"Handling map of files in c++" says no, one shall use std::map<std::string, std::ofstream*>, but this leads to the new and delete actions, which is not so neat.
Since "Is std::ofstream movable? Yes!" and it's possible to "std::map<>::insert using non-copyable objects and uniform initialization", is it possible to handle a collection of ofstream using std::map? so that one won't worry about closing filestreams and delete to release memory.
I can compromise that during using std::map<std::string, std::ofstream>, only create, use (it to write) and close, not to copy it.
Yes it is possible. See sample code below.
I can compromise that during using std::map<std::string, std::ofstream>, only create, use (it to write) and close, not to copy it.
They are not copyable, so in your final comment, you are correct, you will be unable to copy it. You can move assign though, if that's what you want to do.
#include <iostream>
#include <fstream>
#include <map>
int main()
{
std::map<std::string, std::ofstream> map;
map.emplace("foo", std::ofstream("/tmp/foo"));
map.emplace("bar", std::ofstream("/tmp/bar"));
map["foo"] << "test";
map["foo"].flush();
std::ifstream ifs("/tmp/foo");
std::string data;
ifs >> data;
std::cout << data << '\n';
return 0;
}
Output:
test
Related
I am new to boost:asio. I need to pass shared_ptr as argument to handler function.
E.g.
boost::asio::post(std::bind(&::function_x, std::move(some_shared_ptr)));
Is using std::move(some_shared_ptr) correct? or should I use as below,
boost::asio::post(std::bind(&::function_x, some_shared_ptr));
If both are correct, which one is advisable?
Thanks in advance
Regards
Shankar
Bind stores arguments by value.
So both are correct and probably equivalent. Moving the argument into the bind is potentially more efficient if some_argument is not gonna be used after the bind.
Warning: Advanced Use Cases
(just skip this if you want)
Not what you asked: what if function_x took rvalue-reference arguments?
Glad you asked. You can't. However, you can still receive by lvalue reference and just move from that. because:
std::move doesn't move
The rvalue-reference is only there to indicate potentially-moved-from arguments enabling some smart compiler optimizations and diagnostics.
So, as long as you know your bound function is only executed once (!!) then it's safe to move from lvalue parameters.
In the case of shared-pointers there's actually a little bit more leeway, because moving from the shared-ptr doesn't actually move the pointed-to element at all.
So, a little exercise demonstrating it all:
Live On Coliru
#include <boost/asio.hpp>
#include <memory>
#include <iostream>
static void foo(std::shared_ptr<int>& move_me) {
if (!move_me) {
std::cout << "already moved!\n";
} else {
std::cout << "argument: " << *std::move(move_me) << "\n";
move_me.reset();
}
}
int main() {
std::shared_ptr<int> arg = std::make_shared<int>(42);
std::weak_ptr<int> observer = std::weak_ptr(arg);
assert(observer.use_count() == 1);
auto f = std::bind(foo, std::move(arg));
assert(!arg); // moved
assert(observer.use_count() == 1); // so still 1 usage
{
boost::asio::io_context ctx;
post(ctx, f);
ctx.run();
}
assert(observer.use_count() == 1); // so still 1 usage
f(); // still has the shared arg
// but now the last copy was moved from, so it's gone
assert(observer.use_count() == 0); //
f(); // already moved!
}
Prints
argument: 42
argument: 42
already moved!
Why Bother?
Why would you care about the above? Well, since in Asio you have a lot of handlers that are guaranteed to execute precisely ONCE, you can sometimes avoid the overhead of shared pointers (the synchronization, the allocation of the control block, the type erasure of the deleter).
That is, you can use move-only handlers using std::unique_ptr<>:
Live On Coliru
#include <boost/asio.hpp>
#include <memory>
#include <iostream>
static void foo(std::unique_ptr<int>& move_me) {
if (!move_me) {
std::cout << "already moved!\n";
} else {
std::cout << "argument: " << *std::move(move_me) << "\n";
move_me.reset();
}
}
int main() {
auto arg = std::make_unique<int>(42);
auto f = std::bind(foo, std::move(arg)); // this handler is now move-only
assert(!arg); // moved
{
boost::asio::io_context ctx;
post(
ctx,
std::move(f)); // move-only, so move the entire bind (including arg)
ctx.run();
}
f(); // already executed
}
Prints
argument: 42
already moved!
This is going to help a lot in code that uses a lot of composed operations: you can now bind the state of the operation into the handler with zero overhead, even if it's bigger and dynamically allocated.
