twos-complement

问题 How does C represent negative integers? Is it by two\'s complement representation or by using the MSB (most significant bit)? -1 in hexadecimal is ffffffff . So please clarify this for me. 回答1: ISO C ( C99 section 6.2.6.2/2 in this case but it carries forward to later iterations of the standard (a) ) states that an implementation must choose one of three different representations for integral data types, two's complement, ones' complement or sign/magnitude (although it's incredibly likely

问题 I\'m just curious if there\'s a reason why in order to represent -1 in binary, two\'s complement is used: flipping the bits and adding 1? -1 is represented by 11111111 (two\'s complement) rather than (to me more intuitive) 10000001 which is binary 1 with first bit as negative flag. Disclaimer: I don\'t rely on binary arithmetic for my job! 回答1: It's done so that addition doesn't need to have any special logic for dealing with negative numbers. Check out the article on Wikipedia. Say you have

问题 I was using -1 as a flag value for a function whose return type is size_t (an unsigned type). I didn\'t notice it at first, particularly because it wasn\'t causing any errors in my code (I was checking it with x == -1, not x < 0). Are there any subtle reasons I shouldn\'t leave it as is? When might this behave unexpectedly? Is this commonly used? ptrdiff_t is less common, takes longer to type, and anyway it\'s not really the appropriate type since the function returns an index into an array.

问题 Is there a built in function in python which will convert a binary string, for example \'111111111111\', to the two\'s complement integer -1? 回答1: Two's complement subtracts off (1<<bits) if the highest bit is 1. Taking 8 bits for example, this gives a range of 127 to -128. A function for two's complement of an int... def twos_comp(val, bits): """compute the 2's complement of int value val""" if (val & (1 << (bits - 1))) != 0: # if sign bit is set e.g., 8bit: 128-255 val = val - (1 << bits) #

问题 I am trying to understand how Java stores integer internally. I know all java primitive integers are signed, (except short?). That means one less bit available in a byte for the number. My question is, are all integers (positive and negative) stored as two\'s complement or are only negative numbers in two\'s complement? I see that the specs says x bit two\'s complement number . But I often get confused. For instance: int x = 15; // Stored as binary as is? 00000000 00000000 00000000 00001111?

问题 In assembly programming, it\'s fairly common to want to compute something from the low bits of a register that isn\'t guaranteed to have the other bits zeroed. In higher level languages like C, you\'d simply cast your inputs to the small size and let the compiler decide whether it needs to zero the upper bits of each input separately, or whether it can chop off the upper bits of the result after the fact. This is especially common for x86-64 (aka AMD64), for various reasons 1 , some of which

问题 I\'m in a computer systems course and have been struggling , in part, with Two\'s Complement. I want to understand it but everything I\'ve read hasn\'t brought the picture together for me. I\'ve read the wikipedia article and various other articles, including my text book. Hence, I wanted to start this community wiki post to define what Two\'s Complement is, how to use it and how it can affect numbers during operations like casts (from signed to unsigned and vice versa), bit-wise operations