pack

I have some 256-character strings of hexadecimal characters which represent a sequence of bit flags, and I'm trying to convert them back into a bitstring so I can manipulate them with & , | , vec and the like. The hex strings are written in integer-wide big-endian groups, such that a group of 8 bytes like "76543210" should translate to the bitstring "\x10\x32\x54\x76" , i.e. the lowest 8 bits are 00001000 . The problem is that pack 's " h " format works on one byte of input at a time, rather than 8, so the results from just using it directly won't be in the right order. At the moment I'm doing

pack() syntax is (from http://php.net/manual/en/function.pack.php ) string pack ( string $format [, mixed $args [, mixed $... ]] ) so assuming I need to pack three bytes $packed = pack( "c*", 65, 66, 67 ); But what if I have to pack an arbitrary number of bytes? They could coveniently be stored into an array so I naively tried $a = array( 65, 66, 67 ); $packed = pack( "c*", $a ); But it doesn't work. Is there a way to make pack() work with an array ? At little late to the party, but for future reference, you can use the new ... operator (v5.6+) to explode the array inline: $packed = pack("c*",

I have a UUID that I was thinking of packing into a struct using UUID.int, which turns it into a 128-bit integer. But none of the struct format characters are large enough to store it, how to go about doing this? Sample code: s = struct.Struct('L') unique_id = uuid.uuid4() tuple = (unique_id.int) packed = s.pack(*tuple) The problem is, struct format 'L' is only 4 bytes...I need to store 16. Storing it as a 32-char string is a bit much. It is a 128-bit integer, what would you expect it to be turned into? You can split it into several components — e.g. two 64-bit integers: max_int64 =

I have the following struct, from the NRPE daemon code in C: typedef struct packet_struct { int16_t packet_version; int16_t packet_type; uint32_t crc32_value; int16_t result_code; char buffer[1024]; } packet; I want to send this data format to the C daemon from Python. The CRC is calculated when crc32_value is 0 , then it is put into the struct. My Python code to do this is as follows: cmd = '_NRPE_CHECK' pkt = struct.pack('hhIh1024s', 2, 1, 0, 0, cmd) # pkt has length of 1034, as it should checksum = zlib.crc32(pkt) & 0xFFFFFFFF pkt = struct.pack('hhIh1024s', 2, 1, checksum, 0, cmd) socket

I would like to pack all the data in a list into a single buffer to send over a UDP socket. The list is relatively long, so indexing each element in the list is tedious. This is what I have so far: NumElements = len(data) buf = struct.pack('d'*NumElements,data[0],data[1],data[2],data[3],data[4]) But I would like to do something more pythonic that doesn't require I change the call if I added more elements to the list... something like: NumElements = len(data) buf = struct.pack('d'*NumElements,data) # Returns error Is there a good way of doing this?? Yes, you can use the *args calling syntax.

when i change number into hex in struct module of python, >>> import struct >>> struct.pack("i",89) 'Y\x00\x00\x00' >>> struct.pack("i",890) 'z\x03\x00\x00' >>> struct.pack("i",1890) 'b\x07\x00\x00' what is the meaning of "Y,z,b" in the output? oldrinb You're not converting to hex. You're packing the integer as binary data... in this case, little-endian binary data. The first characters are just the corresponding ASCII characters to the raw bytes; e.g. 89 is Y , 122 is z , and 98 is b . The first pack produces '\x59\x00\x00\x00' for 0x00000059 ; '\x59' is 'Y' . The second produces '\x7a\x03

Consider this short example: $a = pack("d",255); print length($a)."\n"; # Prints 8 $aa = pack("ddddd", 255,123,0,45,123); print length($aa)."\n"; # Prints 40 @unparray = unpack("d "x5, $aa); print scalar(@unparray)."\n"; # Prints 5 print length($unparray[0])."\n" # Prints 3 printf "%d\n", $unparray[0] ' # Prints 255 # As a one-liner: # perl -e '$a = pack("d",255); print length($a)."\n"; $aa = pack("dd", 255,123,0,45,123); print length($aa)."\n"; @unparray = unpack("d "x5, $aa); print scalar(@unparray)."\n"; print length($unparray[0])."\n"; printf "%d\n", $unparray[0] ' Now, I'd expect a double

I need to convert the following perl function to php: pack("SSA12AC4L", $id, $loc, $name, 'ar', split(/\./, $get->getIP), time+(60*60); I use the following code (to test) in PHP: echo pack("SSA12AC4L", '25', '00001', '2u7wx6fd94fd', 'f', preg_split('/\./','10.2.1.1', -1, PREG_SPLIT_NO_EMPTY), time()+(60*60)); But I'm getting the following error: Warning: pack() [function.pack]: Type C: too few arguments in D:\wamp\www\test.php on line 8 Any suggestions? Thanks a lot. The problem is that the code is giving to pack() (I am referring the the last arguments) a character, an array, and an integer.