile process function.
Thanks a lot.
template <typename T>
auto process(T arg)
{
// 4. With constexpr if, to enable the compiler to generate different code based on the type the template is instantiated with:
if constexpr (std::is_same<T, bool>::value)
return !arg;
else if constexpr (std::is_integral<T>::value)
return -arg;
else if constexpr (std::is_floating_point<T>::value)
return std::abs(arg);
else
return arg;
}
int main(){
...
{
auto v1 = process(false); // true
auto v2 = process(42); // -42
auto v3 = process(-42.0); // 42.0
auto v4 = process("42"s); // "42"
}
...
return 0;
}
what's the real code compiler for process() is generated after we call above code in main function.
what's the real code compiler for process() is generated after we call above code in main function.
process() is not a function, and no compiled version of it is produced (at least in typical implementations); rather, your program produces four separate functions, namely process<bool>, process<int>, process<double>, and process<std::string>, each of which has its own compiled version.
And that's not specific to if constexpr — it's just how templates work in general in C++.
Those compiled versions can completely omit the branches of the if statement that don't hold for the type argument; so, for example, process<bool> is as if it were defined like this:
template<>
bool process<bool>(bool arg)
{
return !arg;
}
If I have a pure virtual class InterfaceA that consists solely of a pure virtual destructor, why do I have to define the destructor as inline? I I don't I get an error when I try to link it.
Below is an admittedly contrived example, however it illustrates the point. The point does not compile for me using cmake and g++. However, if I change the InterfaceA destructor definition as follows - inline InterfaceA::~InterfaceA(){}; then it compiles.
Why is this? What does the inline keyword do?
// InterfaceA.h, include guards ommitted for clarity
class InterfaceA
{
public:
virtual ~InterfaceA() = 0;
};
InterfaceA::~InterfaceA(){};
// A.h, include guards ommitted for clarity
#include "InterfaceA.h"
class A : public InterfaceA
{
public:
A(int val)
: myVal(val){};
~A(){};
int myVal;
};
// AUser.h, include guards ommitted for clarity
#include "InterfaceA.h"
class AUser
{
public:
AUser(InterfaceA& anA)
: myA(anA){};
~AUser(){};
int getVal() const;
private:
InterfaceA& myA;
};
// AUser.cpp
#include "AUser.h"
#include "A.h"
int AUser::getVal() const
{
A& anA = static_cast<A&>(myA);
return anA.myVal;
}
// main.cpp
#include "AUser.h"
#include "A.h"
#include <iostream>
int main(){
A anA(1);
AUser user(anA);
std::cout << "value = " << user.getVal() << std::endl;
return 0;
}
You have to use the inline keyword when defining functions in header files. If you do not, and the file is included in more than one translation unit, the function will be defined twice (or more times).
The linker error is probably something like "Symbol ... is multiply defined" right?
If you defined the member function in the body of the class, it would be implicitly inline and it would also work.
See this answer
To answer the question "What does the inline keyword do?":
In the old days it would be used to ask the compiler to inline functions i.e. insert the code whenever the function is used instead of adding a function call. Eventually it turned into a simple suggestion since compiler optimizers became more knowledgeable about which functions were inline candidates. These days it is used almost exclusively to define functions in header files that must have external linkage.
inline means that compiler is allowed to add code directly to where the function was called. It also removes function from external linkage, so both your compile units would have local version of.. pure destructor.
// InterfaceA.h, include guards ommitted for clarity
class InterfaceA
{
public:
virtual ~InterfaceA() = 0;
};
You declare destructor virtual, so compiler almost never would make it inline. Why? because virtual functions are called through vtable - a internal working of virtual functions system, vtable most likely implemented as an array of pointers to member functions. If function is inlined, it would have no address, no legal pointer. If attempt to get address of function is taken, then compiler silently disregards inline keyword. The other effect will be still in place: inlined destructor stops to be visible to linker.
It may look like declaring pure virtual destructor looks like oxymoron , but it isn't. The pure destructor is kind of destructor that would be always called without causing UB. Its presence would make class abstract, but the implicit call in sequence of destructor calls would still happen. If you didn't declare destructor body, it would lead to an UB, e.g. purecall exception on Windows.
If you don't need an abstract base class, then inline definition will suffice:
class InterfaceA
{
public:
virtual ~InterfaceA() {}
};
that is treated by compiler as inline as well, but mixing inline definition and pure member declaration is not allowed.
#include <iostream>
struct A
{
virtual void foo(){ std::cout << "A"; };
};
struct B : public A
{
private:
void foo() override { std::cout << "B"; }
};
int main()
{
A *p = new B;
p->foo(); // prints B
// B b;
// b.foo(); // error: foo is private
return 0;
}
// g++ -std=c++11 -Wall -Wextra -Wpedantic main.cpp && ./a.out
So we can call B.foo() polymorphically, but not directly. Are there any use cases, when someone would want to use this functionality?
It sort of depends on the design of the base class. Suppose you have a base class
class Stream {
public:
virtual bool canSeek() = 0;
virtual void seek(int offset) = 0;
};
Note: this example comes from the .NET world, where the base class library Stream class really does have such a virtual CanSeek property. I do not wish to discuss whether this is good design, as I can see valid arguments for both sides. It suffices that such base classes exist in reality.
