Short answer: cout is interpreting the object as a bool due to the volatile qualifier. It's a quirk of overloading for the << operator.
Long answer: A volatile pointer can't be converted to a non-volatile pointer without an explicit cast, so neither the char* nor the void* overload can be used when the << operator is called. There's no volatile qualified overload, and the closest match is the bool overload, thus your array is interpreted as a boolean value rather than an address or a string.
You can fix it a number of ways, but an explicit cast is probably what you wanted:
std::cout<< (char*)str <<std::endl;
p.s. While your compiler may set uninitialized variables to 0 automatically, it's a very bad idea to rely on that behavior because it's not universal. So in your loop, int i; should be int i = 0; .
i am trying to create a library management system which is supposed to be used to store/keep infp for the library and the library members( book title, author, name and picture for borrower and other stuffs). The app has 6 buttons, delete, save, add,previsous,next, and upload( for upload ...