What is the behaviour on converting a negative floating point value into an unsigned int? [duplicate]

Answer 1

In case the negative value is -1.0 or lower, it invokes undefined behavior since the integral part then cannot be represented by an unsigned number. Otherwise, (as in the case of -0.1), if it can be represented by an integer type, it is well-defined behavior. See the C11 standard, ISO 9899:2011:

6.3.1.4

When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero). If the value of the integral part cannot be represented by the integer type, the behavior is undefined. 61)

And then there is a non-normative foot note explaining the above text:

61) The remaindering operation performed when a value of integer type is converted to unsigned type need not be performed when a value of real floating type is converted to unsigned type. Thus, the range of portable real floating values is (−1, Utype_MAX+1).

ISO/IEC 9899:1999 (C99) contains exactly the same text.

Answer 2

Your example, unsigned i = -.1; is well-defined by both C11 and C99, and the result is i == 0.

Quoted from N1570, 6.3.1.4 Real floating and integer:

1. When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero). If the value of the integral part cannot be represented by the integer type, the behavior is undefined.61)

_{61) The remaindering operation performed when a value of integer type is converted to unsigned type need not be performed when a value of real floating type is converted to unsigned type. Thus, the range of portable real floating values is (-1, Utype_MAX+1).}

Quoted from N869, 6.3.1.4 Real floating and integer:

#1

When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero). If the value of the integral part cannot be represented by the integer type, the behavior is undefined.43)

_{43)The remaindering operation performed when a value of integer type is converted to unsigned type need not be performed when a value of real floating type is converted to unsigned type. Thus, the range of portable real floating values is (-1, Utype_MAX+1).}

However, as you can see from the quotations, trying to convert floating-point constants outside the range (-1, Utype_MAX+1) invokes undefined behaviour.

Answer 3

It is undefined behaviour in C99 if the floating point number is less than or equal to -1.0. If it's in the range (-1.0, 0.0), the resulting value will be 0.

From C99, §6.3.1.4, paragraph 1

When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero). If the value of the integral part cannot be represented by the integer type, the behavior is undefined

Footnote 50 clarifies the behaviour for the (-1.0, 0.0) range.