Bit fields

Perspektyva · 11-22-2008, 08:15 AM

the findfirst and findnext functions populate ffblk structure with file info. One of the ffblk struct element's is ff_ftime which is an unsigned short and, according to the help file (Borland builder 5.0), it contains bit fields for reffering the seconds, minutes and hours:

ff_ftime and ff_fdate contain bit fields for referring to the current date and time. The structure of these fields was established by the operating system. Both are 16-bit structures divided into three fields.

ff_ftime:

Bits 0 to 4 The result of seconds divided by 2 (for example 10 here means 20 seconds)
Bits 5 to 10 Minutes
Bits 11 to 15 Hours

Well, yeah there's ff_fdate too but they're pretty much the same. So... I've checked MSDN what are bit fields - seemed pretty clear, but... as I understood they are used to save some space when using integrals in structures and ff_fdate is not a structure (I think) and I'm lost now.

I, basicaly, need to extract a viable time from ff_ftime, but for, what is 21:35:47, I get 44152 and for, what is 15:19:36, I get 31347. Halp.

C_ntua · 11-22-2008, 08:50 AM

I am reading the page you find the information. I believe it doesn't mean bit fields in that sense. Just field of bits, meaning that you can use a mask to take the information you want.
Like if you want seconds:

Code:

ff_ftime t;
unsigned short SEC_MASK = 15; // 15 = 0..01111
unsigned short sec = t & SEC_MASK;
sec *= 2;

try it

Perspektyva · 11-22-2008, 10:38 AM

It works! But how to get the minutes and hours out? or anything that starts not at the beggining?

abachler · 11-22-2008, 10:59 AM

minutes would be (timecode & 2016) / 32;

Perspektyva · 11-22-2008, 11:24 AM

didn't you mean & 2047?

**CornedBee** · 11-22-2008, 11:29 AM

And that's why you should always specify bitmasks in hex.

Perspektyva · 11-22-2008, 11:32 AM

Dude... I'm an amateur, I've started c++ programming 2.5 months ago, before that I only knew only how to create arrays, procedures, functions and write and read from files in pascal. I do not know what bitmasks are, I know what hex is, though. And isn't the time encoded in binary, in this case?

**CornedBee** · 11-22-2008, 11:47 AM

Numbers are always "encoded" in binary, as the computer sees it. That's irrelevant when it comes to the programmer reasoning about them, though. Radix is just a display convention; numbers are pure.

The advantage of hex is that there's a very simple mapping to binary: four binary digits form a single hex digit, always. Thus, it's more compact than binary, and easier to translate than decimal.

Speaking of binary, here's how bitfields and bitmasks work. Suppose you have a 16-bit integer:
0000000000000000
Now, what's stopping you from reinterpreting that? Say, as two 5-bit and one 6-bit integers?
00000 000000 00000
When you do that, you've turned the integer into a bitfield. A bitmask then is an integer that you can use in bitwise operations (bitwise-AND, &, usually) to assist you in using the bitfield. The bitmask for the first block above is
1111100000000000
For the second block
0000011111100000
And so on. It's pretty simple.

Perspektyva · 11-22-2008, 12:20 PM

I see, but wouldn't specifying bitmasks in hex would restrict the masks to be 4*n bits in size, in which case, with time firmats, where numbers don't go higher than 60, would cause empty bits of memory that stay 0 all the time?

Anyways, I can't get abachlers example to work, do i have to put it like this:

Code:

unsigned short sec = (t & 2016) / 32;

Or?...

I mean, I understand how the 1 and 0 are placed and I could brew a method that would extract decimals from specific intervals of a binary number, but... what's the shorter way?

**CornedBee** · 11-22-2008, 12:26 PM

Quote:

Originally Posted by Perspektyva

I see, but wouldn't specifying bitmasks in hex would restrict the masks to be 4*n bits in size, in which case, with time firmats, where numbers don't go higher than 60, would cause empty bits of memory that stay 0 all the time?

