8.5 — Explicit type conversion (casting) and static_cast

In lesson 8.1 -- Implicit type conversion (coercion), we discussed that the compiler can implicitly convert a value from one data type to another through a system called implicit type conversion. When you want to numerically promote a value from one data type to a wider data type, using implicit type conversion is fine.

Many new C++ programmers try something like this:


double d = 10 / 4; // does integer division, initializes d with value 2.0

Because 10 and 4 are both of type int, integer division is performed, and the expression evaluates to int value 2. This value then undergoes numeric conversion to double value 2.0 before being used to initialize variable d. Most likely, this isn’t what was intended.

In the case where you are using literal operands, replacing one or both of the integer literals with double literals will cause floating point division to happen instead:


double d = 10.0 / 4.0; // does floating point division, initializes d with value 2.5

But what if you are using variables instead of literals? Consider this case:


int x { 10 };
int y { 4 };
double d = x / y; // does integer division, initializes d with value 2.0

Because integer division is used here, variable d will end up with the value of 2.0. How do we tell the compiler that we want to use floating point division instead of integer division in this case? Literal suffixes can’t be used with variables. We need some way to convert one (or both) of the variable operands to a floating point type, so that floating point division will be used instead.

Fortunately, C++ comes with a number of different type casting operators (more commonly called casts) that can be used by the programmer to request that the compiler perform a type conversion. Because casts are explicit requests by the programmer, this form of type conversion is often called an explicit type conversion (as opposed to implicit type conversion, where the compiler performs a type conversion automatically).

Type casting

C++ supports 5 different types of casts: C-style casts, static casts, const casts, dynamic casts, and reinterpret casts. The latter four are sometimes referred to as named casts.

We’ll cover C-style casts and static casts in this lesson.

Related content

We discuss dynamic casts in lesson 18.10 -- Dynamic casting, after we’ve covered other prerequisite topics.

Const casts and reinterpret casts should generally be avoided because they are only useful in rare cases and can be harmful if used incorrectly.

Warning

Avoid const casts and reinterpret casts unless you have a very good reason to use them.

C-style casts

In standard C programming, casts are done via the () operator, with the name of the type to convert the value to placed inside the parenthesis. You may still see these used in code (or by programmers) that have been converted from C.

For example:


#include <iostream>

int main()
{
    int x { 10 };
    int y { 4 };

    
    double d { (double)x / y }; // convert x to a double so we get floating point division
    std::cout << d; // prints 2.5

    return 0;
}

In the above program, we use a C-style cast to tell the compiler to convert x to a double. Because the left operand of operator/ now evaluates to a floating point value, the right operator will be converted to a floating point value as well, and the division will be done using floating point division instead of integer division!

C++ will also let you use a C-style cast with a more function-call like syntax:


    double d { double(x) / y }; // convert x to a double so we get floating point division

This performs identically to the prior example, but has the benefit of parenthesizing the value being converted (making it easier to tell what is being converted).

Although a C-style cast appears to be a single cast, it can actually perform a variety of different conversions depending on context. This can include a static cast, a const cast or a reinterpret cast (the latter two of which we mentioned above you should avoid). As a result, C-style casts are at risk for being inadvertently misused and not producing the expected behavior, something which is easily avoidable by using the C++ casts instead.

Related content

If you’re curious, this article has more information on how C-style casts actually work.

Best practice

Avoid using C-style casts.

static_cast

C++ introduces a casting operator called static_cast, which can be used to convert a value of one type to a value of another type.

You’ve previously seen static_cast used to convert a char into an int so that std::cout prints it as an integer instead of a char:


#include <iostream>

int main()
{
    char c { 'a' };
    std::cout << c << ' ' << static_cast<int>(c) << '\n'; // prints a 97

    return 0;
}

The static_cast operator takes an expression as input, and returns the evaluated value converted to the type specified inside the angled brackets. static_cast is best used to convert one fundamental type into another.


#include <iostream>

int main()
{
    int x { 10 };
    int y { 4 };

    // static cast x to a double so we get floating point division
    double d { static_cast<double>(x) / y };  
    std::cout << d; // prints 2.5

    return 0;
}

The main advantage of static_cast is that it provides compile-time type checking, making it harder to make an inadvertent error. static_cast is also (intentionally) less powerful than C-style casts, so you can’t inadvertently remove const or do other things you may not have intended to do.

Best practice

Favor static_cast when you need to convert a value from one type to another type.

Using static_cast to make narrowing conversions explicit

Compilers will often issue warnings when a potentially unsafe (narrowing) implicit type conversion is performed. For example, consider the following program:


int i { 48 };
char ch = i; // implicit narrowing conversion

Casting an int (2 or 4 bytes) to a char (1 byte) is potentially unsafe (as the compiler can’t tell whether the integer value will overflow the range of the char or not), and so the compiler will typically print a warning. If we used list initialization, the compiler would yield an error.

To get around this, we can use a static cast to explicitly convert our integer to a char:


int i { 48 };

// explicit conversion from int to char, so that a char is assigned to variable ch
char ch { static_cast<char>(i) };

When we do this, we’re explicitly telling the compiler that this conversion is intended, and we accept responsibility for the consequences (e.g. overflowing the range of a char if that happens). Since the output of this static_cast is of type char, the initialization of variable ch doesn’t generate any type mismatches, and hence no warnings or errors.

Here’s another example where the compiler will typically complain that converting a double to an int may result in loss of data:


int i { 100 };
i = i / 2.5;

To tell the compiler that we explicitly mean to do this:


int i { 100 };
i = static_cast<int>(i / 2.5);

Quiz time

Question #1

What’s the difference between implicit and explicit type conversion?

Show Solution

guest
Your email address will not be displayed
Avatars from https://gravatar.com/ are connected to your provided email address.
Notify me about replies:  
150 Comments
Newest
Oldest Most Voted
Inline Feedbacks
View all comments