Now, a derived class may specify that
class SpecificStream final : Stream {
private:
virtual bool canSeek() { return false; }
virtual void seek(int offset) { throw "no seek for you"; }
}
In this derived class, the fact that seek is implemented at all is because it is technically required. However, any code that deals with this SpecificStream already knows that the seek function is utterly useless with this class and should not be called. When coding against the base Stream class, it may make sense to check canSeek()'s result and call seek only if the result was true. When coding against the SpecificStream class, it doesn't make sense to check canSeek(), as its result is statically known, and it definitely doesn't make sense to call seek(). If such calls would be a programmer error, it makes sense to help the compiler give useful messages for such calls.
It stops you from calling the method non-polymorphically, that's all: using the scope resolution operator to access a method directly can lead to difficult to maintain code. In an environment where you know that not everybody is an experienced implementer (scientific programmers contributing to a large codebase perhaps), it's worthwhile introducing patterns to protect your code from them!
That said, Java explicitly forbids it as they consider it bad style.
I am trying my C++11 code to see if all recent major compiler supports the features I used, and the following shortened code
#include <valarray>
struct T
{
double vv[3];
};
class V : public std::valarray<T>
{
public:
auto begin()->decltype(std::begin(static_cast<std::valarray<T>>(*this)))
{
return std::begin(static_cast<std::valarray<T>>(*this));
}
};
int main(void)
{
}
would compile with g++ 4.8.1(from Debian sid repository), Intel C++ compiler 13.1.1 20130313, but not Clang 3.3-2(from Debian sid repository).
The given error is:
test.cpp:11:73: error: no viable conversion from 'V' to 'std::valarray<T>'
auto begin()->decltype(std::begin(static_cast<std::valarray<T>>(*this)))
^~~~~
However, code like this
namespace std
{
auto begin(V& vv) -> decltype(std::begin(static_cast<V::parent_t>(vv)))
{
return std::begin(static_cast<V::parent_t>(vv));
}
}
would compile by all three compilers.
My question is: is the code itself allowed by the language standard, just Clang miscompiled it, or it is only supported by g++/icc extension? Or it is undefined behavior?
The code very dangerous and needs to be fixed even for GCC and ICC.
You're doing a static_cast to a value type, not a reference or pointer. That creates a new temporary valarray object, so the const overload of begin gets called (probably not what you intended), and the iterator returned by begin() refers to the temporary which goes out of scope immediately, so the returned iterator is invalid and dereferencing it is undefined behaviour.
The code will compile like this:
auto begin()->decltype(std::begin(std::declval<std::valarray<T>&>()))
{
return std::begin(static_cast<std::valarray<T>&>(*this));
/* cast to reference type! ^^^^^^^^^^^^^^^^^ */
}
The decltype doesn't need to cast this, it just needs to know the type of calling std::begin on a valarray<T>, so it doesn't matter if the type is incomplete because you don't need a cast.
In the body of the function the type is considered complete anyway, so the cast is valid.
I've started playing around with AVX instructions on the new Intel's Sandy Bridge processor. I'm using GCC 4.5.2, TDM-GCC 64bit build of MinGW64.
I want to overload operator<< for ostream to be able to print out the vector types __m256, __m128 etc to the console. But I'm running into an overloading conflict. The 2nd function in the following code produces an error "conflicts with previous declaration void f(__vector(8) float)":
void f(__m128 v) {
cout << 4;
}
void f(__m256 v) {
cout << 8;
}
It seems that the compiler cannot distinguish between the two types and consideres them both f(float __vector).
Is there a way around this? I haven't been able to find anything online. Any help is greatly appreciated.
I accidentally stumbled upon the answer when having a similar problem with function templates. In this case, the GCC error message actually suggested a solution:
add -fabi-version=4 compiler option.
This solves my problem, and hopefully doesn't cause any issues when linking the standard libraries.
One can read more about ABI (Application Binary Interface) and GCC at ABI Policy and Guidelines and ABI specification. ABI specifies how the functions names are mangled when the code is compiled into object files. Apparently, ABI version 3 used by GCC by default cannot distinguish between the various vector types.
I was unsatisfied with the solution of changing compiler ABI flags to solve this, so I went looking for a different solution. It seems they encountered this issue in writing the Eigen library - see this source file for details http://eigen.tuxfamily.org/dox-devel/SSE_2PacketMath_8h_source.html
My solution to this is a slightly tweaked version of theirs:
template <typename T, unsigned RegisterSize>
struct Register
{
using ValueType = T;
enum { Size = RegisterSize };
inline operator T&() { return myValue; }
inline operator const T&() const { return myValue; }
inline Register() {}
inline Register(const T & v) : myValue(v) {} // Not explicit
inline Register & operator=(const T & v)
{
myValue = v;
return *this;
}
T myValue;
};
using Register4 = Register<__m128, 4u>;
using Register8 = Register<__m256, 8u>;
// Could provide more declarations for __m128d, __m128i, etc. if needed
Using the above, you can overload on Register4, Register8, etc. or produce template functions taking Registers without running into linking issues and without changing ABI settings.