module myRegister
(input clk,
input [3:0] write,
input [3:0] read1,
input [3:0] read2,
input [3:0]writedata);
reg[3:0]thereg[7:0];
reg [3:0]readdata1;
reg [3:0]readdata2;
always #(posedge clk) begin
readdata1=thereg[read1];
readdata2=thereg[read2];
end
always #(negedge clk) begin
thereg[write]=writedata;
end
endmodule
I need to force values (1s and 0s) into the "thereg" registry file and read from it twice on the positive edge from the clock and write from it once on the negative edge of the clock. I am unable to force any values however. I click force and nothing happens, however, I am able to force values into everything else. Any ideas?
In testbench, you will get access only to the input/output ports, but not to the wire/reg. Better way for your requirement is to force the values to input port writedata at each negative edges and you will get the same data at successive positive clock edges on the output ports readdata1 & readdata2.
Related
So I have a function I'm using to read data from a file. It works fine if the file is plain text, but when I try to read a binary file, like a png, it returns a different text (diff confirms that). I opened a hex editor to see what was wrong and found out it is putting some c2 bytes along with the file (I don't know if the position is random or if there are other bytes except this c2 one).
This is my function. I just want it to read and save to a variable.
proc read_file {path} {
set channel [open $path r]
fconfigure $channel -translation binary
set return_string "[read $channel]"
close $channel
return "$return_string"
}
To actually print, I'm doing this:
puts -nonewline [read_file file.png]
When you open a file, it defaults to being in text mode . In text mode (which is really a combination of options) the IO layer translates characters from whatever encoding they are in into Tcl's internal encoding, and does the reverse operation on output. The default encoding scheme is platform specific, but in your case it sounds like it is UTF-8. (Tcl uses a complex internal system of encodings; it doesn't expose those to the outside world.)
By contrast, when you put the channel into binary mode, the bytes on the outside are directly mapped to characters in the range 0-255 (and vice versa on output). You get a perfect copy, provided you put both input and output channels in binary mode. (There are other optimisations for binary mode, but they don't matter here.)
When you only put one of the channels in binary mode, you get what looks like corruption. It isn't random though. In particular, when the input is binary but the output is UTF-8, input bytes in the range 128-255 get converted into multiple output bytes, where the first of those bytes is in the sort of range you observed. There are other combinations that mess things up; the whole range of problems is collectively known as mojibake.
tl;dr Don't mix up binary and text data unless you're very careful. The results of getting it wrong are "surprising".
from GNU gawk's page
https://www.gnu.org/software/gawk/manual/html_node/Checking-for-MPFR.html
they have a formula to check arbitrary precision
function adequate_math_precision(n) { return (1 != (1+(1/(2^(n-1))))) }
My question is : wouldn't it be more efficient by staying within integer math domain with a formula such as
( 2^abs(n) - 1 ) % 2 # note 2^(n-1) vs. 2^|n| - 1
Since any power of 2 must also be even, then subtracting 1 must always be odd, then its modulo (%) over 2 becomes indicator function for is_odd() for n >= 0, while the abs(n) handles the cases where it's negative.
Or does the modulo necessitate a casting to float point, thus nullifying any gains ?
Good question. Let's tackle it.
The proposed snippet aims at checking wether gawk was invoked with the -M option.
I'll attach some digression on that option at the bottom.
The argument n of the function is the floating point precision needed for whatever operation you'll have to perform. So, say your script is in a library somewhere and will get called but you have no control over it. You'll run that function at the beginning of the script to promptly throw exception and bail out, suggesting that the end result will be wrong due to lack of bits to store numbers.
Your code stays in the integer realm: a power of two of an integer is an integer. There is no need to use abs(n) here, because there is no point in specifying how many bits you'll need as a negative number in the first place.
Then you subtract one from an even, integer number. Now, unless n=0, in which case 2^0=1 and then your code reads (1 - 1) % 2 = 0, your snippet shall always return 1, because the quotient (%) of an odd number divided by two is 1.
Problem is: you are trying to calculate a potentially stupidly large number in a function that should check if you are able to do so in the first place.
Since any power of 2 must also be even, then subtracting 1 must always
be odd, then its modulo (%) over 2 becomes indicator function for
is_odd() for n >= 0, while the abs(n) handles the cases where it's
negative.
Except when n=0 as we discussed above, you are right. The snippet will tell that any power of 2 is even, and any power of 2, minus 1, is odd. We were discussing another subject entirely thought.
Let's analyze the other function instead:
return (1 != (1+(1/(2^(n-1)))))
Remember that booleans in awk runs like this: 0=false and non zero equal true. So, if 1+x where x is a very small number, typically a large power of two (2^122 in the example page) is mathematically guaranteed to be !=1, in the digital world that's not the case. At one point, floating computation will reach a precision rock bottom, will be rounded down, and x=0 will be suddenly declared. At that point, the arbitrary precision function will return 0: false: 1 is equal 1.
A larger discussion on types and data representation
The page you link explains precision for gawk invoked with the -M option. This sounds like technoblahblah, let's decipher it.
At one point, your OS architecture has to decide how to store data, how to represent it in memory so that it can be accessed again and displayed. Terms like Integer, Float, Double, Unsigned Integer are examples of data representation. We here are addressing Integer representation: how is an integer stored in memory?
A 32-bit system will use 4 bytes to represent and integer, which in turn determines how larger the integer will be. The 32 bits are read from most significative (MSB) to less significative (LSB) and if signed, one bit will represent the sign (the MSB typically, drastically reducing the max size of the integer).
