I'm studying some C++ features, trying to play around with some experiments. However, I stuck in a place where it compiled error:
#include <iostream>
#include <string>
using namespace std;
int main()
{
string str = "some string";
auto &c = str.begin(); // compile error
*c = toupper(*c);
cout << *c << ", str: " << str << endl;
}
I'm not sure why it was not acceptable. My thought was that c had type char * (a pointer to a char), so that's why I had written as above. But why it failed in compiling?
Error C2440 Cannot transform 'std::_String_iteratorstd::_String_val<std::_Simple_types<_Elem>>' to'std::_String_iterator<std::_String_val<std::_Simple_types<_Elem
PS: Another method which I had tried first was OK.
#include <iostream>
#include <string>
using namespace std;
int main()
{
string str = "some string";
auto &c = *str.begin(); // success
c = toupper(c);
cout << c << ", str: " << str << endl;
}
begin() returns an iterator by value, not a reference. You are not allowed to form a non-const lvalue reference.
Making it const would prolong the life of the returned iterator and the program would then compile:
const auto &c = str.begin();
On the other hand, iterators are supposed to be cheap to copy and iterators from contiguous containers are often implemented as pure pointers. The idiomatic approach is:
auto c = str.begin();
In your second example, the idiomatic approach to form a reference to the first element would be:
auto& c = str.front();
Related
I'm trying to create a program that filters through speech text, removes any unwanted characters (",", "?", etc., etc.") and then produces a new speech where the words are jumbled based on what words follow or precede them. So for example, if you had the Gettysburg Address:
Four score and seven years ago our fathers brought forth, on this continent, a new nation, conceived in Liberty, and dedicated to the proposition that all men are created equal.
my program would take that text, put it into a set of strings. i.e. ["Four","score","and","seven",...."continent,"..."Liberty,"..."equal."] Then it would remove any unwanted characters from each string using c++ .erase and c++ .remove, like "," or "." and capitals. After, you'd have a filtered string like ["four","score","and","seven",...."continent"..."liberty"..."equal."]
After that then the words would be rearranged into a new coherent, funnier speech, like:
"Seven years ago our fathers conceived on men...", etc.
That was just so you know the scope of this project. My trouble at the moment has to do with either using my iterator properly or null terminators.
#include <iostream>
#include <fstream>
#include <iomanip>
#include <string>
#include <set>
#include <iterator> //iterates through sets
#include <algorithm>
using namespace std;
int main() {
set <string> speechSet;
set <string> ::iterator itr; //forgot what :: means. Declares iterator as set
int sum = 0;
int x;
string data;
ofstream out;
string setString;
ifstream speechFile; //declare output file stream object. Unknown type name
speechFile.open("./MySpeech");
if (!speechFile) {
cerr << "Unable to open file " << endl;
exit(1);
}
char unwantedCharacters[] = ".";
while (!speechFile.eof()) {
speechFile >> data; //speechFile input into data
for (unsigned int i = 0; i < strlen(unwantedCharacters); ++i) {
data.erase((remove(data.begin(), data.end(),
unwantedCharacters[i]), data.end())); //remove doesn't delete.
data.end() - 1 = '\0'; //Reorganizes
cout << data << endl;
}
speechSet.insert(string(data));
}
//Go through each string (word) one at a time and remove "",?, etc.
/*for(itr = speechSet.begin(); itr != speechSet.end(); ++itr){
if(*itr == ".")//if value pointed to by *itr is equal to '.'
itr = speechSet.erase(itr);//erase the value in the set and leave blank
cout << " " << *itr;//print out the blank
else{
cout << " " << *itr;
}
}*/
speechFile.close();
return (0);
}
I keep getting an error that says error: no viable overloaded '='. At first I thought it might be due to .end() not being a command for a C++ string, but I checked the documentation and it shouldn't be an issue of mismatched data typed. Then I thought it might have to set the iterator itr equal to the end of the data.
iterator itr = data.end() - 1;
and then dereference that pointer and set it equal to the null terminator
itr* = '\0';
That removed the overload error, but I still had another error use of class template 'iterator' requires template arguments. Let me know if any more clarification is needed.
In the for loop, use auto for iterator so you don't have to specify its type like:
for(auto itr = speechSet.begin(); itr != speechSet.end(); ++itr){
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'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.
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().
I want to pass out some parameter from the lambda function, so I bind a reference parameter to a lambda function. However, the outer variable is not changed after calling the function. If I bind the lambda function with the pointer of outer variable, the result is correct.
I show the test program as follows and want to know why the outer variable does not change, given that I have already defined the pass-by-reference lambda function [&]?
#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>
using namespace std;
int main(int argc, char* argv[])
{
string acc_ref, acc_ptr;
// define function via reference
auto fProcRefEx = [&](string& acc, const string& s) -> void
{
acc += s;
};
auto fProcRef = bind(fProcRefEx, acc_ref, placeholders::_1);
// define function via pointer
auto fProcPtrEx = [&](string* pacc, const string& s) -> void
{
(*pacc) += s;
};
auto fProcPtr = bind(fProcPtrEx, &acc_ptr, placeholders::_1);
// test
vector<string> v = {"abc", "def"};
for_each(v.begin(), v.end(), fProcRef);
cout << "acc_ref: " << acc_ref << endl; // acc_ref is empty, wrong
for_each(v.begin(), v.end(), fProcPtr);
cout << "acc_ptr: " << acc_ptr << endl; // acc_ptr is "abcdef", correct
return 0;
}
I think std::bind will decay the reference to just a plain value type when it stores acc_ref. ie In the unspecified object instance returned from bind, it will have a member string acc_ref, not string& acc_ref. You have to use std::ref to make it really store a reference:
auto fProcRef = bind(fProcRefEx, ref(acc_ref), placeholders::_1);