Consider:
9223372034707292159. 1 um/mod
For example, it raises a division-by-zero error on my machine.
Why? The divisor is 1 (not 0!) and 9223372034707292159 fits within an unsigned double.
The same goes for fm/mod and sm/rem.
I'm using Gforth, but these are core Forth words.
See the specification for UM/MOD ( ud u1 -- u2 u3 ):
An ambiguous condition exists if u1 is zero or if the quotient lies
outside the range of a single-cell unsigned integer.
If you use a Forth system having a cell size of 32 bits, then in the given case the quotient lies outside the rage of a single-cell unsigned integer, and then it's an ambiguous condition and the behavior of a system is not specified by the standard.
Certainly, a better error message would be "integer overflow" than "division by zero" in this case.
[update] As it later appeared, in Gforth "the -fast engine does make some compromises speed vs. accuracy of error messages".
Related
I am working with the mips assembly language but am confused on the overflow aspect of arithmetic here.
Say I am subtracting 25 from 20 and end up with -5. Would this result in an overflow?
I understand that with addition if you add 2 positive numbers or 2 negative numbers and the output is the opposite sign then there is overflow but am lost when it comes to subtraction.
Find the examples at the extreme, let's do it in 8 bits signed to make it simple but the same principles holds in 32 bits
minuend: the smallest possible positive (non-negative) number, 0 and
subtrahend: the largest possible number, 127
When we do 0 - 127, the answer is -127 and that indeed fits in 8 bits signed. There is a borrow. In any processor the effect of the borrow is to propagate 1's though out the upper bits, making it a negative number of the proper magnitude.
Different processors set flags differently based on this borrow, MIPS doesn't have flags, while x86 will set flags to indicate borrow, and other processors will set the flags to indicate carry.
In 8 bit signed numbers:
minuend: the smallest possible positive (non-negative) number, 0 and
subtrahend: the largest possible number, -128
When we do 0 - -128, the answer should be 128, but that cannot be represented in 8 bit signed format, so this is the example of overflow. 0 - -127 = 127 and that can be represented, so no overflow there.
If we do it in 8 bits unsigned, your example of 25-20 = -5, but -5 cannot be represented in an unsigned format, so this is indeed overflow (or modular arithmetic, if you like that).
Short answer: Yes, as the 32-bit representation of -5 is FFFFFFFB.
Long answer: Depends on what you mean by "overflow".
There's signed overflow, which is when you cross the 7FFFFFFF-80000000 boundary.
And there's unsigned overflow where you cross the FFFFFFFF-00000000 boundary.
For signed arithmetic, signed overflow is undeniably a bad thing (and is considered undefined behavior in C and other languages). However, unsigned overflow is not necessarily a problem. Usually it is, but many procedures rely on it to work.
For example, imagine you have a "frame timer" variable, i.e. a 32-bit counter variable that increments by 1 during an interrupt service routine. This interrupt is tied to a real-time clock running at 60 hertz, so every 1/60th of a second the variable's value increases by 1.
Now, this variable will overflow eventually. But do we really care? No. It just wraps around back to zero again. For our purposes, it's fine, since we really don't need to know that accurately how long our program has been running since it began. We probably have events that occur every n ticks of the timer, but we can just use a bitmask for that. Effectively in this case we're using unsigned overflow to say "if this value equals FFFFFFFF and we're about to add 1 to it, reset it to zero instead." Which thanks to overflow we can easily implement without any additional condition checking.
The reason I bring this up is so that you understand that overflow is not always a bad thing, if it's the unsigned variety. It depends entirely on what your data is intended to represent (which is something you can't explain even to a C compiler.)
Let's say I have a fixed point value in my VHDL - Code which is defined as std_logic_vector. I know that my last 4-bit are the decimals.
When I use the simulator it will of course not see the last 4 bits as decimals, is there any possibility to change it in the simulation, so that the simulation knows that the 3rd bit has the value of 0.5 , the 2nd the value of 0.25 and so on ?
It's possible in Vivado to show the result in the simulator as a fixed point representation.
