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Re:Issues with porting C++ code to a 64-bit platform by Arelius on Monday April 01, 2013 @03:24PM (#5475280)
> 64 bits is 64 bits he said, right... well yes

That's the problem right there. 64 bits is not always 64 bits. While a 64 bit int gives you a 64 bit number (Actually a 63 bit number and a 1 bit sign flag). An IEEE double precision float only gives you a 52 bit number, and a separate 11 bit number as an exponent. So the problem is that declaring that 64 bits is 64 bits is actually precisely the issue here.

Re:Issues with porting C++ code to a 64-bit platform by MichaelSalib on Monday April 01, 2013 @03:36PM (#5475348)
Actually a 63 bit number and a 1 bit sign flag

Not true. Modern architectures use 2s-complement rather than an explicit sign bit to represent integers. If you use a sign bit, there are two different zeros (positive and negative) but the range of positive numbers is exactly as large as the range of negative numbers. With 2s-complement, there's only one zero and the range of negative numbers is larger (by one) than the range for positive numbers.

Re:Issues with porting C++ code to a 64-bit platform by dschatz on Monday April 01, 2013 @03:55PM (#5475467)
In 2s complement, the most significant bit is the sign bit. In fact all valid C integer representations have a sign bit.
Re:Issues with porting C++ code to a 64-bit platform by cube13 on Monday April 01, 2013 @03:57PM (#5475485)
Signed integer representations. Unsigned ints do not, and you're given the full range of the integer as valid data.
Re:Issues with porting C++ code to a 64-bit platform by dschatz on Monday April 01, 2013 @04:09PM (#5475566)
Yes you're right. I should have been more specific that all signed integers have a sign bit.
Re:Issues with porting C++ code to a 64-bit platform by MichaelSalib on Monday April 01, 2013 @04:36PM (#5475749)
This is wrong.

Take an integer value x. Flip the sign bit. Do you now have the value -x? On a 2s complement architecture, you do not. In 2s complement, you have to flip all the bits and add one to get -x.

Re:Issues with porting C++ code to a 64-bit platform by dschatz on Thursday April 04, 2013 @10:02AM (#5492679)
Sign bit means one bit represents the sign of the integer which is the case with the most significant bit in 2s complement.

If you flip the sign bit, the sign changes. It does not mean that if you flip the sign, the value is negated.

Re:Issues with porting C++ code to a 64-bit platform by to3m on Monday April 01, 2013 @04:20PM (#5475630)
I think his point is that changing a sign bit doesn't affect the absolute value, just its sign. A sign bit has no value itself - it's just a flag. If the representation has a sign bit, you'd have a negative zero.

But the top bit in 2s complement has a value - it's just a negative one (that's large enough to make any value with it set negative). That's not a sign bit! If you change it, the absolute value most definitely changes, and quite substantially.

Re:Issues with porting C++ code to a 64-bit platform by xentronium on Monday April 01, 2013 @04:06PM (#5475545)
This is a technicality. Leftmost bit still denotes a sign and modulo of the number is still capped at n-1 bits.
Re:Issues with porting C++ code to a 64-bit platform by MichaelSalib on Monday April 01, 2013 @04:33PM (#5475728)
The differences between the representations matter. In 2s complement, -INT_MIN is not INT_MAX; in fact, -INT_MIN is undefined and a C compiler is justified in deleting all your code if you ever cause it to calculate -INT_MIN.

I tell you what: if you think modern architectures use an explicit sign bit to represent integers, how do you explain the fact that INT_MAX+1==-INT_MIN? If there's a sign bit, how could you possibly represent more negative integers than positive integers?

Re:Issues with porting C++ code to a 64-bit platform by Arelius on Monday April 01, 2013 @11:58PM (#5477617)
Are you arguing that given such, I have more than 63 bits for the number, or do you just have a problem with the usage of the words "sign flag"?