It seems that if I write
#include <random>
std::minstd_rand engine(1);
std::cout << engine;
then this prints out the internal state of the engine (which is a linear congruential generator). Right now the state equals the seed (1), but if I call a random number and print out engine, it returns some large number, which is probably the state.
How do I actually get the state, in a variable?
Use a string stream instead of stdout. Example:
#include <sstream>
...
std::ostringstream os;
os << engine;
string mystate = os.str();
The o in ostringstream is for output.
The state should be last random number generated, which is why there is not an easier way to do this. It's not as ideal as something like int a; a << engine, but it'll have to do. If you need it that often, make the stringstream operation a function (Including perhaps a conversion from string to integer). You can also typedef a pair of engine/integer with the integer being the state, and make a couple of methods so it's autoset every generation call if you need the performance.
If you don't care about the state, and just want it for the future, do
int engineState = engine();
Now you have the state. Though it's not the same as what it was before, it might not matter depending on your use case.
Output from linear congruential RNG is the state. Or, as alreadynoted, use operator<< to output and convert state
Code
#include <random>
#include <iostream>
#include <sstream>
int main() {
auto engine = std::minstd_rand{ 1 };
auto q = engine();
auto os = std::ostringstream{};
os << engine;
auto r = std::stoul(os.str()); // use ul to fit output
std::cout << q << " " << os.str() << " " << r << '\n';
return 0;
}
prints
48271 48271 48271
Alternative might be if particular implementation implements discard properly in O(log2(N)) time, according to paper by F.Brown https://laws.lanl.gov/vhosts/mcnp.lanl.gov/pdf_files/anl-rn-arb-stride.pdf. In such case you could move one position back, call RNG again and get your state as output.
Compiler and library I use - Visual C++ 2017 15.7 - has not implemented discard in such way, and useless for moving back.
LCGs consist of a simple state that is represented by a single integer.
This means you can treat this pointer as a pointer to an integer.
Below, I have provided an example of a template function that gets
the state (seed) of an engine and even works for classes deriving LCGs.
#include <random>
template <class T, T... v>
T getSeed(std::linear_congruential_engine<T, v...>& rand) {
static_assert(sizeof(rand) == sizeof(T));
return *reinterpret_cast<T*>(&rand);
}
#include <iostream>
int main() {
std::minstd_rand engine(19937);
auto seed = getSeed(engine);
std::cout << sizeof(engine);
std::cout << '\t' << seed;
}
^ This method is way more efficient (x320 times) than serializing through a stream,
or by creating a dummy ostream and specializing std::operator<< for every case.
template<class T, T... v>
using LCG = std::linear_congruential_engine<T, v...>;
#define DummyRandSpec32 uint_fast32_t, 0xDEADBEEF, 0xCAFE, 0xFFFFFFFF
typedef LCG<DummyRandSpec32> DummyRand32; // the same engine type
template<class T, class R>
T* getSeed(R& rand) // getSeed 70:1 nextInt
{ // creating stream is heavy operation
// return rand._M_x; // cannot access private
__dummy_ostream<T> dumdum; // workaround
auto& didey = *reinterpret_cast<DummyRand32*>(&rand);
std::operator<<(dumdum, didey); // specialized
return dumdum.retrieve(); // pointer to state
}
int main() {
std::minstd_rand engine(19937);
std::cout << *getSeed<uint_fast32_t>(engine);
std::cout << std::endl << engine << std::endl;
}
^ Here is ill-coded my first attempt at a solution, if you want to compare.
It is worth mentioning that a field name of the state is implementation-specific.
Purposefully left out std::operator<< and __dummy_ostream.
I want to read a binary file in 16 bit words. Right now, I'm using an std::ifstream to read into a 2 character array c.
#include <iostream>
#include <fstream>
#include <stdint.h>
int main() {
std::ifstream file("./tetris.rom", std::ios::in | std::ios::binary);
char c[2];
while (file.read(c, 2)) {
uint16_t word = (static_cast<uint8_t>(c[0]) << 8) | static_cast<uint8_t>(c[1]);
std::cout << "word\t" << std::hex << word << std::endl;
}
}
This works for me, but is there a better (either safer or faster) way of doing this in C++11?
There are no new APIs of reading files in C++11.
If the file fits into your RAM, the most optimal way is to map it into memory and access it as a byte array. However, the C++ standard library does not provide an API for that. You can do that with Boost though, see Boost.Interprocess Memory Mapped Files.
The usual advice stands though: start with your simple and correctly working code, benchmark and see if file reading is the bottleneck.
