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To initialize the const value using constructor, we have to use the initialize list. This initializer list is used to initialize the data member of a class. The list of members, that will be initialized, will be present after the constructor after colon. members will be separated using comma.
A non-const reference cannot point to a literal. You cannot bind a literal to a reference to non-const (because modifying the value of a literal is not an operation that makes sense) and only l-values can be bound to references to non-const.
There are three steps to initializing a reference variable from scratch: declaring the reference variable; using the new operator to build an object and create a reference to the object; and. storing the reference in the variable.
You can declare the test function as: int test(gadget const &g); In this case, parameter g has type “reference to const gadget .” This lets you write the call as test(x) , as if it were passing by value, but it yields the exact same performance as if it were passing by address.
How does the following code work in C++? Is it logical?
const int &ref = 9; const int &another_ref = ref + 6; Why does C++ allow literal initialization for const references when the same is not permitted for non-const references? E.g.:
const int days_of_week = 7; int &dof = days_of_week; //error: non const reference to a const object This can be explained by the fact that, a non-const reference can be used to change the value of the variable it is referring to. Hence, C++ does not permit a non-const reference to a const variable.
Could this be a possible explanation? C++ does not allow:
int &ref = 7; Because that is not logical, but:
const int &ref = 7; Is almost equivalent to:
const int val = 7; So literal initialization is permitted for const variables.
P.S.: I'm currently studying Lippman's C++ Primer.
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