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In this response:
https://stackoverflow.com/a/14382318/1676605
this program is given:
std::vector<int> vi{ 0, 2, 4 };
std::vector<std::string> vs{ "1", "3", "5", "7" };
for (auto i : redi::zip(vi, vs))
std::cout << i.get<0>() << ' ' << i.get<1>() << ' ';
I have no idea what the type of auto i is, making it harder to reuse expertise and learn from examples. Here is what changing auto i into char i returns
In function ‘int main()’:|
/data/cbworkspace/TestZip/TestZip.cpp|14|error: cannot convert ‘boost::iterator_facade<boost::zip_iterator<boost::tuples::tuple<__gnu_cxx::__normal_iterator<int*, std::vector<int> >, __gnu_cxx::__normal_iterator<int*, std::vector<int> > > >, boost::tuples::cons<int&, boost::tuples::cons<int&, boost::tuples::null_type> >, boost::random_access_traversal_tag, boost::tuples::cons<int&, boost::tuples::cons<int&, boost::tuples::null_type> >, long int>::reference {aka boost::tuples::cons<int&, boost::tuples::cons<int&, boost::tuples::null_type> >}’ to ‘char’ in initialization|
/data/cbworkspace/TestZip/TestZip.cpp|14|warning: unused variable ‘i’ [-Wunused-variable]|
||=== Build finished: 1 errors, 1 warnings (0 minutes, 0 seconds) ===|
Try to figure out the type from that.
Is there a way to figure out what the type a variable of an auto is in C++11? To be more clear, I have a struct like this:
struct EventData
{
// return value from redi::zip<std::vector<PriceQuote>, std::vector<PriceQuote>> what goes here????? So REDI::Zip is zipping PriceQuote, PriceQuote of bids and asks.
};
struct PriceQuote
{
double price;
double size;
};
333
In computer programming, an automatic variable is a local variable which is allocated and deallocated automatically when program flow enters and leaves the variable's scope. The scope is the lexical context, particularly the function or block in which a variable is defined.
To get the datatype of variable, use typeid(x). name() of typeinfo library. It returns the type name of the variable as a string.
The following code fragment initializes variable x to type int , variable y to a reference to type const int , and variable fp to a pointer to a function that returns type int . int f(int x) { return x; } int main() { auto x = f(0); const auto& y = f(1); int (*p)(int x); p = f; auto fp = p; //... }
The auto keyword can not only be used as the return type for functions but also as the data type of the parameters declared in the function.
Try to change auto into a char and read the error message.
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