int main(){
auto func1 = [](int y) {
cout << y << " ";
};
auto func2 = [](int y) {
cout << y * y << " ";
};
cout << "func1 is : " << typeid(func1).name() << endl;
cout << "func2 is : " << typeid(func2).name() << endl;
cout << "main is : " << typeid(main).name() << endl;
}
OSX output:
func1 is : Z4mainE3$_0
func2 is : Z4mainE3$_1
main is : FivE
Can someone explain the output ??
Thanks, I am just exploring some c++11 features.
What you are seeing here is the name mangled name of each of these symbols. This is what typeid.name() for your compiler implementation returns. There is no requirement that the mangled name precisely relates directly to your code. Using the mangled symbol names already present in the object files for linking is a convenient implementation choice.
You can unmangle names using the c++filt tool:
Thus:
$ c++filt _FivE
yields
int ()
In other words, a function returning an int. Remember that what you are asking for here is the type of the function and not its name.
If you were to apply this to a class
class foo
{
};
cout << "foo is : " << typeid(foo).name() << endl;
You will find output is 3foo and the unmanged name foo.
The two lambdas don't unmangle. This is because they are anonymous functions, so don't need an external name.
Furthermore, compiler generates a functor class for each lambda. The first of which would look like
struct Z4mainE3
{
void operator()(int y)
{
cout << y << " ";
}
}
This means that each one is a distinct type. The name is synthetic, and generated by the compiler such that is won't collide with anything else.
The typeid operator will operate on the functor struct and not the apparent return and argument type of the lambda itself, hence the two of them are a different type despite apparently being functions having the same signature.
The long-standing advice about typeid().name() operator is that it is not portable; you should not rely on the values returned.
Related
Here is a code snippet I have :
struct PairHasher {
size_t operator()(const std::pair<std::string_view, std::string_view>& stop_stop) const {
return hasher(stop_stop.first) + 37*hasher(stop_stop.second);
}
std::hash<std::string_view> hasher;
};
BOOST_FIXTURE_TEST_CASE(unordered_map_string_view_pair_must_be_ok, TestCaseStartStopMessager)
{
const std::vector<std::string> from_stops = {"from_0", "from_1", "from_2"};
const std::vector<std::string> to_stops = {"to_0", "to_1", "to_2"};
std::unordered_map<std::pair<std::string_view, std::string_view>, std::int32_t, TransportCatalogue::PairHasher> distance_between_stops;
for ( std::size_t idx = 0; idx < from_stops.size(); ++idx) {
std::cout << from_stops[idx] << " : " << to_stops[idx] << std::endl;
distance_between_stops[std::pair(from_stops[idx], to_stops[idx])] = idx;
}
std::cout << "MAP CONTENT :" << std::endl;
for (auto const& x : distance_between_stops)
{
std::cout << x.first.first << " : " << x.first.second << std::endl;
}
}
I expect to see 3 pairs inside the container, but there is only 1 concerning to the output :
MAP CONTENT :
from_2 : to_2
So, where are two more pair lost? What am I doing wrong?
Moving my comment to an answer.
This is pretty sneaky. I noticed in Compiler Explorer that changing:
distance_between_stops[std::pair(from_stops[idx], to_stops[idx])] = idx;
to
distance_between_stops[std::pair(std::string_view{from_stops[idx]}, std::string_view{to_stops[idx]})] = idx;
fixes the bug. This hints that the problem lies in some implicit string -> string_view conversion. And indeed that is the case, but it is hidden behind one extra layer.
std::pair(from_stops[idx], to_stops[idx]) creates a std::pair<std::string, std::string>, but distance_between_stops requires a std::pair<std::string_view, std::string_view>. When we insert values into the map, this conversion happens implicitly via overload #5 here:
template <class U1, class U2>
constexpr pair(pair<U1, U2>&& p);
Initializes first with std::forward<U1>(p.first) and second with std::forward<U2>(p.second).
This constructor participates in overload resolution if and only if std::is_constructible_v<first_type, U1&&> and std::is_constructible_v<second_type, U2&&> are both true.
This constructor is explicit if and only if std::is_convertible_v<U1&&, first_type> is false or std::is_convertible_v<U2&&, second_type> is false.
