signed

I am working with some old data imports and came across a bunch of data from an external source that reports financial numbers with a signed overpunch . I've seen alot, but this is before my time. Before I go about creating a function to parse these strangers, I wanted to check to see if there was a standard way to handle these. I guess my question is, does the .Net framework provide a standard facility for converting signed overpunch strings? If not .NET, are there any third party tools I can use so I don't reinvent the wheel? Over-punched numeric ( Zoned-Decimal in Cobol) comes from the old

This is what I have, currently. Is there any nicer way to do this? import struct def int32_to_uint32(i): return struct.unpack_from("I", struct.pack("i", i))[0] Not sure if it's "nicer" or not... import ctypes def int32_to_uint32(i): return ctypes.c_uint32(i).value using numpy for example: import numpy result = numpy.uint32( numpy.int32(myval) ) or even on arrays, arr = numpy.array(range(10)) result = numpy.uint32( numpy.int32(arr) ) 来源： https://stackoverflow.com/questions/16452232/how-to-convert-signed-32-bit-int-to-unsigned-32-bit-int

When I was working on string::npos I noticed something and I couldn't find any explanation for it on the web. (string::npos == ULONG_MAX) and (string::npos == -1) are true. So I tried this: (18446744073709551615 == -1) which is also true. How can it be possible? Is it because of binary conversation? Evan Carroll 18,446,744,073,709,551,615 This number mentioned, 18,446,744,073,709,551,615 , is actually 2^64 − 1 . The important thing here is that 2^64-1 is essentially 0-based 2^64 . The first digit of an unsigned integer is 0 , not 1 . So if the maximum value is 1 , it has two possible values: 0

On left shift of (char) 0xff by 8 and casting it to int we get -256 or 0xffffff00. Can somebody explain why this should happen? #include <stdio.h> int main (void) { char c = 0xff; printf("%d %x\n", (int)(c<<8),(int)(c<<8)); return 0; } Output is -256 ffffff00 char can be signed or unsigned - it's implementation-defined. You see these results because char is signed by default on your compiler. For the signed char 0xFF corresponds to −1 (that's how two's complement work). When you try to shift it it is first promoted to an int and then shifted - you effectively get multiplication by 256. So it

Assuming you have a machine instruction udive that does a special case 64 by 32 unsigned division by taking a (32bit dividend << 32) / 32bit divisor, we can do a full 64 by 32 division using the following: // assume: a / b guaranteed not to overflow a = 64bit dividend, a.h & a.l are hi & lo 32bits respectively b = 32bit divisor q1 = udive(a.h, b) // (a.h << 32) / b r1 = -(q1 * b) // remainder of the above, shortcut since a.h & 0xffffffff == 0 q2 = a.l / b // a.l / b using regular unsigned division r2 = a.l - (q2 * b) // remainder of the above q = q1 + q2 r = r1 + r2 // r < r2, r overflowed and

I'm more than half way through learning assembly and I'm familiar with the concept of how signed and unsigned integers are presented in bits, I know that it might seem a weird question of which the answer would be pretty obvious, but I'm wondering if using an arithmetic operation like addition makes sense for a pair of numbers that one of them is considered signed and the other one unsigned, I've thought of multiple examples like below that will yield a correct result: 10000001 (1-byte integer and considered unsigned, equivalent to 129) + 11111111 (1-byte integer and considered signed(two's

I am a perl newbie, Can I simply use 64-bit arithmetic in Perl? For example $operand1 = 0xFFFFFFFFFFFF; # 48 bit value $operand2 = 0xFFFFFFFFFFFF; # 48 bit value $Result = $operand1 * $operand2; I am basically looking for a replacement for the int64_t in perl. Is there any way to mention, if the variable is signed or unsigned? Yes, however you need to have Perl compiled with 64-bit support. Sinan Ünür See bigint : Transparent BigInteger support for Perl... All operators (including basic math operations) except the range operator .. are overloaded. Integer constants are created as proper