When you rightclick in the simulator on the signal you want to show in fixed point, click radix --> real settings. There you get the following window and you can select fixed point.
Real settings window
I doubt about that this is possible unfortunately. Specifically for std_logic the types are the following:
'U': Uninitialized. This signal hasn't been set yet.
'X': Unknown Impossible to determine this value/result.
'0': Logic 0
'1': Logic 1
'Z': High Impedance
'W': Weak signal, can't tell if it should be 0 or 1.
'L': Weak signal that should probably go to 0
'H': Weak signal that should probably go to 1
'-': Don't care.
As a result the simulator is recognizing ONLY the above symbols, anything else will result in error, including a floating point number to describe a bit. Here's an example from me trying to put a floating point value to be displayed:
add_force {/test/a[27]} -radix unsigned {0.4 0ns}
ERROR: [Simtcl 6-179] Couldn't add force for the following reason:
Illegal value '0.4': Could not convert value '0.4' to a decimal
number.
I also observed Vivado as tagged so I guess you use the integrated simulator. In my example the closest to a floating point number was the decimal number. There is no built in variable at Vivado to support floating points for display in the simulation. Below you see the radix it supports so you are tightly restricted only to those choices, which all result in error except ASCII but I don't think that's the behavior you want.
In many algorithm i have seen people are using two different way to get middle point.
(LOW+HIGH)/2
LOW+(HIGH-LOW)/2
Mostly i have seen 2nd method for example in QuickSort. What is the best way to find middle between two number and why ?
It entirely depends on the context, but I'll make a case for case number 2 and explain why.
Let's first assume you go for case number 1, which is this:
(LOW + HIGH) / 2
This looks entirely reasonable, and mathematically, it is. Let's plug in two numbers and look at the results:
(12345 + 56789) / 2
The result is 34567. Looks OK doesn't it?
Now, the problem is that in computers it's not as simple as that. You also got something called data types to contend with. Usually this is denoted with such things as number of bits. In other words, you might have a 32-bit number, or a 16-bit number, or a 64-bit number, and so on.
All of these have what is known as a legal range of values, ie. "what kind of values will these types hold". An 8-bit number, unsigned (which means it cannot be negative) can hold 2 to the power of 8 different values, or 256. A 16-bit unsigned value can hold 65536 values, or a range from 0 to 65535. If the values are signed they go from -half to +half-1, meaning it will go from -128 to +127 for a 8-bit signed value, and from -32768 to +32767 for a signed 16-bit value.
So now we go back to the original formula. What if the data type we're using for the calculation isn't enough to hold LOW + HIGH ?
For instance, let's say we used 16-bit signed values for this, and we still got this expression:
(12345 + 56789) / 2
12345 can be held in a 16-bit value (it is less than 65536), same with 56789, but what about the result? The result of adding 12345 and 56789 is 69134 which is more than 65535 (the highest unsigned 16-bit value).
So what will happen to it? There's two outcomes:
It will overflow, which means it will start back at 0 and count upwards, which means it will actually end up with the result of 3598 (123456 + 56789) - 65536
It will throw an exception or similar, crashing the program with an overflow problem.
If we get the first result, then (12345 + 56789)/2 becomes 3598/2 or 1799. Clearly incorrect.
So, then what if we used the other approach:
12345 + (56789-12345)/2
First, let's do the parenthesis: 56789-12345 is equal to 44444, a number that can be held in a 16-bit data type.
Adding 12345 + 44444 gives us 56789, a number that can also be held in a 16-bit data type.
Dividing 56789 by 2 gives us 28934.5. Since we're probably dealing with "integers" here we get 28934 (typically, unless your particular world rounds up).
So the reason the second expression is chosen above the first is that it doesn't have to handle overflow the same way and is more resilient to this kind of problem.
In fact, if you think about it, the maximum second value you can have is the maximum legal value you can have for your data type, so this kind of expression:
X + (Y-X)
... assuming both X and Y are the same data type, can at most be the maximum of that data type. Basically, it will not have to contend with an overflow at all.
2nd method used, for avoid int overflow during computation.