(For reference, std::is_constructible_v<std::string_view, std::string&&> and std::is_convertible_v<std::string&&, std::string_view> are both true, so we know this overload is viable and implicit.)
See the problem yet? When we use the map's operator[], it has to do an implicit conversion to create a key with the proper type. This implicit conversion constructs a pair of string_views that are viewing the temporary memory from the local pair of strings, not the underlying strings in the vector. In other words, it is conceptually similar to:
std::string_view foo(const std::string& s) {
std::string temp = s + " foo";
return temp;
}
int main() {
std::string_view sv = foo("hello");
std::cout << sv << "\n";
}
Clang emits a warning for this small example, but not OP's full example, which is unfortunate:
warning: address of stack memory associated with local variable 'temp' returned [-Wreturn-stack-address]
return temp;
^~~~
Imagine you have the following code where logDebug() is expensive or is not appropriate to call more than once:
QDebug d = logDebug();
d << __FUNCTION__ << ":";
d << "positions separated with \" --- \":";
for (const auto& str : positions)
{
d << "---" << str;
}
A macro (just to replace the function name correctly) already exists which replaces the first 2 lines:
#define LOG_FUNCTION this->logDebug() << __FUNCTION__ << ":"
It creates the local variable by calling logDebug(). Once called, you can only use the operator<< onto the macro.
The problem is you can't attach the for loop body to logger.
Q: Is there a way I could use the macro for pasting all the positions (without calling logDebug again?
I would guess this should be possible using lambdas, but I quite don't know how to.
Please help, the shortest answer wins!
Q: Is there a way I could use the macro for pasting all the positions (without calling logDebug again? I would guess this should be possible using lambdas, but I quite don't know how to.
I suppose it's possible with something as follows (used std::cout instead of logDebug())
#include <iostream>
#define LOG_FUNCTION std::cout << __FUNCTION__ << ": "
#define LOG_DEB(ps) \
[](auto & s, auto const & _ps) { for ( auto const & p : _ps ) s << p; } \
(LOG_FUNCTION, ps)
int main ()
{
int a[] { 0, 1, 2, 3, 4 };
LOG_DEB(a);
}
I've used a couple of auto as types of the lambda arguments and this works only starting from C++14.
In C++11 you have to replace they with the correct types.
Well the macro can be coerced to return your debug object:
#define LOG_FUNCTION() this->logDebug() << __FUNCTION__ << ":"
Then use it like this:
auto& d = LOG_FUNCTION();
d << "positions separated with \" --- \":";
for (const auto& str : positions)
{
d << "---" << str;
}
I think I might have missed the subtlety in move construction because when I change the line Foo copy(*this); to decltype(*this) copy(*this);, I am thoroughly surprised by the output.
I checked it against, clang++-3.5 and g++-4.9, with the same behavior.
Would really appreciate a quick tip from the C++11 guru.
Update: Just forced the compiler to print the type of decltype(*this), it is actually a reference type i.e. Foo&.
class Foo {
public:
Foo(int a): val(a) {}
operator int() { return val; }
auto& operator++() {
val++;
return *this;
}
auto operator++(int) {
//Foo copy(*this);
decltype(*this) copy(*this);
++(*this);
return copy;
}
private:
int val;
};
int main()
{
Foo foo=1;
cout << "foo++ = " << foo++ << "\n";
cout << "foo++ = " << foo++ << "\n";
cout << "foo = " << foo << "\n";
return 0;
}
The output
foo++ = 2
foo++ = 3
foo = 3
There seems to be a confusion as to why decltyp(*this) is Foo& and not Foo in your case. Firstly, think about dereferencing a pointer always resulting in a reference to the pointed to object.
temp = *ptr // this would work if dereferencing returned by value or by reference
*ptr = expr // this would only work if dereferencing results in a reference.
Now decltype(expr) always gives you exactly the same type as the expr. For you *this is of type Foo&.
If you want type deduction without it resulting in a reference use auto instead of decltype, so:
auto copy(*this);
instead of
decltype(*this) copy(*this);
Also I don't know why your question is talking about move construction so much as there is no move involved anywhere.
When advancing the state of an object, use of std::swap works well for simple objects and pointer swaps. For other in place actions, Boost.ScopeExit works rather well, but it's not terribly elegant if you want to share exit handlers across functions. Is there a C++11 native way to accomplish something similar to Boost.ScopeExit but allow for better code reuse?
