6.15 — Implicit type conversion (coercion)

Previously, you learned that a value of a variable is stored as a sequence of bits, and the data type of the variable tells the compiler how to interpret those bits into meaningful values. Different data types may represent the “same” number differently -- for example, the integer value 3 and the float value 3.0 are stored as completely different binary patterns.

So what happens when we do something like this?

In such a case, the compiler can’t just copy the bits representing the int value 3 into float variable f. Instead, it needs to convert the integer value 3 to a floating point number, which can then be assigned to float variable f.

The process of converting a value from one data type to another is called a type conversion. Type conversions can happen in many different cases:

  • When assigning to or initializing a variable with a value of a different data type:

  • When passing a value to a function where the function parameter is of a different data type:

  • When returning a value from a function where the function return type is of a different data type:

  • Using a binary operator with operands of different types:

In all of these cases (and quite a few others), C++ will use type conversion to convert one data type to another data type.

There are two basic types of type conversion: implicit type conversion, where the compiler automatically transforms one data type into another, and explicit type conversion, where the developer uses a casting operator to direct the conversion.

We’ll cover implicit type conversion in this lesson, and explicit type conversion in the next.

Implicit type conversion

Implicit type conversion (also called automatic type conversion or coercion) is performed whenever one data type is expected, but a different data type is supplied. If the compiler can figure out how to do the conversion between the two types, it will. If it doesn’t know how, then it will fail with a compile error.

All of the above examples are cases where implicit type conversion will be used.

There are two basic types of implicit type conversion: promotions and conversions.

Numeric promotion

Whenever a value from one fundamental data type is converted into a value of a larger fundamental data type from the same family, this is called a numeric promotion (or widening, though this term is usually reserved for integers). For example, an int can be widened into a long, or a float promoted into a double:

While the term numeric promotion covers any type of promotion, there are two other terms with specific meanings in C++:

  • Integral promotion involves the conversion of integer types narrower than int (which includes bool, char, unsigned char, signed char, unsigned short, and signed short) to an int (if possible) or an unsigned int (otherwise).
  • Floating point promotion involves the conversion of a float to a double.

Integral promotion and floating point promotion are used in specific cases to convert smaller data types to int/unsigned int or double, because int and double are generally the most performant types to perform operations on.

The important thing to remember about promotions is that they are always safe, and no data loss will result.

For advanced readers

Under the hood, promotions generally involve extending the binary representation of a number (e.g. for integers, adding leading 0s).

Numeric conversions

When we convert a value from a larger type to a similar smaller type, or between different types, this is called a numeric conversion. For example:

Unlike promotions, which are always safe, conversions may or may not result in a loss of data. Because of this, code that causes an implicit conversion to be performed will often cause the compiler to issue a warning.

The rules for conversions are complicated and numerous, so we’ll just cover the common cases here.

In all cases, converting a value into a type that doesn’t have a large enough range to support the value will lead to unexpected results. For example:

In this example, we’ve assigned a large integer to a char (that has range -128 to 127). This causes the char to overflow, and produces an unexpected result:


However, converting from a larger integral or floating point type to a smaller similar type will generally work so long as the value fits in the range of the smaller type. For example:

This produces the expected result:


In the case of floating point values, some rounding may occur due to a loss of precision in the smaller type. For example:

In this case, we see a loss of precision because the float can’t hold as much precision as a double:


Converting from an integer to a floating point number generally works as long as the value fits within the range of the floating type. For example:

This produces the expected result:


Converting from a floating point to an integer works as long as the value fits within the range of the integer, but any fractional values are lost. For example:

In this example, the fractional value (.5) is lost, leaving the following result:


Conversions that could cause loss of information, e.g. floating point to integer, are called narrowing conversions. Since information loss is generally undesirable, brace initialization doesn’t allow narrowing conversions.

Evaluating arithmetic expressions

When evaluating expressions, the compiler breaks each expression down into individual subexpressions. The arithmetic operators require their operands to be of the same type. To ensure this, the compiler uses the following rules:

  • If an operand is an integer that is narrower than an int, it undergoes integral promotion (as described above) to int or unsigned int.
  • If the operands still do not match, then the compiler finds the highest priority operand and implicitly converts the other operand to match.

The priority of operands is as follows:

  • long double (highest)
  • double
  • float
  • unsigned long long
  • long long
  • unsigned long
  • long
  • unsigned int
  • int (lowest)

We can see the usual arithmetic conversion take place via use of the typeid operator (included in the <typeinfo> header), which can be used to show the resulting type of an expression.

In the following example, we add two shorts:

Because shorts are integers, they undergo integral promotion to ints before being added. The result of adding two ints is an int, as you would expect:

int 9

Note: Your compiler may display something slightly different, as the format of is left up to the compiler.

Let’s take a look at another case:

In this case, the short undergoes integral promotion to an int. However, the int and double still do not match. Since double is higher on the hierarchy of types, the integer 2 gets converted to the double 2.0, and the doubles are added to produce a double result.

double 6.0

This hierarchy can cause some interesting issues. For example, take a look at the following code:

you might expect the expression 5u - 10 to evaluate to -5 since 5 - 10 = -5. But here’s what actually happens:


In this case, the signed integer 10 is promoted to an unsigned integer (which has higher priority), and the expression is evaluated as an unsigned int. Since -5 can’t be stored in an unsigned int, the calculation wraps around, and we get an answer we don’t expect.

This is one of many good reasons to avoid unsigned integers in general.

Quiz time

Question #1

What’s the difference between a numeric promotion and a numeric conversion?

Show Solution

6.16 -- Explicit type conversion (casting) and static_cast
6.14 -- The auto keyword

145 comments to 6.15 — Implicit type conversion (coercion)

  • Chayim

    Why does this code not work?

  • Forhad Rahman

    My compiler is showing 'i' instead of 'int'. That's okay. But I didn't include <typeinfo> in header. Despite, it's showing the type.

    Is it normal?

  • Gwennan

    "promotions generally involve extending the binary representation of a number (e.g. for integers, adding leading 0s)"

    Just wanted to  point out that it's true that promotion for unsigned integers is adding leading 0s, but for signed I'd say we need to add leading 1s in order to make the promotion when the number is negative, and 0s if it's non-negative.

  • Alan LAI

    I find an good example of numeric conversions is converting int to char

  • Muscaria


    I'm guessing that they do that in order to ensure that when these values are operated on the result will have sufficient space. It isn't really checking the size of what the short ints contain. It's just a safety feature. Again, that is just my speculation.

  • bob

    I think a note should be added in regards to the "." operator instead of just thrusting it in our faces without assuming we've magically heard of it before when it hasn't even been mentioned yet.

  • SuperNoob

    Wanted to resolve input type in the runtime but failed hilariously :D Cause once the auto type is initialized with a literal, that "auto" type gains the type of that literal. So there's no way to resolve input type in this SuperNoob method :')

    Sample Output:

    Enter anything: asdf

  • J34NP3T3R

    in this example ;

    int main()
        short a{ 4 };
        short b{ 5 };
        std::cout << typeid(a + b).name() << ' ' << a + b << '\n'; // show us the type of a + b

        return 0;
    Because shorts are integers, they undergo integral promotion to ints before being added. The result of adding two ints is an int, as you would expect:

    why does integral promotion happen when both a and b are of the same type and the result is also within range of "short" ?

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