Imagine, you used just 1-byte unsigned integer, so overflow happening if value reached 256. Imagine, we have low=100 and high=200. See computing:
1. (lo + hi) / 2 = (100 + 200) / 2 = 300 / 2; // 300 > 256, int overflow
2. lo + (hi - lo) / 2 = 100 + (200 - 100) / 2 = 150; // no overflow
There isn't a best one, but they are clearly different.
(LOW+HIGH)/2 causes trouble if the addition overflows, both in the signed and unsigned case. Doesn't have to be an issue, if you can assume that it never happens it's a perfectly reasonable thing to do.
LOW+(HIGH-LOW)/2 can still cause trouble with overflows if LOW is allowed to be arbitrarily negative. For non-negative input it's fine though. Costs an extra operation.
(LOW+HIGH) >>> 1 as seen in Java. For non-negative but signed input, overflow is non-destructive. The sign bit is used as an extra bit of space that the addition can carry into, the shift then divides the unsigned result by two. Does not work at all if the result is supposed to negative, because by construction the result is non-negative. For array indexes that isn't really a consideration. Doesn't help if you were working with unsigned indexes though.
MOV MID, LOW \ ADD MID, HIGH \ RCR MID in pseudo x86 asm, explicitly uses an extra bit of space and therefore works for all unsigned input no matter what, but cannot be used in most languages.
They all have their upsides and downsides. There is no winner.
The best way is dependent on what you're trying to accomplish. The first is clearly faster (best for performance), while the second one is used to avoid overflow (best for correctness). So the answer to your question is dependent on your definition of "best".
I'm using VBScript in ASP on IIS and I cannot seem to get the annoying error to go anyway so this makes debugging so much harder not to mention it does not tell me the exact error except the same error message so I can only assume what's wrong in my code.
My question is: Using the logical operators AND,XOR,NOT,OR in VBScript , is there a limit on the range to what the operands can be? I have implemented a bit shift right function and used the mod operator and I didn't notice until now that my function was causing the error.
My right shift function
function rshift(number,n)
'Shifts a number's bits n bits to the right
for i=1 to n
if number mod 2 = 0 then
number = number / 2
else
number = (number - 1) / 2
end if
next
rshift = number
end function
'Fails
rshift(1125899906842624,2)
I think for values larger than 2^32 ( or 31) - 1 that the operators do not work. Tried googling for the range of the operands but couldn't find anything helpful. I saw someone posted a topic about logical operators not working on large values but I can't seem to find that anymore.
Can someone verify this ?
Edit: Found a topic which gives more information on using the mod operator on signed 32 bit integers http://blogs.msdn.com/b/ericlippert/archive/2004/12/01/integer-arithmetic-in-vbscript-part-one.aspx
In VBScript there are several subtypes for integers including integer and long. VBScript will attempt to determine what type of value you are using and use the appropriate subtype.
Integer can store a value between -32,768 to 32,767 and long can store a value between -2,147,483,648 to 2,147,483,647. That value that you are using is greater than this and will result in an overflow.
You can use the VarType function to see what type your number is interpreted as. You may use Double to represent larger values.
This answer looks interesting. Maybe you can use it as part of your function.
I'm programming a PLC with some legacy software (RSLogix 500, don't ask) and it does not natively support a modulus operation, but I need one. I do not have access to: modulus, integer division, local variables, a truncate operation (though I can hack it with rounding). Furthermore, all variables available to me are laid out in tables sorted by data type. Finally, it should work for floating point decimals, for example 12345.678 MOD 10000 = 2345.678.
If we make our equation:
dividend / divisor = integer quotient, remainder
There are two obvious implementations.
Implementation 1:
Perform floating point division: dividend / divisor = decimal quotient. Then hack together a truncation operation so you find the integer quotient. Multiply it by the divisor and find the difference between the dividend and that, which results in the remainder.
I don't like this because it involves a bunch of variables of different types. I can't 'pass' variables to a subroutine, so I just have to allocate some of the global variables located in multiple different variable tables, and it's difficult to follow. Unfortunately, 'difficult to follow' counts, because it needs to be simple enough for a maintenance worker to mess with.
