I have problem with this piece of code:
#include <string>
#include <iostream>
struct A{
template<class UT>
A(UT &&s) : internal(std::forward<std::string>(s)){
}
std::string internal;
};
int main(){
const std::string &s = "hello";
A a1{ s };
std::cout << "s = " << s << std::endl;
}
This current example does not compiles and if I change s to be non const, it moves the string.
I have similar codes that works OK, but in this case there is something wrong I can not see.
You are using a forwarding reference incorrectly. You don't give std::forward the destination type. You give it the template it deduced:
A(UT &&s) : internal(std::forward<UT>(s)){
Related
Scenario
I have a C++ function which intakes a parameter as std::chrono::milliseconds. It is basically a timeout value. And, it is a default parameter set to some value by default.
Code
#include <iostream>
#include <chrono>
void Fun(const std::chrono::milliseconds someTimeout = std::chrono::milliseconds(100)) {
if (someTimeout > 0) {
std::cout << "someNumberInMillis is: " << someNumberInMillis.count() << std::endl;
}
}
int main() {
unsigned int someValue = 500;
Fun(std::chrono::milliseconds(someValue))
}
Issue
All of above is okay but, when I call Fun with a value then fails to compile and I get the following error:
No viable conversion from 'bool' to 'std::chrono::milliseconds' (aka
'duration >')
Question:
What am I doing wrong here? I want the caller of Fun to be explicitly aware that it is using std::chrono::milliseconds when it invokes Fun. But the compiler doesn't seem to allow using std::chrono::milliseconds as a parameter?
How use std::chrono::milliseconds as a default parameter?
Environment
Compiler used is clang on macOS High Sierra
With the other syntax errors fixed, this compiles without warnings in GCC 9:
#include <iostream>
#include <chrono>
void Fun(const std::chrono::milliseconds someNumberInMillis
= std::chrono::milliseconds(100))
{
if (someNumberInMillis > std::chrono::milliseconds{0}) {
std::cout << "someNumberInMillis is: " << someNumberInMillis.count()
<< std::endl;
}
}
int main()
{
unsigned int someValue = 500;
Fun(std::chrono::milliseconds(someValue));
}
In the following code, as none of the arguments is const, i can't understand why the second overload is called in the 3 following cases.
#include <iostream>
#include <algorithm>
using namespace std;
void ToLower( std::string& ioValue )
{
std::transform( ioValue.begin(), ioValue.end(), ioValue.begin(), ::tolower );
}
std::string ToLower( const std::string& ioValue )
{
std::string aValue = ioValue;
ToLower(aValue);
return aValue;
}
int main()
{
string test = "test";
cout<<"Hello World" << endl;
// case 1
cout << ToLower("test") << endl;
// case 2
cout << ToLower(static_cast<string>(test)) << endl;
// case 3
cout << ToLower(string(test)) << endl;
}
In all 3 cases you are creating a temporary std::string, this is an unnamed object, an R-value. R-values aren't allowed to bind to non-const l-value references (T&) and so only the overload taking const std::string& ioValue is valid.
The reasoning is the return type is std::string for the second function but void for the first. std::cout << (void) << std::endl is not a valid set of operations. std::cout << (std::string) << std::endl is. If you return a std::string& from the first function you'd probably see #2 & #3 probably use your first function call.
I've got the following test.cpp file
#include <string>
#include <functional>
#include <unordered_map>
#include <iostream>
class Mystuff {
public:
std::string key1;
int key2;
public:
Mystuff(std::string _key1, int _key2)
: key1(_key1)
, key2(_key2)
{}
};
namespace std {
template<>
struct hash<Mystuff *> {
size_t operator()(Mystuff * const& any) const {
size_t hashres = std::hash<std::string>()(any->key1);
hashres ^= std::hash<int>()(any->key2);
std::cout << "Hash for find/insert is [" << hashres << "]" << std::endl;
return (hashres);
}
};
}; /* eof namespace std */
typedef std::unordered_map<Mystuff *, Mystuff *>mystuff_map_t;
mystuff_map_t map;
int insert_if_not_there(Mystuff * stuff) {
std::cout << "Trying insert for " << stuff->key1 << std::endl;
if (map.find(stuff) != map.end()) {
std::cout << "It's there already..." << std::endl;
return (-1);
} else {
map[stuff] = stuff;
std::cout << "Worked..." << std::endl;
}
return (0);
}
int main(){
Mystuff first("first", 1);
Mystuff second("second", 2);
Mystuff third("third", 3);
Mystuff third_duplicate("third", 3);
insert_if_not_there(&first);
insert_if_not_there(&second);
insert_if_not_there(&third);
insert_if_not_there(&third_duplicate);
}
You can compile with g++ -o test test.cpp -std=gnu++11.
I don't get what I'm doing wrong with it: the hash keying algorithm is definitely working, but for some reason (which is obviously in the - bad - way I'm doing something), third_duplicate is inserted as well in the map, while I'd wish it wasn't.