(Ab)use std::unique_ptr's custom Deleters as a ScopeExitVisitor or Post Condition. Scroll down to ~7th line of main() to see how this is actually used at the call site. The following example allows for either std::function or lambdas for Deleter/ScopeExitVisitor's that don't require any parameters, and a nested class if you do need to pass a parameter to the Deleter/ScopeExitVisitor.
#include <iostream>
#include <memory>
class A {
public:
using Type = A;
using Ptr = Type*;
using ScopeExitVisitorFunc = std::function<void(Ptr)>;
using ScopeExitVisitor = std::unique_ptr<Type, ScopeExitVisitorFunc>;
// Deleters that can change A's private members. Note: Even though these
// are used as std::unique_ptr<> Deleters, these Deleters don't delete
// since they are merely visitors and the unique_ptr calling this Deleter
// doesn't actually own the object (hence the label ScopeExitVisitor).
static void ScopeExitVisitorVar1(Ptr aPtr) {
std::cout << "Mutating " << aPtr << ".var1. Before: " << aPtr->var1;
++aPtr->var1;
std::cout << ", after: " << aPtr->var1 << "\n";
}
// ScopeExitVisitor accessing var2_, a private member.
static void ScopeExitVisitorVar2(Ptr aPtr) {
std::cout << "Mutating " << aPtr << ".var2. Before: " << aPtr->var2_;
++aPtr->var2_;
std::cout << ", after: " << aPtr->var2_ << "\n";
}
int var1 = 10;
int var2() const { return var2_; }
// Forward declare a class used as a closure to forward Deleter parameters
class ScopeExitVisitorParamVar2;
private:
int var2_ = 20;
};
// Define ScopeExitVisitor closure. Note: closures nested inside of class A
// still have access to private variables contained inside of A.
class A::ScopeExitVisitorParamVar2 {
public:
ScopeExitVisitorParamVar2(int incr) : incr_{incr} {}
void operator()(Ptr aPtr) {
std::cout << "Mutating " << aPtr << ".var2 by " << incr_ << ". Before: " << aPtr->var2_;
aPtr->var2_ += incr_;
std::cout << ", after: " << aPtr->var2_ << "\n";
}
private:
int incr_ = 0;
};
// Can also use lambdas, but in this case, you can't access private
// variables.
//
static auto changeStateVar1Handler = [](A::Ptr aPtr) {
std::cout << "Mutating " << aPtr << ".var1 " << aPtr->var1 << " before\n";
aPtr->var1 += 2;
};
int main() {
A a;
std::cout << "a: " << &a << "\n";
std::cout << "a.var1: " << a.var1 << "\n";
std::cout << "a.var2: " << a.var2() << "\n";
{ // Limit scope of the unique_ptr handlers. The stack is unwound in
// reverse order (i.e. Deleter var2 is executed before var1's Deleter).
A::ScopeExitVisitor scopeExitVisitorVar1(nullptr, A::ScopeExitVisitorVar1);
A::ScopeExitVisitor scopeExitVisitorVar1Lambda(&a, changeStateVar1Handler);
A::ScopeExitVisitor scopeExitVisitorVar2(&a, A::ScopeExitVisitorVar2);
A::ScopeExitVisitor scopeExitVisitorVar2Param(nullptr, A::ScopeExitVisitorParamVar2(5));
// Based on the control of a function and required set of ScopeExitVisitors that
// need to fire use release() or reset() to control which visitors are used.
// Imagine unwinding a failed but complex API call.
scopeExitVisitorVar1.reset(&a);
scopeExitVisitorVar2.release(); // Initialized in ctor. Use release() before reset().
scopeExitVisitorVar2.reset(&a);
scopeExitVisitorVar2Param.reset(&a);
std::cout << "a.var1: " << a.var1 << "\n";
std::cout << "a.var2: " << a.var2() << "\n";
std::cout << "a.var2: " << a.var2() << "\n";
}
std::cout << "a.var1: " << a.var1 << "\n";
std::cout << "a.var2: " << a.var2() << "\n";
}
Which produces:
a: 0x7fff5ebfc280
a.var1: 10
a.var2: 20
a.var1: 10
a.var2: 20
a.var2: 20
Mutating 0x7fff5ebfc280.var2 by 5. Before: 20, after: 25
Mutating 0x7fff5ebfc280.var2. Before: 25, after: 26
Mutating 0x7fff5ebfc280.var1 10 before
Mutating 0x7fff5ebfc280.var1. Before: 12, after: 13
a.var1: 13
a.var2: 26
On the plus side, this trick is nice because:
Code used in the Deleters can access private variables
Deleter code is able to be centralized
Using lambdas is still possible, though they can only access pubic members.