Implementation 2:
Create a loop such that while dividend > divisor divisor = dividend - divisor. This is very clean, but it violates one of the big rules of PLC programming, which is to never use loops, since if someone inadvertently modifies an index counter you could get stuck in an infinite loop and machinery would go crazy or irrecoverably fault. Plus loops are hard for maintenance to troubleshoot. Plus, I don't even have looping instructions, I have to use labels and jumps. Eww.
So I'm wondering if anyone has any clever math hacks or smarter implementations of modulus than either of these. I have access to + - * /, exponents, sqrt, trig functions, log, abs value, and AND/OR/NOT/XOR.
How many bits are you dealing with? You could do something like:
if dividend > 32 * divisor dividend -= 32 * divisor
if dividend > 16 * divisor dividend -= 16 * divisor
if dividend > 8 * divisor dividend -= 8 * divisor
if dividend > 4 * divisor dividend -= 4 * divisor
if dividend > 2 * divisor dividend -= 2 * divisor
if dividend > 1 * divisor dividend -= 1 * divisor
quotient = dividend
Just unroll as many times as there are bits in dividend. Make sure to be careful about those multiplies overflowing. This is just like your #2 except it takes log(n) instead of n iterations, so it is feasible to unroll completely.
If you don't mind overly complicating things and wasting computer time you can calculate modulus with periodic trig functions:
atan(tan(( 12345.678 -5000)*pi/10000))*10000/pi+5000 = 2345.678
Seriously though, subtracting 10000 once or twice (your "implementation 2") is better. The usual algorithms for general floating point modulus require a number of bit-level manipulations that are probably unfeasible for you. See for example http://www.netlib.org/fdlibm/e_fmod.c (The algorithm is simple but the code is complex because of special cases and because it is written for IEEE 754 double precision numbers assuming there is no 64-bit integer type)
This all seems completely overcomplicated. You have an encoder index that rolls over at 10000 and objects rolling along the line whose positions you are tracking at any given point. If you need to forward project stop points or action points along the line, just add however many inches you need and immediately subtract 10000 if your target result is greater than 10000.
Alternatively, or in addition, you always get a new encoder value every PLC scan. In the case where the difference between the current value and last value is negative you can energize a working contact to flag the wrap event and make appropriate corrections for any calculations on that scan. (**or increment a secondary counter as below)
Without knowing more about the actual problem it is hard to suggest a more specific solution but there are certainly better solutions. I don't see a need for MOD here at all. Furthermore, the guys on the floor will thank you for not filling up the machine with obfuscated wizard stuff.
I quote :
Finally, it has to work for floating point decimals, for example
12345.678 MOD 10000 = 2345.678
There is a brilliant function that exists to do this - it's a subtraction. Why does it need to be more complicated than that? If your conveyor line is actually longer than 833 feet then roll a second counter that increments on a primary index roll-over until you've got enough distance to cover the ground you need.
For example, if you need 100000 inches of conveyor memory you can have a secondary counter that rolls over at 10. Primary encoder rollovers can be easily detected as above and you increment the secondary counter each time. Your working encoder position, then, is 10000 times the counter value plus the current encoder value. Work in the extended units only and make the secondary counter roll over at whatever value you require to not lose any parts. The problem, again, then reduces to a simple subtraction (as above).
I use this technique with a planetary geared rotational part holder, for example. I have an encoder that rolls over once per primary rotation while the planetary geared satellite parts (which themselves rotate around a stator gear) require 43 primary rotations to return to an identical starting orientation. With a simple counter that increments (or decrements, depending on direction) at the primary encoder rollover point it gives you a fully absolute measure of where the parts are at. In this case, the secondary counter rolls over at 43.
This would work identically for a linear conveyor with the only difference being that a linear conveyor can go on for an infinite distance. The problem then only needs to be limited by the longest linear path taken by the worst-case part on the line.
