I came across this line in the book 'The Go Programming Languague' on page 112.
fmt.Printf("#%-5d %9.9s %.55s\n", item.Number, item.User.Login, item.Title)
What do %9.9s and %.55s mean?
From go doc fmt:
Width is specified by an optional decimal number immediately preceding the verb. If absent, the width is whatever is necessary to represent the value. ....
For strings, byte slices and byte arrays, however, precision limits the length of the input to be formatted (not the size of the output), truncating if necessary.
Thus, %.9.9s means minimal width 9 runes with input truncated at 9, and thus exactly length 9. Similar %.55s means no minimal width but input truncated at 55 which means output is at most 55 runes.
Related
I am trying to solve the following problem using Ruby:
I have a requirement to generate strings with variable bits length which contain only alphanumeric characters.
Here is what I have already found:
Digest::SHA2.new(bitlen = 256).to_s
# => "e3b0c44298fc1c149afbf4c8996fb92427ae41e4649b934ca495991b7852b855"
It does exactly what I need, but it accepts only 256, 384, and 512 as bitlen.
Does anybody aware of any alternatives?
Thanks in advance.
Update
One byte = collection of 8 bits.
Every alphanumeric character occupies 1 byte according to String#bytesize.
('a'..'z').chain('A'..'Z').chain('0'..'9').map(&:bytesize).uniq
# => [1]
Based on the facts mentioned above, we can suppose that
SecureRandom.alphanumeric(1) generates an alphanumeric string with 8 bits length.
SecureRandom.alphanumeric(2) generates an alphanumeric string with 16 bits length.
SecureRandom.alphanumeric(3) generates an alphanumeric string with 24 bits length.
And so on...
As a result, #anothermh's answer can be considered as an acceptable solution.
Use SecureRandom.
First, make sure you require it:
require 'securerandom'
Then you can generate values:
SecureRandom.alphanumeric(10)
=> "hxYolwzk0P"
Change 10 to whatever length you require.
It's worth pointing out that the example you used was returning not alphanumeric but hexadecimal values. If you specifically require hex then you can use:
SecureRandom.hex(10)
=> "470eb1d8daebacd20920"
I've been looking into this but searching seems to lead to nothing.
It might be too simple to be described, but here I am, scratching my head...
Any help would be appreciated.
Verilog knows about "strings".
A single ASCII character requires 8 bits. Thus to store 8 characters you need 64 bits:
wire [63:0] string8;
assign string8 = "12345678";
There are some gotchas:
There is no End-Of-String character (like the C null-character)
The most RHS character is in bits 7:0.
Thus string8[7:0] will hold 8h'38. ("8").
To walk through a string you have to use e.g.: string[ index +: 8];
As with all Verilog vector assignments: unused bits are set to zero thus
assign string8 = "ABCD"; // MS bit63:32 are zero
You can not use two dimensional arrays:
wire [7:0] string5 [0:4]; assign string5 = "Wrong";
You are probably mislead by a misconception about characters. There are no such thing as a character in hardware. There are only sets of bits or codes. The only thing which converts binary codes to characters is your terminal. It interprets codes in a certain way and forming letters for you to se. So, all the printfs in 'c' and $display in verilog only send the codes to the terminal (or to a file).
The thing which converts characters to the codes is your keyboard, which you also use to type in the program. The compiler then interprets your program. Verilog (as well as the 'c') compiler represents strings in double quotes (which you typed in) as a set of bytes directly. Verilog, as well as 'c' use ascii-8 encoding for such character strings, meaning that the code for 'a' is decimal 97 and 'b' is 98, .... Every character is 8-bit wide and the quoted string forms a concatenation of bytes of ascii codes.
So, answering you question, you can convert an ascii codes to characters by sending them to the terminal via $display (or other) function, using the %s modifier.
