Signed and unsigned integers with logical operators

int x = /* some integer */; 

unsigned int ux = (unsigned) x;

we have

x >= 0 || x < ux

we know that in x < ux the first x is cast implicitly to unsigned but is the first x in x >= 0 (1) cast to unsigned implicitly?

edited Jan 2 at 18:33

bolov

33.3k877141

asked Jan 2 at 18:29

durnovv

546

Detail: x is not cast - there is no cast in x < ux. x is converted to an unsigned.

– chux
Jan 2 at 22:09

add a comment |

int x = /* some integer */; 

unsigned int ux = (unsigned) x;

we have

x >= 0 || x < ux

we know that in x < ux the first x is cast implicitly to unsigned but is the first x in x >= 0 (1) cast to unsigned implicitly?

edited Jan 2 at 18:33

bolov

33.3k877141

asked Jan 2 at 18:29

durnovv

546

Detail: x is not cast - there is no cast in x < ux. x is converted to an unsigned.

– chux
Jan 2 at 22:09

add a comment |

int x = /* some integer */; 

unsigned int ux = (unsigned) x;

we have

x >= 0 || x < ux

we know that in x < ux the first x is cast implicitly to unsigned but is the first x in x >= 0 (1) cast to unsigned implicitly?

edited Jan 2 at 18:33

bolov

33.3k877141

asked Jan 2 at 18:29

durnovv

546

int x = /* some integer */; 

unsigned int ux = (unsigned) x;

we have

x >= 0 || x < ux

we know that in x < ux the first x is cast implicitly to unsigned but is the first x in x >= 0 (1) cast to unsigned implicitly?

c casting operators

edited Jan 2 at 18:33

bolov

33.3k877141

asked Jan 2 at 18:29

durnovv

546

edited Jan 2 at 18:33

bolov

33.3k877141

asked Jan 2 at 18:29

durnovv

546

edited Jan 2 at 18:33

bolov

33.3k877141

edited Jan 2 at 18:33

bolov

33.3k877141

edited Jan 2 at 18:33

bolov

33.3k877141

asked Jan 2 at 18:29

durnovv

546

asked Jan 2 at 18:29

durnovv

546

asked Jan 2 at 18:29

durnovv

546

Detail: x is not cast - there is no cast in x < ux. x is converted to an unsigned.

– chux
Jan 2 at 22:09

add a comment |

Detail: x is not cast - there is no cast in x < ux. x is converted to an unsigned.

– chux
Jan 2 at 22:09

Detail: x is not cast - there is no cast in x < ux. x is converted to an unsigned.

– chux
Jan 2 at 22:09

add a comment |

3 Answers
3

active

oldest

votes

No. It happens operator by operator.

x >= 0 || x < ux

is naturally

(x >= 0) || (x < ux)

Since x and 0 are both ints, there is no need for any (usual arithmetic) conversions...

And even though x is converted to unsigned in x < ux, the value of the expression x < ux is of type int - either 0 or 1 (just like on the the left-hand side).

edited Jan 2 at 19:59

answered Jan 2 at 18:32

Antti Haapala

85.5k16162204

add a comment |

No it isn’t.

This is because x >= 0 is an expression. (Formally 0 is an octal constant of type int.)

Try 1 / 2 * 1.0 for a more pernicious example. This is grouped as (1 / 2) * 1.0 and is zero since the integers in the expression 1 / 2 are not promoted to floating point.

edited Jan 2 at 18:35

answered Jan 2 at 18:33

Bathsheba

181k27256384

Thanks, makes sense

– durnovv
Jan 2 at 18:35

@durnovv It’s a good question though!

– Bathsheba
Jan 2 at 18:35

add a comment |

No, 0 is an int, so there are no promotions in the x >= 0 part of your expression.

answered Jan 2 at 18:32

Carl Norum

177k22347424

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

No. It happens operator by operator.

x >= 0 || x < ux

is naturally

(x >= 0) || (x < ux)

Since x and 0 are both ints, there is no need for any (usual arithmetic) conversions...

And even though x is converted to unsigned in x < ux, the value of the expression x < ux is of type int - either 0 or 1 (just like on the the left-hand side).

edited Jan 2 at 19:59

answered Jan 2 at 18:32

Antti Haapala

85.5k16162204

add a comment |

No. It happens operator by operator.

x >= 0 || x < ux

is naturally

(x >= 0) || (x < ux)

Since x and 0 are both ints, there is no need for any (usual arithmetic) conversions...

