1'hF or 1'h1?
I am guessing 1'hF is not correct, 4'hF is required because 1 bit is not enough to hold a hexadecimal?
in the 1'h1,1'h or 1'b or 1'd All said number of bits,as well as "4'h"show four binary system,so hexadecimal "f" convert binary is "1111" ,so we need use "4'hf" show this
The LRM section 5.7.1 Integer literal constants say this about any based literal constant
If the size of the unsigned number is larger than the size specified
for the literal constant, the unsigned number shall be truncated from
the left
Most tools also generate a warning if it has to truncate.
Related
I have a text file called "test.txt" containing multiple lines with the fields separated by a semicolon. I'm trying to take the value of field3 > strip out everything but the numbers in the field > compare it to the value of field 3 in the previous line > if the value is unique, redirect the field 3 value and the difference between it and the last value to a file called "differences.txt".
so far, i have the following code:
awk -F';' '
BEGIN{d=0} {gsub(/^.*=/,"",$3);
if(d>0 && $3-d>0){print $3,$3-d} d=$3}
' test.txt > differences.txt
This works absolutely fine when i try to run in the following text:
field1=xxx;field2=xxx;field3=111222222;field4=xxx;field5=xxx
field1=xxx;field2=xxx;field3=111222222;field4=xxx;field5=xxx
field1=xxx;field2=xxx;field3=111222333;field4=xxx;field5=xxx
field1=xxx;field2=xxx;field3=111222444;field4=xxx;field5=xxx
field1=xxx;field2=xxx;field3=111222555;field4=xxx;field5=xxx
field1=xxx;field2=xxx;field3=111222555;field4=xxx;field5=xxx
field1=xxx;field2=xxx;field3=111222777;field4=xxx;field5=xxx
field1=xxx;field2=xxx;field3=111222888;field4=xxx;field5=xxx
output, as expected:
111222333 111
111222444 111
111222555 111
111222777 222
111222888 111
however, when i try and run the following text in, i get completely different, unexpected numbers - I'm not sure if it's due to the increased length of the field or something??
test:
test=none;test=20170606;test=1111111111111111111;
test=none;test=20170606;test=2222222222222222222;
test=none;test=20170606;test=3333333333333333333;
test=none;test=20170606;test=4444444444444444444;
test=none;test=20170606;test=5555555555555555555;
test=none;test=20170606;test=5555555555555555555;
test=none;test=20170606;test=6666666666666666666;
test=none;test=20170606;test=7777777777777777777;
test=none;test=20170606;test=8888888888888888888;
test=none;test=20170606;test=9999999999999999999;
test=none;test=20170606;test=100000000000000000000;
test=none;test=20170606;test=11111111111111111111;
Output, with unexpected values:
2222222222222222222 1111111111111111168
3333333333333333333 1111111111111111168
4444444444444444444 1111111111111111168
5555555555555555555 1111111111111110656
6666666666666666666 1111111111111111680
7777777777777777777 1111111111111110656
8888888888888888888 1111111111111111680
9999999999999999999 1111111111111110656
100000000000000000000 90000000000000000000
Can anyone see where I'm going wrong, as I'm obviously missing something... and it's driving me mental!!
Many thanks! :)
The numbers in the second example input are too large.
Although the logic of the program is correct,
there's a loss of precision when doing computations with very large integers, such as 2222222222222222222 - 1111111111111111111 resulting in 1111111111111111168 instead of the expected 1111111111111111111.
See a detailed explanation in The GNU Awk User’s Guide:
As has been mentioned already, awk uses hardware double precision with 64-bit IEEE binary floating-point representation for numbers on most systems. A large integer like 9,007,199,254,740,997 has a binary representation that, although finite, is more than 53 bits long; it must also be rounded to 53 bits. The biggest integer that can be stored in a C double is usually the same as the largest possible value of a double. If your system double is an IEEE 64-bit double, this largest possible value is an integer and can be represented precisely. What more should one know about integers?