With the caveat that I've never used RSLogix, here is the general idea (I've used generic symbols here and my syntax is probably a bit wrong but you should get the idea)
With the above, you end up with a value ENC_EXT which has essentially transformed your encoder from a 10k inch one to a 100k inch one. I don't know if your conveyor can run in reverse, if it can you would need to handle the down count also. If the entire rest of your program only works with the ENC_EXT value then you don't even have to worry about the fact that your encoder only goes to 10k. It now goes to 100k (or whatever you want) and the wraparound can be handled with a subtraction instead of a modulus.
Afterword :
PLCs are first and foremost state machines. The best solutions for PLC programs are usually those that are in harmony with this idea. If your hardware is not sufficient to fully represent the state of the machine then the PLC program should do its best to fill in the gaps for that missing state information with the information it has. The above solution does this - it takes the insufficient 10000 inches of state information and extends it to suit the requirements of the process.
The benefit of this approach is that you now have preserved absolute state information, not just for the conveyor, but also for any parts on the line. You can track them forward and backward for troubleshooting and debugging and you have a much simpler and clearer coordinate system to work with for future extensions. With a modulus calculation you are throwing away state information and trying to solve individual problems in a functional way - this is often not the best way to work with PLCs. You kind of have to forget what you know from other programming languages and work in a different way. PLCs are a different beast and they work best when treated as such.
You can use a subroutine to do exactly what you are talking about. You can tuck the tricky code away so the maintenance techs will never encounter it. It's almost certainly the easiest for you and your maintenance crew to understand.
It's been a while since I used RSLogix500, so I might get a couple of terms wrong, but you'll get the point.
Define a Data File each for your floating points and integers, and give them symbols something along the lines of MOD_F and MOD_N. If you make these intimidating enough, maintenance techs leave them alone, and all you need them for is passing parameters and workspace during your math.
If you really worried about them messing up the data tables, there are ways to protect them, but I have forgotten what they are on a SLC/500.
Next, defined a subroutine, far away numerically from the ones in use now, if possible. Name it something like MODULUS. Again, maintenance guys almost always stay out of SBRs if they sound like programming names.
In the rungs immediately before your JSR instruction, load the variables you want to process into the MOD_N and MOD_F Data Files. Comment these rungs with instructions that they load data for MODULUS SBR. Make the comments clear to anyone with a programming background.
Call your JSR conditionally, only when you need to. Maintenance techs do not bother troubleshooting non-executing logic, so if your JSR is not active, they will rarely look at it.
Now you have your own little walled garden where you can write your loop without maintenance getting involved with it. Only use those Data Files, and don't assume the state of anything but those files is what you expect. In other words, you cannot trust indirect addressing. Indexed addressing is OK, as long as you define the index within your MODULUS JSR. Do not trust any incoming index. It's pretty easy to write a FOR loop with one word from your MOD_N file, a jump and a label. Your whole Implementation #2 should be less than ten rungs or so. I would consider using an expression instruction or something...the one that lets you just type in an expression. Might need a 504 or 505 for that instruction. Works well for combined float/integer math. Check the results though to make sure the rounding doesn't kill you.
After you are done, validate your code, perfectly if possible. If this code ever causes a math overflow and faults the processor, you will never hear the end of it. Run it on a simulator if you have one, with weird values (in case they somehow mess up the loading of the function inputs), and make sure the PLC does not fault.
If you do all that, no one will ever even realize you used regular programming techniques in the PLC, and you will be fine. AS LONG AS IT WORKS.
This is a loop based on the answer by #Keith Randall, but it also maintains the result of the division by substraction. I kept the printf's for clarity.
#include <stdio.h>
#include <limits.h>
#define NBIT (CHAR_BIT * sizeof (unsigned int))
unsigned modulo(unsigned dividend, unsigned divisor)
{
unsigned quotient, bit;
printf("%u / %u:", dividend, divisor);
for (bit = NBIT, quotient=0; bit-- && dividend >= divisor; ) {
if (dividend < (1ul << bit) * divisor) continue;
dividend -= (1ul << bit) * divisor;
quotient += (1ul << bit);
}
printf("%u, %u\n", quotient, dividend);
return dividend; // the remainder *is* the modulo
}
int main(void)
{
modulo( 13,5);
modulo( 33,11);
return 0;
}