And even though x is converted to unsigned in x < ux, the value of the expression x < ux is of type int - either 0 or 1 (just like on the the left-hand side).

edited Jan 2 at 19:59

answered Jan 2 at 18:32

Antti Haapala

85.5k16162204

add a comment |

No. It happens operator by operator.

x >= 0 || x < ux

is naturally

(x >= 0) || (x < ux)

Since x and 0 are both ints, there is no need for any (usual arithmetic) conversions...

And even though x is converted to unsigned in x < ux, the value of the expression x < ux is of type int - either 0 or 1 (just like on the the left-hand side).

edited Jan 2 at 19:59

answered Jan 2 at 18:32

Antti Haapala

85.5k16162204

No. It happens operator by operator.

x >= 0 || x < ux

is naturally

(x >= 0) || (x < ux)

Since x and 0 are both ints, there is no need for any (usual arithmetic) conversions...

And even though x is converted to unsigned in x < ux, the value of the expression x < ux is of type int - either 0 or 1 (just like on the the left-hand side).

edited Jan 2 at 19:59

answered Jan 2 at 18:32

Antti Haapala

85.5k16162204

edited Jan 2 at 19:59

answered Jan 2 at 18:32

Antti Haapala

85.5k16162204

answered Jan 2 at 18:32

Antti Haapala

85.5k16162204

answered Jan 2 at 18:32

Antti Haapala

85.5k16162204

add a comment |

No it isn’t.

This is because x >= 0 is an expression. (Formally 0 is an octal constant of type int.)

Try 1 / 2 * 1.0 for a more pernicious example. This is grouped as (1 / 2) * 1.0 and is zero since the integers in the expression 1 / 2 are not promoted to floating point.

edited Jan 2 at 18:35

answered Jan 2 at 18:33

Bathsheba

181k27256384

Thanks, makes sense

– durnovv
Jan 2 at 18:35

@durnovv It’s a good question though!

– Bathsheba
Jan 2 at 18:35

add a comment |

No it isn’t.

This is because x >= 0 is an expression. (Formally 0 is an octal constant of type int.)

Try 1 / 2 * 1.0 for a more pernicious example. This is grouped as (1 / 2) * 1.0 and is zero since the integers in the expression 1 / 2 are not promoted to floating point.

edited Jan 2 at 18:35

answered Jan 2 at 18:33

Bathsheba

181k27256384

Thanks, makes sense

– durnovv
Jan 2 at 18:35

@durnovv It’s a good question though!

– Bathsheba
Jan 2 at 18:35

add a comment |

No it isn’t.

This is because x >= 0 is an expression. (Formally 0 is an octal constant of type int.)

Try 1 / 2 * 1.0 for a more pernicious example. This is grouped as (1 / 2) * 1.0 and is zero since the integers in the expression 1 / 2 are not promoted to floating point.

edited Jan 2 at 18:35

answered Jan 2 at 18:33

Bathsheba

181k27256384

No it isn’t.

This is because x >= 0 is an expression. (Formally 0 is an octal constant of type int.)

Try 1 / 2 * 1.0 for a more pernicious example. This is grouped as (1 / 2) * 1.0 and is zero since the integers in the expression 1 / 2 are not promoted to floating point.

edited Jan 2 at 18:35

answered Jan 2 at 18:33

Bathsheba

181k27256384

edited Jan 2 at 18:35

answered Jan 2 at 18:33

Bathsheba

181k27256384

answered Jan 2 at 18:33

Bathsheba

181k27256384

answered Jan 2 at 18:33

Bathsheba

181k27256384

Thanks, makes sense

– durnovv
Jan 2 at 18:35

@durnovv It’s a good question though!

– Bathsheba
Jan 2 at 18:35

add a comment |

Thanks, makes sense

– durnovv
Jan 2 at 18:35

@durnovv It’s a good question though!

– Bathsheba
Jan 2 at 18:35

Thanks, makes sense

– durnovv
Jan 2 at 18:35

@durnovv It’s a good question though!

– Bathsheba
Jan 2 at 18:35

add a comment |

No, 0 is an int, so there are no promotions in the x >= 0 part of your expression.

answered Jan 2 at 18:32

Carl Norum

177k22347424

add a comment |

No, 0 is an int, so there are no promotions in the x >= 0 part of your expression.

answered Jan 2 at 18:32

Carl Norum

177k22347424

add a comment |

No, 0 is an int, so there are no promotions in the x >= 0 part of your expression.

answered Jan 2 at 18:32

Carl Norum

177k22347424

No, 0 is an int, so there are no promotions in the x >= 0 part of your expression.

answered Jan 2 at 18:32

Carl Norum

177k22347424

answered Jan 2 at 18:32

Carl Norum

177k22347424

answered Jan 2 at 18:32

Carl Norum

177k22347424

answered Jan 2 at 18:32

Carl Norum

177k22347424

add a comment |

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