If you want to know what is the largest integer, such that it and all smaller integers can be stored in 64-bit doubles without losing precision, then the answer is 2^53. The next representable number is the even number 2^53 + 2, meaning it is unlikely that you will be able to make gawk print 2^53 + 1 in integer format. The range of integers exactly representable by a 64-bit double is [-2^53, 2^53]. If you ever see an integer outside this range in awk using 64-bit doubles, you have reason to be very suspicious about the accuracy of the output.
As #EdMorton pointed out in a comment,
you can have arbitrary-precision arithmetic if your Awk was compiled with MPFR support and you specify the -M flag.
For more details, see 15.3 Arbitrary-Precision Arithmetic Features.
julia> typeof(-0b111)
Uint64
julia> typeof(-0x7)
Uint64
julia> typeof(-7)
Int64
I find this result a bit surprising. Why does the numeric base of the number determine signed or unsgined-ness?
Looks like this is expected behavior:
This behavior is based on the observation that when one uses unsigned
hex literals for integer values, one typically is using them to
represent a fixed numeric byte sequence, rather than just an integer
value.
http://docs.julialang.org/en/latest/manual/integers-and-floating-point-numbers/#integers
...seems like a bit of an odd choice.
This is a subjective call, but I think it's worked out pretty well. In my experience when you use hex or binary, you're interested in a specific pattern of bits – and you generally want it to be unsigned. When you're just interested a numeric value you use decimal because that's what we're most familiar with. In addition, when you're using hex or binary, the number of digits you use for input is typically significant, whereas in decimal, it isn't. So that's how literals work in Julia: decimal gives you a signed integer of a type that the value fits in, while hex and binary give you an unsigned value whose storage size is determined by the number of digits.
I'm migrating some code from a compiler with 32-bit integers to 64-bit. I found some old code that assumes the highest possible integer is 2147483647 and the lowest possible is -947483647.
I understand the highest (maximum signed 32-bit integer), but does anyone know what makes the lowest special? There is nothing in the business logic that suggests that this integer (used for an ID) can't be below that number.
Searching google turns up very little except some other code where someone used 947483646 in a variable called INF (infinity/highest possible number? In signed, two's-complement representation that would make "negative infinity" -947483647).
It may be just a meaningless number, but there are also a few other hits using that exact number for other integers like monster HP in a video game (while searching for other close-by numbers turn up no results) that makes me think there's something behind it.
Int32 datatype doesn't have a value for infinity, so one can randomly chose magic number to denote INF.
It's naturally to select maximal integer value (2^31-1) for +INF.
But why (-2^31) was not chosen for -INF?
Probably, because both +INF and -INF must have equal length when printed (10 chars each).
I want to convert decimal number to hexadecimal in Embedded C.
Actually in my project input to controller is decimal which will be subtracted from hexadecimal value, so I need a conversion.
I tried to write it but after converting a number (75 for example) to its hexadecimal equivalent it will be (411). The trouble is that I did not know how to convert a number like 11 to b in hexadecimal as you know that there is no 11 in hexadecimal, it is b; so please help.
I want to save the converted value in a flag (for subtracting), not for printing. I can print a hex value by simple put a condition like:
(if (a > 10) printf("b"))
but this is not a solution for Embedded.
So please give me a complete solution.
I am not sure what you mean but your integer is just the "type of interpretation". In your memory it's just a bunch of 0 and 1, and therefore you can also presentate that "data" in an decimal, hexadecimal way.
If you need it as input for a register, you can just pass it into it.
Or do I missunderstand your problem?
just learning as3 for flex. i am trying to do this:
var someNumber:String = "10150125903517628"; //this is the actual number i noticed the issue with
var result:String = String(Number(someNumber) + 1);
I've tried different ways of putting the expression together and no matter what i seem to do the result is always equal to 10150125903517628 rather than 10150125903517629
Anyone have any ideas??! thanks!
All numbers in JavaScript/ActionScript are effectively double-precision IEEE-754 floats. These use a 64-bit binary number to represent your decimal, and have a precision of roughly 16 or 17 decimal digits.
You've run up against the limit of that format with your 17-digit number. The internal binary representation of 10150125903517628 is no different to that of 10150125903517629 which is why you're not seeing any difference when you add 1.
If, however, you add 2 then you will (should?) see the result as 10150125903517630 because that's enough of a "step" that the internal binary representation will change.