(Q) Static typing

Question 1

* Changing the active project:

Now you have two projects open in Qt Creator at the same time. One of the projects is the active one (the name of the project is written in bold in Projects window). If you click compile or run buttons, the action concerns the active project. How can you switch the active project?

By double-clicking the name of the unactive project

From the menu that will be opened, when clicking the name of the unactive project with the right mouse button

From File menu

Question 2

* Type conversion 1:

Write the following lines to your program:

bool b = 1 + 2;
cout << b << endl;

the latter of which just makes easier to study the topic in more detail. What happens, when you try to compile and run the program?

The program cannot be compiled due to a type error (it is not possible to assign an integer to a boolean variable).

Compilation succeeds, but it gives a warning about an implicit type conversion from an integer to a truth value.

Compilation succeeds without warnings, although the compiler makes an implicit type conversion from an integer to a truth value.

Question 3

* Type conversion 2:

Write the following lines to your program:

if ( 1 + 2 ) {
    cout << "Hip" << endl;
}

What happens, when you try to compile and run the program?

The program cannot be compiled due to a type error (an integer cannot be used as a condition).

Compilation succeeds, but it gives a warning about an implicit type conversion from an integer to a truth value.

Compilation succeeds without warnings, although the compiler makes an implicit type conversion from an integer to a truth value.

Question 4

* Type conversion 3:

Write the following line to your program:

string s = 3;

What happens, when you try to compile and run the program?

The program cannot be compiled due to a type error (it is not possible to assign an integer to a string).

Compilation succeeds, but it gives a warning about an implicit type conversion from an integer to a string.

Compilation succeeds without warnings, although the compiler makes an implicit type conversion from an integer to a string.

Question 5

* Type conversion 4:

Write the following lines to your program:

int i = -1;
unsigned int ui = 1;
cout << ui << endl;

if ( ui == i ) {
    cout << "Hello" << endl;
}

What happens, when you try to compile and run the program?

The program cannot be compiled due to a type error (it is not possible to compare between unsigned int and int).

Compilation succeeds, but it gives a warning about an implicit type conversion when comparing unsigned int and int.

Compilation succeeds without warnings, although the compiler makes an implicit type conversion when comparing unsigned int and int.

Question 6

* Type conversion 5:

Write the following lines to your program:

int i;

if ( i = 0 ) {
    cout << "Hurrah" << endl;
}

Note that the condition of the if structure has not relational operator == but assignment operator =. So, we are doing something stupid.

In addition, you need to know that the assignment operator is indeed an operator in C++, as said previously. This means that it is possible to chain assignments as:

i = j = k = 0;

Previous line first assigns k = 0, the result of which is 0. Next the result of assignment k = 0 is assigned to variable j, and so on. What happens, when you try to compile and run the program?

The program cannot be compiled due to a type error (it is not possible to use the integer that is the result of an assignment as a condition).

Compilation succeeds, but it gives a warning about an implicit type conversion from an integer to a truth value.

Compilation succeeds, but it gives a warning, because assigning in a condition is stupid. After the warning, the compiler makes an implicit type conversion from an integer to a truth value.

Compilation succeeds without warnings, although the compiler makes an implicit type conversion from an integer to a truth value.

Question 7

* Static typing:

So, what is meant with static typing in practice?

The type of all variables and literals is determined already at compile-time.

The value that is assigned to a variable must be of the same type as the variable itself.

The checkings made by the compiler prevent type errors concerning types of variables.

(Q) Static typing¶