8.3 — Numeric conversions

In the previous lesson (8.2 -- Floating-point and integral promotion), we covered numeric promotions, which are conversions of specific narrower numeric types to wider numeric types (typically int or double) that can be processed efficiently.

C++ supports another category of numeric type conversions, called numeric conversions, that cover additional type conversions not covered by the numeric promotion rules.

There are five basic types of numeric conversions.

1. Converting an integral type to any other integral type (excluding integral promotions):

short s = 3; // convert int to short
long l = 3; // convert int to long
char ch = s; // convert short to char

2. Converting a floating point type to any other floating point type (excluding floating point promotions):

float f = 3.0; // convert double to float
long double ld = 3.0; // convert double to long double

3. Converting a floating point type to any integral type:

int i = 3.5; // convert double to int

4. Converting an integral type to any floating point type:

double d = 3; // convert int to double

5. Converting an integral type or a floating point type to a bool:

bool b1 = 3; // convert int to bool
bool b2 = 3.0; // convert double to bool

Narrowing conversions

Unlike a numeric promotion (which is always safe), a numeric conversion may (or may not) result in the loss of data or precision.

Some numeric conversions are always safe (such as int to long, or int to double). Other numeric conversions, such as double to int, may result in the loss of data (depending on the specific value being converted and/or the range of the underlying types):

int i1 = 3.5; // the 0.5 is dropped, resulting in lost data
int i2 = 3.0; // okay, will be converted to value 3, so no data is lost

In C++, a narrowing conversion is a numeric conversion that may result in the loss of data. Such narrowing conversions include:

• From a floating point type to an integral type.
• From a wider floating point type to a narrower floating point type, unless the value being converted is constexpr and is in range of the destination type (even if the narrower type doesn’t have the precision to store the whole number).
• From an integral to a floating point type, unless the value being converted is constexpr and is in range of the destination type and can be converted back into the original type without data loss.
• From a wider integral type to a narrower integral type, unless the value being converted is constexpr and after integral promotion will fit into the destination type.

The good news is that you don’t need to remember these. Your compiler will usually issue a warning (or error) when it determines that an implicit narrowing conversion is required.

For example, when compiling the following program:

int main()
{
    int i = 3.5;
}

Visual Studio produces the following warning:

warning C4244: 'initializing': conversion from 'double' to 'int', possible loss of data

In general, narrowing conversions should be avoided, but there are situational cases where you might need to do one. In such cases, you should make the implicit narrowing conversion explicit by using static_cast. For example:

void someFcn(int i)
{
}

int main()
{
    double d{ 5.0 };
    
    someFcn(d); // bad: will generate compiler warning about narrowing conversion
    someFcn(static_cast<int>(d)); // good: we're explicitly telling the compiler this narrowing conversion is expected, no warning generated
    
    return 0;
}

Brace initialization disallows narrowing conversions

Narrowing conversions are strictly disallowed when using brace initialization (which is one of the primary reasons this initialization form is preferred):

int main()
{
    int i { 3.5 }; // won't compile
}

Visual Studio produces the following error:

error C2397: conversion from 'double' to 'int' requires a narrowing conversion

More on numeric conversions

The specific rules for numeric conversions are complicated and numerous, so here are the most important things to remember.

In all cases, converting a value into a type whose range doesn’t support that value will lead to results that are probably unexpected. For example:

int main()
{
    int i{ 30000 };
    char c = i; // chars have range -128 to 127

    std::cout << static_cast<int>(c);

    return 0;
}

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

48

Converting from a larger integral or floating point type to a smaller type from the same family will generally work so long as the value fits in the range of the smaller type. For example:

    int i{ 2 };
    short s = i; // convert from int to short
    std::cout << s << '\n';

    double d{ 0.1234 };
    float f = d;
    std::cout << f << '\n';

This produces the expected result:

2
0.1234

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

    float f = 0.123456789; // double value 0.123456789 has 9 significant digits, but float can only support about 7
    std::cout << std::setprecision(9) << f << '\n'; // std::setprecision defined in iomanip header

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

0.123456791

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

    int i{ 10 };
    float f = i;
    std::cout << f;

This produces the expected result:

10

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:

    int i = 3.5;
    std::cout << i << '\n';

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

3

While the numeric conversion rules might seem scary, in reality the compiler will generally warn you if you try to do something dangerous (excluding some signed/unsigned conversions).