T F The structure pointer operator does not automatically dereference the structure pointer on its left.

Structure pointers are variables that hold the address of a structure. They are used when we need to access and manipulate data in a structure using the address where the structure is stored in memory. The structure pointer operator is used to access members of a structure using a pointer to the structure. The structure pointer operator is represented by -> in programming languages like C and C++. The left-hand side of the operator has a pointer to the structure, and the right-hand side of the operator has the member name we want to access.

Dereferencing a pointer means getting the value stored at the memory address to which the pointer is pointing. In programming languages like C and C++, the dereference operator (*) is used to access the value pointed to by a pointer.

Now, let us evaluate the given statement "The structure pointer operator does not automatically dereference the structure pointer on its left". When we use the structure pointer operator (->), it automatically dereferences the pointer on the left side of the operator and accesses the member on the right side of the operator. In other words, the structure pointer operator (->) takes care of both dereferencing the pointer and accessing the structure member. For example, consider the following structure and pointer declaration in C++: ```cpp struct Example { int data; }; Example e1 = {5}; Example* ePtr = &e1 ``` Now, if we want to access the 'data' member of the structure using the structure pointer ePtr, we write: ```cpp int value = ePtr->data; cout << value << endl; // Output is 5 ``` In this example, the structure pointer operator (->) dereferenced the pointer 'ePtr' and accessed the 'data' member of the structure.

Based on our understanding and the example shown above, it is clear that the structure pointer operator does automatically dereference the structure pointer on its left. Thus, the given statement is False.