# 7.1 Ap0010: self-assessment, primitive types  (Page 5/9)

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## Miscellaneous

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Housekeeping material
• Module name: Ap0010: Self-assessment, Primitive types
• File: Ap0010.htm
• Originally published: December 17, 2001
• Published at cnx.org: 12/01/12
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D. 0.3333333333333333

## Explanation 10

Divide floating type by integer type

This program divides the literal floating value of 1.0 by the literal integer value of 3 (no decimal point is specified in the integer literal value) .

Automatic conversion from narrow to wider type

To begin with, whenever division is performed between a floating type and an integer type, the integer type is automatically converted (sometimes called promoted) to a floating type and floating arithmetic is performed.

What is the actual floating type, float or double?

The real question here is, what is the type of the literal shown by 1.0 (with a decimal point separating the 1 and the 0) . Is it a double or a float ?

Type double is the default

By default, a literal floating value is treated as a double .

The result is type double

Consequently, this program divides a double type by an integer type, producing a result of type double . This is somewhat evident in the output, which shows about 17 digits plus a decimal point in the result. (Recall that the maximum value for a float shown earlier had only about eight digits plus thedecimal point and the exponent.)

How can you create literals of type float?

What if you don't want your literal floating value to be treated as a double , but would prefer that it be treated as a float instead.

You can usually force this to be the case by adding a suffix of either F or f to the end of the literal value (as in 1.0F) . If you were to modify this program to cause it to divide 1.0F by 3, the output would be 0.33333334 withonly nine digits in the result.

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