You can only divide bitfields at bit boundaries, obviously, so to represent a second or minute value (0-59) you'll have to use 6 bits (0-63), wasting 4 values.
Now, the entire mask is restricted to be of an actually supported integer type, as every other integer. I don't know what you mean by the other part. There's no number you can represent in binary that you can't represent in hex.

Perspektyva · 11-22-2008, 12:37 PM

Quote:

Originally Posted by CornedBee

You can only divide bitfields at bit boundaries, obviously, so to represent a second or minute value (0-59) you'll have to use 6 bits (0-63), wasting 4 values.
Now, the entire mask is restricted to be of an actually supported integer type, as every other integer. I don't know what you mean by the other part. There's no number you can represent in binary that you can't represent in hex.

We're either not on the same page (with me being on a page that doesn't actually exist) or my knowledge backroud is simply too thin for me to understand this stuff. So, maybe, you know how can I extract a decimal value of minutes which resides in the 6th to 11th bit of a 16bit integer... using masks, or some other quicker way.
A code snippet which would fit nicely in C_ntua's one would be nice.

**CornedBee** · 11-22-2008, 12:43 PM

Assuming 0-based counting of bits. You either mask first and then shift, or you shift first and then mask. Shifting first is easier, because the mask always starts at the 0th bit.

Code:

unsigned int packed_time = get_a_packed_time();
unsigned minutes = (packed_time
    >> 6)   // Shift so that 6th to 11th bit are now 0th to 5th
    & 0x1F; // Mask out everything beyond the 5th bit. 0x1F is 0000000000011111

Perspektyva · 11-22-2008, 01:03 PM

Quote:

Originally Posted by CornedBee

Assuming 0-based counting of bits. You either mask first and then shift, or you shift first and then mask. Shifting first is easier, because the mask always starts at the 0th bit.

Code:

unsigned int packed_time = get_a_packed_time();
unsigned minutes = (packed_time
    >> 6)   // Shift so that 6th to 11th bit are now 0th to 5th
    & 0x1F; // Mask out everything beyond the 5th bit. 0x1F is 0000000000011111

Hehe, it works, I was thinking how to shift binary numbers left and right and I was thinking that you'd have to assign a mask as an interval or smth... anyways It works now and I understand it. Thanks!

abachler · 11-22-2008, 02:38 PM

Quote:

Originally Posted by Perspektyva

didn't you mean & 2047?

no, I didn't, although that woudl work also, I specifically used the number 0000001111100000, although im sure 0000001111111111 woudl also work sicne you are just shifting after that anyway.

11-22-2008, 08:15 AM	#1
Perspektyva Registered User Join Date: Oct 2008 Posts: 10	Bit fields the findfirst and findnext functions populate ffblk structure with file info. One of the ffblk struct element's is ff_ftime which is an unsigned short and, according to the help file (Borland builder 5.0), it contains bit fields for reffering the seconds, minutes and hours: ff_ftime and ff_fdate contain bit fields for referring to the current date and time. The structure of these fields was established by the operating system. Both are 16-bit structures divided into three fields. ff_ftime: Bits 0 to 4 The result of seconds divided by 2 (for example 10 here means 20 seconds) Bits 5 to 10 Minutes Bits 11 to 15 Hours Well, yeah there's ff_fdate too but they're pretty much the same. So... I've checked MSDN what are bit fields - seemed pretty clear, but... as I understood they are used to save some space when using integrals in structures and ff_fdate is not a structure (I think) and I'm lost now. I, basicaly, need to extract a viable time from ff_ftime, but for, what is 21:35:47, I get 44152 and for, what is 15:19:36, I get 31347. Halp.
Perspektyva is offline

11-22-2008, 08:50 AM	#2
C_ntua Registered User Join Date: Jun 2008 Posts: 1,135	I am reading the page you find the information. I believe it doesn't mean bit fields in that sense. Just field of bits, meaning that you can use a mask to take the information you want. Like if you want seconds: Code: ff_ftime t; unsigned short SEC_MASK = 15; // 15 = 0..01111 unsigned short sec = t & SEC_MASK; sec *= 2; try it
C_ntua is offline