What am I doing wrong?
IIRC unordered containers need operator== as well as std::hash. Without it, I'd expect a compilation error. Except that your key is actually MyStuff* - the pointer, not the value.
That means you get the duplicate key stored as a separate item because it's actually not, to unordered_map, a real duplicate - it has a different address, and address equality is how unordered_map is judging equality.
Simple solution - use std::unordered_map<Mystuff,Mystuff> instead. You will need to overload operator== (or there's IIRC some alternative template, similar to std::hash, that you can specialize). You'll also need to change your std::hash to also accept the value rather than the pointer.
Don't over-use pointers in C++, especially not raw pointers. For pass-by-reference, prefer references to pointers (that's a C++-specific meaning of "reference" vs. "pointer"). For containers, the normal default is to use the type directly for content, though there are cases where you might want a pointer (or a smart pointer) instead.
I haven't thoroughly checked your code - there may be more issues than I caught.
The following is not possible for any boost output archive:
int foo(){
return 4;
}
ar << static_cast<unsigned int>(foo());
Is there an alternative without out creating a local temporary x=foo().
and why is the underlying archive operator <<(T & t) not const reference , for an output archive such that the above would work?
This seems to work, and I think this is why:
... To help detect such cases, output archive operators expect to be
passed const reference arguments.
It seems worth noting that in your example ar << foo(); does not work either (i.e. it doesn't have to do with your cast).
#include <fstream>
#include <iostream>
#include <boost/serialization/serialization.hpp>
#include <boost/archive/text_iarchive.hpp>
#include <boost/archive/text_oarchive.hpp>
unsigned int foo(){
return 4;
}
int main()
{
{
std::ofstream outputStream("someFile.txt");
boost::archive::text_oarchive outputArchive(outputStream);
outputArchive << static_cast<const int&>(foo());
}
std::ifstream inputStream("someFile.txt");
boost::archive::text_iarchive inputArchive(inputStream);
int readBack;
inputArchive >> readBack;
std::cout << "Read back: " << readBack << std::endl;
return 0;
}
I want to move a stringstream, in the real world application I have some stringstream class data member, which I want to reuse for different string's during operation.
stringstream does not have a copy-assignment or copy constructor, which makes sense. However, according to cppreference.com and cplusplus.com std::stringstream should have a move assignment and swap operation defined. I tried both, and both fail.
Move assignment
#include <string> // std::string
#include <iostream> // std::cout
#include <sstream> // std::stringstream
int main () {
std::stringstream stream("1234");
//stream = std::move(std::stringstream("5678"));
stream.operator=(std::move(std::stringstream("5678")));
//stream.operator=(std::stringstream("5678"));
return 0;
}
source: http://ideone.com/Izyanb
prog.cpp:11:56: error: use of deleted function ‘std::basic_stringstream<char>& std::basic_stringstream<char>::operator=(const std::basic_stringstream<char>&)’
stream.operator=(std::move(std::stringstream("5678")));
The compiler states that there is no copy assignment for all three statements, which is true. However, I fail to see why it is not using the move-assignment, especially since std::move is supposed to return a rvalue reference. Stringstream should have a move assignment, as shown here: http://en.cppreference.com/w/cpp/io/basic_stringstream/operator%3D
PS: I'm working with c++11, hence rvalue-references are part of the 'world'.
Swap
This I found really strange, I copied example code from cplusplus.com and it failed:
// swapping stringstream objects
#include <string> // std::string
#include <iostream> // std::cout
#include <sstream> // std::stringstream
int main () {
std::stringstream foo;
std::stringstream bar;
foo << 100;
bar << 200;
foo.swap(bar);
int val;
foo >> val; std::cout << "foo: " << val << '\n';
bar >> val; std::cout << "bar: " << val << '\n';
return 0;
}
source: http://ideone.com/NI0xMS
cplusplus.com source: http://www.cplusplus.com/reference/sstream/stringstream/swap/
prog.cpp: In function ‘int main()’:
prog.cpp:14:7: error: ‘std::stringstream’ has no member named ‘swap’
foo.swap(bar);
What am I missing? Why can't I move or swap a stringstream? How should I swap or move a stringstream?
This is a missing feature on GCC : see bug 54316 , it has been fixed (you can thank Jonathan Wakely) for the next versions (gcc 5)
Clang with libc++ compiles this code :
int main () {
std::stringstream stream("1234");
std::stringstream stream2 = std::move(std::stringstream("5678"));
return 0;
}
Live demo
And it also compiles the example with std::stringstream::swap
I have an alternative to moving or swapping, one can also clear and set a stringstream to a new string:
#include <string> // std::string
#include <iostream> // std::cout
#include <sstream> // std::stringstream
int main () {
std::stringstream ss("1234");
ss.clear();
ss.str("5678");
int val;
ss >> val; std::cout << "val: " << val << '\n';
return 0;
}
It's a clean work around that does not require one to refactor code, except for the localized section where the swap is changed to a clear() and str().