unsigned int
in [Visual C]
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From: John Keenan on 20 May 2008 10:13 Given the following example: int i20 = 20; unsigned int ui21 = 21; int i = ( i20 - ui21 ); // i is set to -1 double d = ( i20 - ui21 ); // d is set to 4294967295.0 Are the results shown correct? What can I search for to find the standard behavior of calculations involving unsigned int? Since i20 is an int I was expecting the expression ( i20 - ui21 ) to be an int... but that expectation was based on gut-feel as opposed to knowledge of the standard. John
From: David Connet on 20 May 2008 10:19 "John Keenan" <john.removeme.keenan(a)optimapowerware.com> wrote in news:Q%AYj.4619$7k7.866(a)flpi150.ffdc.sbc.com: > Given the following example: > > int i20 = 20; > unsigned int ui21 = 21; > int i = ( i20 - ui21 ); // i is set to -1 > double d = ( i20 - ui21 ); // d is set to 4294967295.0 > > Are the results shown correct? What can I search for to find the > standard behavior of calculations involving unsigned int? Since i20 is > an int I was expecting the expression ( i20 - ui21 ) to be an int... > but that expectation was based on gut-feel as opposed to knowledge of > the standard. That "is" -1. 4294967295 == 0xFFFFFFFF When you see funny numbers like that, windows 'calc' (in scientific mode) is great for switching between hex/dec/oct/bin. Dave Connet
From: Victor Bazarov on 20 May 2008 10:37 John Keenan wrote: > Given the following example: > > int i20 = 20; > unsigned int ui21 = 21; > int i = ( i20 - ui21 ); // i is set to -1 > double d = ( i20 - ui21 ); // d is set to 4294967295.0 > > Are the results shown correct? What can I search for to find the standard > behavior of calculations involving unsigned int? Since i20 is an int I was > expecting the expression ( i20 - ui21 ) to be an int... but that expectation > was based on gut-feel as opposed to knowledge of the standard. Since ui21 is 'unsigned', the Standard says that 'i20' shall be converted to 'unsigned', so the subexpression 'i20 - ui21' shall have the type 'unsigned'. The value of it is (2^32 - 1) because arithmetic operations on unsigned are performed modulo 2^b where 'b' is the number of bits in the representation. The next subexpression is the assignment from 'unsigned' to 'double'. Luckily you don't lose any bits here. See Standard, [expr]/9, to see what's converted to what when expressions are evaluated. V -- Please remove capital 'A's when replying by e-mail I do not respond to top-posted replies, please don't ask
From: Vincent Fatica on 20 May 2008 10:44 On Tue, 20 May 2008 14:19:19 GMT, David Connet <stuff(a)agilityrecordbook.com> wrote: >"John Keenan" <john.removeme.keenan(a)optimapowerware.com> wrote in >news:Q%AYj.4619$7k7.866(a)flpi150.ffdc.sbc.com: > >> Given the following example: >> >> int i20 = 20; >> unsigned int ui21 = 21; >> int i = ( i20 - ui21 ); // i is set to -1 >> double d = ( i20 - ui21 ); // d is set to 4294967295.0 >> >> Are the results shown correct? What can I search for to find the >> standard behavior of calculations involving unsigned int? Since i20 is >> an int I was expecting the expression ( i20 - ui21 ) to be an int... >> but that expectation was based on gut-feel as opposed to knowledge of >> the standard. > >That "is" -1. 4294967295 == 0xFFFFFFFF The double 4294967295.0 is not -1. -- - Vince
From: Pavel A. on 20 May 2008 15:51 "John Keenan" <john.removeme.keenan(a)optimapowerware.com> wrote in message news:Q%AYj.4619$7k7.866(a)flpi150.ffdc.sbc.com... > Given the following example: > > int i20 = 20; > unsigned int ui21 = 21; > int i = ( i20 - ui21 ); // i is set to -1 > double d = ( i20 - ui21 ); // d is set to 4294967295.0 > > Are the results shown correct? What can I search for to find the standard > behavior of calculations involving unsigned int? Since i20 is an int I was > expecting the expression ( i20 - ui21 ) to be an int... but that > expectation > was based on gut-feel as opposed to knowledge of the standard. > > John By C rules, if signed and unsigned types are mixed in one expression, the signed operands are cast to the unsigned type. So in your example you get unsigned (positive) ~0 rather than signed -1. Was this an interview question? :) --PA
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