If asked to compute a large number, a machine will try to fit in in the max number available. If the end result is larger than that, you have overflow and end up with a wrong result or an error. Many online challenges typically ask you to write code for arbitrary long loops or large sums, then test it with inputs that will break the 64bit barrier, to see if you master proper types for indexes.
AWK is not a strongly typed language. Meaning, any variable can store data, regardless of the type. The data type can change and it is determined at runtime by the interpreter, so that the developer doesn't need to care. For instance:
$awk '{a="this is text"; print a; a=2; print a; print a+3.0*2}'
-| this is text
-| 2
-| 8
In the example, a is text, then is an integer and can be summed to a floating point number and printed as integer without any special type handling.
The Arbitrary Precision Page presents the following snippet:
$ gawk -M 'BEGIN {
> s = 2.0
> for (i = 1; i <= 7; i++)
> s = s * (s - 1) + 1
> print s
> }'
-| 113423713055421845118910464
There is some math voodoo behind, we will skip that. Since s is interpreted as a floating point number, the end result is computed as floating point.
Try to input that number on Windows calculator as decimal, and it will fail. Although you can compute it as a binary. You'll need the programmer setting and to add up to 53 bits to be able to fit it as unsigned integer.
53 is a magic number here: with the -M option, gawk uses arbitrary precision for numbers. In other words, it commandeers how many bits are necessary, track them and breaks free of the native OS architecture. The default option says that gawk will allocate 53 bits for any given arbitrary number. Fun fact, the actual result of that snippet is wrong, and it would take up to 100 bits to compute correctly.
To implement arbitrary large numbers handling, gawk relies on an external library called MPFR. Provided with an arbitrary large number, MPFR will handle the memory allocation and bit requisition to store it. However, the interface between gawk and MPFR is not perfect, and gawk can't always control the type that MPFR will use. In case of integers, that's not an issue. For floating point numbers, that will result in rounding errors.
This brings us back to the snippet at the beginning: if gawk was called with the -M option, numbers up to 2^53 can be stored as integers. Floating points will be smaller than that (you'll need to make the comma disappear somehow, or rather represent it spending some of the bits allocated for that number, just like the sign). Following the example of the page, and asking an arbitrary precision larger than 32, the snippet will return TRUE only if the -M option was passed, otherwise 1/2^(n-1) will be rounded down to be 0.
I am wondering if it is possible to use Yosys in simplifying logic equations.
For example:
module top
(
output [31:0] cipher,
input [31:0] plain,
input [63:0] key
);
wire tmp = key[31:0];
wire tmp2 = key[63:32] & 0;
assign cipher = (tmp & plain) | tmp2;
endmodule
When I use the command "show" it plots the circuit:
I tried using "opt" and "freduce" commands but it did not reduce the equation.
You probably want to use opt -fine which does finer-grain optimisations rather than optimising whole words at a time. This gives a single 1-bit $and gate as expected.
Alternatively techmap; abc will produce an optimised gate-level circuit.
I was working in a project and I can´t use the bitwise negation with a U32 bits (Unsigned 32 bits) because when I tried to use the negation operator for example I have 1 and the negation (according to this function) was the biggest number possible with U32 and I expected the zero. My idea was working with a binary number like (110010) and I need to only negate the bits after the first 1-bit(001101). There is a way to do that in LabVIEW?
This computes the value you are looking for.
1110 --> 0001 (aka, 1)
1010 --> 0101 (aka, 101)
111 --> 000 (aka, 0) [indeed, all patterns that are all "1" will become 0]
0 --> 0 [because there are no bits to negate... maybe you want to special case this as "1"?)
Note: This is a VI Snippet. Save the .png file to your disk then drag the image from your OS into LabVIEW and it will generate the block diagram (I wrote it in LV 2016, so it works for 2016 or later). Sometimes dragging directly from browser to diagram works, but most browsers seem to strip out the EXIF data that makes that work.
Here's an alternative solution without a loop. It formats the input into its string representation (without leading zeros) to figure out how many bits to negate - call this n - and then XOR's the input with 2^n - 1.
Note that this version will return an output of 1 for an input of 0.
Using the string functions feels a bit hacky... but it doesn't use a loop!!
Obviously we could instead try to get the 'bit length' of the input using its base-2 log, but I haven't sat down and worked out how to correctly ensure that there are no rounding issues when the input has only its most significant bit set, and therefore the base-2 log should be exactly an integer, but might come out a fraction smaller.
Here's a solution without strings/loops using the conversion to float (a common method of computing floor(log_2(x))). This won't work on unsigned types.
what is the method by which I can read the input of the user, say the input is "500"
then store this number in a variable?
The only method I know would be to store them character by character with possibly the need of register offsets.
Is there any other way, preferably storing the number directly?
i.e. something like:
mov var1, inbuffer
Details on environment:
32 bit Assembly w/ DGJPP
Thank you.
Ahhh... DJGPP, that'd be dos I guess. Look into int 21h/0Ah (0Ah in ah). Or you might be better off with the readfile subfunction (3Fh ???) on stdin. Look it up in Ralf Brown's Interrupt list.
In any case, what you're going to get is the characters '5', '0', and '0' - 35h, 30h, 30h. It will take some processing to get the number 500 out of this. If you're reading numbers from left to right, zero up a register to use as "result so far". Read a character from your input buffer. If it's a valid decimal digit, subtract '0' to convert character to number, multiply "result so far" by ten, and add in your new number. Repeat until you run out of characters.