11-22-2008, 10:38 AM	#3
Perspektyva Registered User Join Date: Oct 2008 Posts: 10	It works! But how to get the minutes and hours out? or anything that starts not at the beggining?
Perspektyva is offline

11-22-2008, 10:59 AM	#4
abachler Rampaging 35 Stone Welsh Join Date: Apr 2007 Posts: 2,929	minutes would be (timecode & 2016) / 32; __________________ He is free, you say. Ah! That is his misfortune… These men… [have] the most terrible, the most imperious of masters, that is, need. … They must therefore find someone to hire them, or die of hunger. Is that to be free? - Simon Linguet
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11-22-2008, 11:24 AM	#5
Perspektyva Registered User Join Date: Oct 2008 Posts: 10	didn't you mean & 2047?
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11-22-2008, 11:29 AM	#6
CornedBee Cat without Hat Join Date: Apr 2003 Posts: 8,439	And that's why you should always specify bitmasks in hex. __________________ All the buzzt! CornedBee "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code." - Flon's Law
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11-22-2008, 11:32 AM	#7
Perspektyva Registered User Join Date: Oct 2008 Posts: 10	Dude... I'm an amateur, I've started c++ programming 2.5 months ago, before that I only knew only how to create arrays, procedures, functions and write and read from files in pascal. I do not know what bitmasks are, I know what hex is, though. And isn't the time encoded in binary, in this case?
Perspektyva is offline

11-22-2008, 11:47 AM	#8
CornedBee Cat without Hat Join Date: Apr 2003 Posts: 8,439	Numbers are always "encoded" in binary, as the computer sees it. That's irrelevant when it comes to the programmer reasoning about them, though. Radix is just a display convention; numbers are pure. The advantage of hex is that there's a very simple mapping to binary: four binary digits form a single hex digit, always. Thus, it's more compact than binary, and easier to translate than decimal. Speaking of binary, here's how bitfields and bitmasks work. Suppose you have a 16-bit integer: 0000000000000000 Now, what's stopping you from reinterpreting that? Say, as two 5-bit and one 6-bit integers? 00000 000000 00000 When you do that, you've turned the integer into a bitfield. A bitmask then is an integer that you can use in bitwise operations (bitwise-AND, &, usually) to assist you in using the bitfield. The bitmask for the first block above is 1111100000000000 For the second block 0000011111100000 And so on. It's pretty simple. __________________ All the buzzt! CornedBee "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code." - Flon's Law
CornedBee is offline

11-22-2008, 12:20 PM	#9
Perspektyva Registered User Join Date: Oct 2008 Posts: 10	I see, but wouldn't specifying bitmasks in hex would restrict the masks to be 4*n bits in size, in which case, with time firmats, where numbers don't go higher than 60, would cause empty bits of memory that stay 0 all the time? Anyways, I can't get abachlers example to work, do i have to put it like this: Code: unsigned short sec = (t & 2016) / 32; Or?... I mean, I understand how the 1 and 0 are placed and I could brew a method that would extract decimals from specific intervals of a binary number, but... what's the shorter way?
Perspektyva is offline

11-22-2008, 12:43 PM	#12
CornedBee Cat without Hat Join Date: Apr 2003 Posts: 8,439	Assuming 0-based counting of bits. You either mask first and then shift, or you shift first and then mask. Shifting first is easier, because the mask always starts at the 0th bit. Code: unsigned int packed_time = get_a_packed_time(); unsigned minutes = (packed_time >> 6) // Shift so that 6th to 11th bit are now 0th to 5th & 0x1F; // Mask out everything beyond the 5th bit. 0x1F is 0000000000011111 __________________ All the buzzt! CornedBee "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code." - Flon's Law
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