So, an example:
module A;
reg[8*5-1:0] hello;
reg[8*3 - 1: 0] bye;
initial begin
hello = "hello"; // 5 bytes of characters
bye = {8'd98, 8'd121, 8'd101}; // 3 bytes 'b' 'y' 'e'
$display("hello=%s bye=%s", hello, bye);
end
endmodule
I am a tad confused about what I see here:
a = [ "a", "b", "c" ]
n = [ 65, 66, 67 ]
a.pack("A3A3A3") #=> "a b c "
a.pack("a3a3a3") #=> "a\000\000b\000\000c\000\000"
n.pack("ccc") #=> "ABC"
From the docs:
Packs the contents of arr into a binary sequence according to the directives in aTemplateString (see the table below) Directives “A,'' “a,'' and “Z'' may be followed by a count, which gives the width of the resulting field.
Here are the directives:
So we're using the A directive 3 times it seems? What does it mean to pack the string a into an arbitrary binary string (space padded, count is width?) Can you help me understand the output? Why are there so many 0s?
In the first case, you're printing "a" but padding its length to 3 with spaces, hence the two spaces to get the total length to 3.
In the second case, you're doing the same but padding with null bytes instead (ASCII value 0). Null bytes in Ruby are printed (and can be read) using the escape syntax \000 (this is one character), so \000\000 is actually just two null bytes.
The variable n is irrelevant, so you can ignore it.
In the pack statements, the bytes "a", "b" and "c" are concatenated ("packed") into a single string, with padding between them. The padding is such that the number of bytes (the width) taken up by the contents plus the padding equals the number provided.
So in the first pack statement, the "a" is padded with two spaces to make these three bytes: "a.." where I've put a . in place of the spaces to make it clear. That is concatenated with the "b" and the "c" similarly padded, to produce "a..b..c..".
In the second pack statement, null characters ('\000') are used instead of spaces. The \xxx notation (called an "escape sequence") means the byte with octal value xxx. It's used when there isn't a useful ASCII character (like 'a' or ' ') to show. A null character has no useful ASCII character, so the \xxx notation is used instead.
I have a data structure that I'd like to convert back and forth from hex to binary in Ruby. The simplest approach for a binary to hex is '0010'.to_i(2).to_s(16) - unfortunately this does not preserve leading zeroes (due to the to_i call), as one may need with data structures like cryptographic keys (which also vary with the number of leading zeroes).
Is there an easy built in way to do this?
I think you should have a firm idea of how many bits are in your cryptographic key. That should be stored in some constant or variable in your program, not inside individual strings representing the key:
KEY_BITS = 16
The most natural way to represent a key is as an integer, so if you receive a key in a hex format you can convert it like this (leading zeros in the string do not matter):
key = 'a0a0'.to_i(16)
If you receive a key in a (ASCII) binary format, you can convert it like this (leading zeros in the string do not matter):
key = '101011'.to_i(2)
If you need to output a key in hex with the right number of leading zeros:
key.to_s(16).rjust((KEY_BITS+3)/4, '0')
If you need to output a key in binary with the right number of leading zeros:
key.to_s(2).rjust(KEY_BITS, '0')
If you really do want to figure out how many bits might be in a key based on a (ASCII) binary or hex string, you can do:
key_bits = binary_str.length
key_bits = hex_str.length * 4
The truth is, leading zeros are not part of the integer value. I mean, it's a little detail related to representation of this value, not the value itself. So if you want to preserve properties of representation, it may be best not to get to underlying values at all.
Luckily, hex<->binary conversion has one neat property: each hexadecimal digit exactly corresponds to 4 binary digits. So assuming you only get binary numbers that have number of digits divisible by 4 you can just construct two dictionaries for constructing back and forth:
# Hexadecimal part is easy
hex = [*'0'..'9', *'A'..'F']
# Binary... not much longer, but a bit trickier
bin = (0..15).map { |i| '%04b' % i }
Note the use of String#% operator, that formats the given value interpreting the string as printf-style format string.
Okay, so these are lists of "digits", 16 each. Now for the dictionaries:
hex2bin = hex.zip(bin).to_h
bin2hex = bin.zip(hex).to_h
Converting hex to bin with these is straightforward:
"DEADBEEF".each_char.map { |d| hex2bin[d] }.join
Converting back is not that trivial. I assume we have a "good number" that can be split into groups of 4 binary digits each. I haven't found a cleaner way than using String#scan with a "match every 4 characters" regex:
"10111110".scan(/.{4}/).map { |d| bin2hex[d] }.join
The procedure is mostly similar.