Parameters can be passed to the Deleter via nested classes acting as closures
Not all std::unique_ptr instances need to have an object assigned to them (e.g. it's perfectly acceptable to leave unneeded Deleters set to nullptr)
Changing behavior at runtime is simply a matter of calling reset() or release()
Based on the way you build your stack it's possible at compile time to change the safety guarantees on an object when the scope of the std::unique_ptr(s) go out of scope
Lastly, using Boost.ScopeExit you can forward calls to a helper function or use a conditional similar to what the Boost.ScopeExit docs suggest with bool commit = ...;. Something similar to:
#include <iostream>
#include <boost/scope_exit.hpp>
int main() {
bool commitVar1 = false;
bool commitVar2 = false;
BOOST_SCOPE_EXIT_ALL(&) {
if (commitVar1)
std::cout << "Committing var1\n"
if (commitVar2)
std::cout << "Committing var2\n"
};
commitVar1 = true;
}
and there's nothing wrong with that, but like was asked in the original question, how do you share code without proxying the call someplace else? Use std::unique_ptr's Deleters as ScopeExitVisitors.
I have a simple struct
// in namespace client
struct UnaryExpression
{
std::string key;
SomeEnums::CompareType op;
};
SomeEnums::CompareType is an enum where I define a symbol table as such:
struct UnaryOps : bsq::symbols<char, SomeEnums::CompareType>
{
UnaryOps() : bsq::symbols<char, SomeEnums::CompareType>(std::string("UnaryOps"))
{
add("exists", SomeEnums::Exists)
("nexists", SomeEnums::NotExists);
}
};
I have two different ways I want to parse the struct, which I asked about in another thread and got to work (mostly).
My grammar looks as follows:
template<typename Iterator>
struct test_parser : bsq::grammar<Iterator, client::UnaryExpression(), bsq::ascii::space_type>
{
test_parser()
: test_parser::base_type(unaryExp, std::string("Test"))
{
using bsq::no_case;
key %= bsq::lexeme[bsq::alnum >> +(bsq::alnum | bsq::char_('.'))];
unaryExp %= unaryE | unaryF;
unaryE %= key >> no_case[unaryOps];
unaryF %= no_case[unaryOps] >> '(' >> key >> ')';
};
UnaryOps unaryOps;
bsq::rule<Iterator, std::string(), bsq::ascii::space_type> key;
bsq::rule<Iterator, client::UnaryExpression(), bsq::ascii::space_type> unaryExp;
bsq::rule<Iterator, client::UnaryExpression(), bsq::ascii::space_type> unaryE;
bsq::rule<Iterator, client::UnaryFunction(), bsq::ascii::space_type> unaryF;
};
And I'm parsing the code using the following logic:
bool r = phrase_parse(iter, end, parser, bsq::ascii::space, exp);
if (r && iter == end)
{
std::cout << "-------------------------\n";
std::cout << "Parsing succeeded\n";
std::cout << "key: " << exp.key << "\n";
std::cout << "op : " << exp.op << "\n";
std::cout << "-------------------------\n";
}
This all works fine if I do the input like foo exists and exp.key equals "foo" and exp.op equals the corresponding enum value (in this case 0). Something like foo1 nexists also works.
However, that second rule doesn't work like I expect. If I give it input of nexists(foo) then I get the following output:
-------------------------
Parsing succeeded
key: nexistsfoo
op : 1
-------------------------
It seems that the enum value is getting set appropriately but I can't figure out why the "nexsts" is getting prepended to the key string. Can someone please tell me how I can fix my rule so that the key would equal just 'foo' with the second rule?
I have posted a copy of the stripped down code that illustrates my problem here: http://pastebin.com/402M9iTS