#include<iostream>
using namespace std;
int main()
{
char a[2][2] = {"A","B"};
cout << a << endl;
}
Since a[] stores the address of the first index it have to print A and B. But it is printing
some address. What is the wrong in the program to print A and B ?
The problem lies in the overload of operator<< for std::ostream. The specialization that prints ascii characters uses const char *, and you are trying to print char[][], which will decay into a type that's going to be printed as a "generic pointer" (void const * to be precise) - char(*)[2], or "a pointer to two-element array of char (thanks to @Wintermute for that).
std::array<std::string, 2> a { "A", "B" }; is probably the most idiomatic solution here.
If you don't want to modify the contents, std::array<const char*, 2> will also do just fine; const char* is the type of string literals. If you store them that way, you're actually storing the addresses to where the compiler puts them when building the binary. This memory isn't modifiable by you, that's why you have to use const. The string variant copies the data, when that place of code is encountered, to a part of memory you have write access to.
Of course, when you choose the proper way, the actual printing needs to be changed, too!
for (auto const& str : a) {
cout << str << endl;
}
Oh and also, you probably don't want to use using namespace std;.
char* is printed as a string, not as a pointer. Nor does the array decay to char*; it is not one-dimensional.char[]* insteadchar(*)[2], which is a pointer to an array of two characters. The ultimately used operator<< is the one for void const *, because char(*)[2] is convertible to it but not to char const *.[] and [n] :S. Thanks for your comments!char *(*)[2], which is a pointer to an array of two pointers to char. This is still convertible to void const * but not to char const *.In C++, just make yourself a favor and use some convenient container classes like a proper combination of std::vector and std::string: std::vector<std::string>.
std::vector<std::string> a{"A", "B"};
is much simpler and (easier to extend, e.g. if you want to dynamically add more strings to the vector) than the raw-C-like solution.
coutwith an array will not work to print all elements of the array, FYI.