I have a vector of unique_ptrs and want to filter it into a new vector of the same type.
vector<unique_ptr<Thing>> filter_things(const vector<unique_ptr<Thing>> &things) {
vector<unique_ptr<Thing>> things;
// i want the above line to be something like: vector<const unique_ptr<Thing> &>
// but I don't think this is valid
for (const unique_ptr<Thing> &thing : things) {
if (check(thing)) {
filtered.push_back(thing); // this part shouldn't work since it
// would duplicate a unique_ptr
}
}
return filtered;
}
I want the caller to maintain ownership of all the Things. I want the return value of this function to be purely read only (const), and I don't want to make copies as it is very expensive to copy a Thing.
What is the best way to accomplish this?
Is this possible with unique_ptrs?
In some sense, we are creating multiple references by returning a new vector of references, so unique_ptr may not make sense. However, it is purely read only! So there should be some way to make this work. The lifetime of ``things'' is guaranteed to be larger than the filtered things.
Note that the caller owns the parameter supplied.
You can use reference_wrapper from <functional>
#include <memory>
#include <functional>
#include <vector>
#include <iostream>
using namespace std;
struct Thing {};
using PThing = unique_ptr<Thing>;
using RefThing = reference_wrapper<const PThing>;
vector<RefThing> filter_things( const vector<PThing>& things )
{
vector<RefThing> filtered;
int i = 0;
for( auto&& thing : things )
{
if( i++%2 )
filtered.push_back( ref(thing) );
}
return filtered;
}
int main()
{
vector<PThing> vec;
vector<RefThing> flt;
vec.resize(25);
flt = filter_things(vec);
cout << flt.size() << endl;
}
If what you want is getting a filtered set of element not an actual container containing them, boost::range can be a good solution.
auto filtered_range(const std::vector<std::unique_ptr<Thing>> &things) {
return things | boost::adaptors::filtered([](const auto& thing) {
return check(thing);
});
}
I used some of c++14 syntax but I don't think it's hard to make it to c++11.
You can use it like this.
std::vector<std::unique_ptr<Thing> > things;
for(const auto& thing : filtered_range(things)) {
// do whatever you want with things satisfying 'check()'
}
One of disadvantages is that the range itself is not a container so if you traverse the range more than once, every 'thing' will be checked if it satisfies check().
If a container storing the checked things AND controlling the lifetime of things are what you really want, I would prefer using std::vector<std::shared_ptr<Thing> > and returning std::vector<std::weak_ptr<Thing> >. You can check if it's really one and the only ptr to a thing with std::shared_ptr::unique() before deleting it from things.
How can one convert a shared_ptr that points to a const object to a shared_ptr that points to a non-const object.
I am trying to do the following :
boost::shared_ptr<const A> Ckk(new A(4));
boost::shared_ptr<A> kk=const_cast< boost::shared_ptr<A> > Ckk;
But it does not work.
'boost::const_pointer_cast' will do what you're asking for, but the obligatory second half of the answer is that you probably shouldn't use it. 99% of the time when it seems like you need to cast away the const property of a variable, it means that you have a design flaw. Const is sometimes more than just window dressing and casting it away may lead to unexpected bugs.
Without knowing more details of your situation one can't say for certain. But no discussion of const-cast is complete without mentioning this fact.
use boost::const_pointer_cast, documentation.
the proper way should be this
boost::shared_ptr<A> kk (boost::const_pointer_cast<A>(Ckk));
std::const_cast_pointer makes a second managed pointer. After the cast you have a writable pointer and the original const-pointer. The pointee remains the same. The reference count has been increased by 1.
Note that const_cast is a builtin keyword, but const_pointer_cast is a template function in namespace std.
The writable pointer can then be used to change the value from under the shared_ptr<const T>. IMHO the writable pointer should only persist temporarily on the stack; otherwise there must be a design flaw.
I once wrote a small test program to make this clear to myself which I adapted for this thread:
#include <memory>
#include <iostream>
#include <cassert>
using namespace std;
typedef shared_ptr<int> int_ptr;
typedef shared_ptr<const int> const_int_ptr;
int main(void)
{
const_int_ptr Ckk(new int(1));
assert(Ckk.use_count() == 1);
cout << "Ckk = " << *Ckk << endl;
int_ptr kk = const_pointer_cast<int>(Ckk); // obtain a 2nd reference
*kk = 2; // change value under the const pointer
assert(Ckk.use_count() == 2);
cout << "Ckk = " << *Ckk << endl; // prints 3
}
Under UNIX or Windows/Cygwin, compile with
g++ -std=c++0x -lm const_pointer_cast.cpp