Bonus task: implement the same conversion disregarding my assumption of having only "good binary numbers", i. e. "110101".
"I-should-have-read-the-docs" remark: there is Hash#invert that returns a hash with all key-value pairs inverted.
This is the most straightforward solution I found that preserves leading zeros. To convert from hexadecimal to binary:
['DEADBEEF'].pack('H*').unpack('B*').first # => "11011110101011011011111011101111"
And from binary to hexadecimal:
['11011110101011011011111011101111'].pack('B*').unpack1('H*') # => "deadbeef"
Here you can find more information:
Array#pack: https://ruby-doc.org/core-2.7.1/Array.html#method-i-pack
String#unpack1 (similar to unpack): https://ruby-doc.org/core-2.7.1/String.html#method-i-unpack1
I was trying to understand BSON Notations from the site BSON Site. However I was unable to understand the reason behind the correlations.
I also referred the following questions, but am not convinced because of the following reasons.
Question 1 : Not familiar with ruby implementations
Question 2 : I understood the byte allocation. But unsure about the notations.
I would like to know how the bson object is formed for the following examples in BSON Site
1.{"hello": "world"}
2.{"BSON": ["awesome", 5.05, 1986]}
{"hello": "world"}
\x16\x00\x00\x00
\x02 hello\x00 \x06\x00\x00\x00 world\x00
\x00
(overall: 22 bytes)
The first four bytes contain the overall length as a 32-bit little-endian integer.
\x16\x00\x00\x00 => That's 22 in decimal.
Now comes the first element. The first byte gives the type of data.
\x02 => That's a UTF-8 string.
Then comes the name of the first element, as a null-terminated string.
hello\x00
Next comes the data of the element in the previously given type, in this case a string.
For scannability (so you can quickly skip when you don't need them), strings start with their length, and are null-terminated.
\x06\x00\x00\x00 => That's length 6.
world\x00
Now would come subsequent elements, if there were any. The whole thing is terminated with a null byte.
\x00
{"BSON": ["awesome", 5.05, 1986]}
\x31\x00\x00\x00
\x04 BSON\x00 \x26\x00\x00\x00
\x02 0\x00 \x08\x00\x00\x00 awesome\x00
\x01 1\x00 \x33\x33\x33\x33\x33\x33\x14\x40
\x10 2\x00 \xc2\x07\x00\x00
\x00
\x00
(overall: 49 bytes, array: 38 bytes)
The first four bytes contain the overall length as a 32-bit little-endian integer.
\x31\x00\x00\x00 => That's 49 in decimal.
Now comes the first element. The first byte gives the type of data.
\x04 => That's an array.
Then comes the name of the first element, as a null-terminated string.
BSON\x00
Next comes the data of the element in the previously given type, in this case an array.
[Quote: "The document for an array is a normal BSON document with
integers for the keys, starting with 0 (..)"]
For scannability and because they form document in their own right, arrays start with their length, and are null-terminated.
\x26\x00\x00\x00 => That's 38 in decimal.
Now comes the first element of the array. The first byte gives the type of data.
\x02 => That's a UTF-8 string.
Then comes the name of the first element of the array, null-terminated.
0\x00 => That's key 0.
Next comes the data of the element in the previously given type, in this case a string.
Strings start with their length, and are null-terminated.
\x08\x00\x00\x00 => length 8
awesome\x00
Now comes the second element of the array. The first byte gives the type of data.
\x01 => That's a double floating point number.
Then comes the name of the second element of the array, null-terminated.
1\x00 => That's key 1.
Next comes the data of the element in the previously given type, in this case a double floating point number.
\x33\x33\x33\x33\x33\x33\x14\x40 => That's 5.5.
Now comes the third element of the array. The first byte gives the type of data.
\x10 => That's a 32-bit integer.
Then comes the name of the third element of the array, null-terminated.
2\x00 => That's key 2.
Next comes the data of the element in the previously given type, in this case a 32-bit integer.
\xc2\x07\x00\x00 => That's 1986.
The array is terminated with a null byte.
\x00
The whole thing is terminated with a null byte.
\x00