I tried using attribute const and pure to try to make GCC enforce that a function is not allowed to access any global variables but this compiles just fine:
#include <stdlib.h>
#include <stdio.h>
int data;
__attribute__((const))
int test_func(int val){
data = val;
return data;
}
int main(){
scanf("%d", &data);
printf("test: %d\n", test_func(data));
return 0;
}
This should not compile.
Is there a way to enforce the rule that neither the function itself nor any other functions it calls can access any variables besides their arguments?
GCC manual says that __attribute__((const)) is used by the programmer to indicate the behavior of the function so that it can be optimized better.
Calls to functions whose return value is not affected by changes to the observable state of the program and that have no observable effects on such state other than to return a value may lend themselves to optimizations such as common subexpression elimination. Declaring such functions with the const attribute allows GCC to avoid emitting some calls in repeated invocations of the function with the same argument values.
In the documentation there are some cases for which a compile error is promised, and wrong behavior (accessing globals within the function) is not one of those. Therefore you are responsible to not use this attribute for functions that use globals.
Related
Using C++11, g++ (GCC) 4.4.7 20120313 (Red Hat 4.4.7-18).
Lets pretend I have a templated function (pardon my terminology if it isn't quite right).
I want to perform a "general" algorithm based on what was supposed to be compile-time instances of "field". Where the only things that really changed are these constants which I moved into trait classes (only added one here but imagine there are more). Originally I was declaring it as
constexpr field FIELD1{1};
However in C++11, non-type template params need to have external linkage (unlike C++14 which can have internal and external linkage?). So because not's in the same translation unit I needed to use extern in order to give it external linkage (sorry if I butchered that explanation also). But by defining it extern I can't define it using constexpr and it seems that losing that constexpr constructor this field is no longer a valid constant expression to qualify as a non-type template param.
Any suggestions if there is some way I can get around this? Open to a new method of doing things. Below is a simplified (incomplete, and non-compiling version to get the gist of the organization).
So the error I am seeing is along the lines of
error: the value of ‘FIELD1’ is not usable in a constant expression
note: ‘FIELD1’ was not declared ‘constexpr’
extern const field FIELD1;
Not quite sure what could be a best alternative.
I can get rid of the second error by removing the constexpr from the constructor. But then I don't know how to approach the constant expression issue.
field.H
struct field
{
int thingone;
constexpr field(int i):thingone(i){}
};
extern const field FIELD1;
field.C
#include "field.H"
const field FIELD1{0};
field_traits.H
#include "field.H"
template< const field& T >
class fieldTraits;
template< >
class fieldTraits<FIELD1>
{
public:
// Let's say I have common field names
// with different constants that I want to plug
// into the "function_name" algorithm
static constexpr size_t field_val = 1;
};
function.H
#include "field.H"
template< const field& T, typename TT = fieldTraits<T> >
void function_name()
{
// Let's pretend I'm doing something useful with that data
std::cout << T.thingone << std::endl;
std::cout << TT::field_val << std::endl;
}
So because not's in the same translation unit I needed to use extern in order to give it external linkage (sorry if I butchered that explanation also). But by defining it extern I can't define it using constexpr [...]
Per my comment, you can. It wouldn't work for you, but it's a step that helps in coming up with something that would work:
extern constexpr int i = 10;
This is perfectly valid, gives i external linkage, and makes i usable in constant expressions.
But it doesn't allow multiple definitions, so it can't work in a header file which is included in multiple translation units.
Ordinarily, the way around that is with inline:
extern inline constexpr int i = 10;
But variables cannot be declared inline in C++11.
Except... when they don't need to be declared inline because the effect has already been achieved implicitly:
struct S {
static constexpr int i = 10;
};
Now, S::i has external linkage and is usable in constant expressions!
You may not even need to define your own class for this, depending on the constant's type: consider std::integral_constant. You can write
using i = std::integral_constant<int, 10>;
and now i::value will do exactly what you want.
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).
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>
#include <functional>
using namespace std;
std::function<void(void)> makeLambda(int param)
{
return [¶m](){cout << param << endl;};
}
int main()
{
auto callback = makeLambda(5);
callback();
}
Based on lambda description as following, it looks the program will cause an undefined behavior because when callback is invoked, the captured var, function parameter, is out-of-scope. But I see it always can print 5.
My g++ version is gcc-4.9.1.
Dangling references
If an entity is captured by reference, implicitly or explicitly, and
the function call operator of the closure object is invoked after the
entity's lifetime has ended, undefined behavior occurs. The C++
closures do not extend the lifetimes of the captured references.
Same applies to the lifetime of the object pointed to by the captured
this pointer.
Why can it work?
As you note, this is undefined behaviour. Anything can happen, including appearing to work. If you switch compiler, change flags, forget to do the dishes, or get up an hour later, you could get completely different results.
As an example, Clang prints 32767 for some version and set of flags.
recently I've tried to take advantage of C++0x constexpr under MSVC 2015 and my objective was to achieve compile-time hash strings. I wrote a simple FNV-1a hash algorithm as a constexpr function using, as required, a single return statement (ternary operator) and calling only constexpr functions, here it is:
template <size_t N>
constexpr U32 StringID_FNV1a_32(const char(&str)[N], I32 charIndex = 0, U32 hash = 2166136261U)
{
return charIndex < N-1 ? StringID_FNV1a_32(str, charIndex +1, (hash ^ str[charIndex]) * 16777619U) : hash;
}
I also made a little macro to be able to change the algorithm under the hood without any trouble:
#define STRING_ID(str) core::utility::StringID_FNV1a_32(str)
then I used this macro in my code, carefully checking if any breakpoint was hit and, also, the generated assembly code. Here's the little scenario:
//1. normal variable test
U32 hash1 = STRING_ID("abc");
//2. enum test
enum {
hash2 = STRING_ID("abc")
};
//3. constexpr variable test
constexpr U32 hash3 = STRING_ID("abc");
And here the facts:
first test was called at run time
second test was performed at compile time
third test was called at run time
As you can imagine I'm a little confused about the first and the third attempt.
Why in the third scenario is the compiler allowed to call the function at runtime? even though the msdn says clearly "The primary difference between const and constexpr variables is that the initialization of a const variable can be deferred until run time whereas a constexpr variable must be initialized at compile time." [https://msdn.microsoft.com/it-it/library/dn956974.aspx#Anchor_3]
Can be related to the fact that I'm in debug mode with all the optimizations turned off? and what about the first test?, is there any way to force the compiler to perform the hash at compile time?
MSVC's behavior can be quite strange, however it is possible to force it to make constexpr functions run at compile time.
#define COMPILE_TIME(value) ((decltype(value))CompileTime<decltype(value), value>::ValueHolder::VALUE)
template<typename T, T Value>
struct CompileTime
{
enum class ValueHolder : T
{
VALUE = Value
};
};
This forces the value to be passed as a template argument + an enumeration value, thus making it strictly compile-time only. Also please note that this works only for integer types.
You can use it simply by putting the call to the constexpr function as a parameter to the COMPILE_TIME macro:
constexpr U32 hash = COMPILE_TIME(STRING_ID("abc"));