twos-complement

问题 This is what I see in java, and it puzzles me. Long.toHexString(0xFFFFFFFF) returns ffffffffffffffff Similarly, 0xFFFFFFFF and Long.parseLong("FFFFFFFF", 16) are unequal. 回答1: As others have said, 0xFFFFFFFF evaluates to the int value -1 , which is promoted to a long . To get the result you were expecting, qualify the constant with the L suffix to indicate it should be treated as a long , i.e. Long.toHexString(0xFFFFFFFFL) . 回答2: This: Long.toHexString(0xFFFFFFFF) is equivalent to: Long

问题 This is what I see in java, and it puzzles me. Long.toHexString(0xFFFFFFFF) returns ffffffffffffffff Similarly, 0xFFFFFFFF and Long.parseLong("FFFFFFFF", 16) are unequal. 回答1: As others have said, 0xFFFFFFFF evaluates to the int value -1 , which is promoted to a long . To get the result you were expecting, qualify the constant with the L suffix to indicate it should be treated as a long , i.e. Long.toHexString(0xFFFFFFFFL) . 回答2: This: Long.toHexString(0xFFFFFFFF) is equivalent to: Long

问题 This question already has answers here : C++ literal integer type (2 answers) Closed 12 days ago . Why doesn't this code print same number? : long long a, b; a = 2147483647 + 1; b = 2147483648; printf("%lld\n", a); printf("%lld\n", b); I know that int variable's maximum number is 2147483647 because int variable is 4 byte. But as I know, long long variable is 8 byte, but why does that code act like that? 回答1: 2147483647 + 1 is executed as the sum of two ints and therefore overflows. 2147483648

问题 This is most probably the dumbest question anyone would ask, but regardless I hope I will find a clear answer for this. My question is - How is an integer stored in computer memory? In c# an integer is of size 32 bit. MSDN says we can store numbers from -2,147,483,648 to 2,147,483,647 inside an integer variable. As per my understanding a bit can store only 2 values i.e 0 & 1. If I can store only 0 or 1 in a bit, how will I be able to store numbers 2 to 9 inside a bit? More precisely, say I

问题 This is most probably the dumbest question anyone would ask, but regardless I hope I will find a clear answer for this. My question is - How is an integer stored in computer memory? In c# an integer is of size 32 bit. MSDN says we can store numbers from -2,147,483,648 to 2,147,483,647 inside an integer variable. As per my understanding a bit can store only 2 values i.e 0 & 1. If I can store only 0 or 1 in a bit, how will I be able to store numbers 2 to 9 inside a bit? More precisely, say I

问题 This is most probably the dumbest question anyone would ask, but regardless I hope I will find a clear answer for this. My question is - How is an integer stored in computer memory? In c# an integer is of size 32 bit. MSDN says we can store numbers from -2,147,483,648 to 2,147,483,647 inside an integer variable. As per my understanding a bit can store only 2 values i.e 0 & 1. If I can store only 0 or 1 in a bit, how will I be able to store numbers 2 to 9 inside a bit? More precisely, say I

问题 I've decided to do it this way flip numbers 0=1, 1=0 add 1 to LSB if carry, loop until array[i]==0 But I'm stuck on the last point; how can I say that in a conditional loop? 回答1: You are talking about extended arithmetic. Most processors have carry-out and overflow results from every addition operation, but C does not provide access to them. Your problem is that numbers get longer as they get bigger. If you're at the last bit you have, and you need to carry out, you need another bit! That

问题 I have a negative integer (4 bytes) of which I would like to have the hexadecimal form of its two's complement representation. >>> i = int("-312367") >>> "{0}".format(i) '-312367' >>> "{0:x}".format(i) '-4c42f' But I would like to see "FF..." 回答1: Here's a way (for 16 bit numbers): >>> x=-123 >>> hex(((abs(x) ^ 0xffff) + 1) & 0xffff) '0xff85' (Might not be the most elegant way, though) 回答2: >>> x = -123 >>> bits = 16 >>> hex((1 << bits) + x) '0xff85' >>> bits = 32 >>> hex((1 << bits) + x)

问题 Since SE 7 Java allows to specify values as binary literal. The documentation tells me 'byte' is a type that can hold 8 Bit of information, the values -128 to 127. Now i dont know why but i cannot define 8 bits but only 7 if i try to assign a binary literal to a byte in Java as follows: byte b = 0b000_0000; //solves to the value 0 byte b1 = 0b000_0001; //solves to the value 1 byte b3 = 0b000_0010; //solves to the value 2 byte b4 = 0b000_0011; //solves to the value 3 And so on till we get to

问题 I understand what the Two's complement is and what it is useful for. What I'd like to know is how does the computer decide that the number is in its Two's complement form? How and when does it decide that 1111 1110 is -2 and not 254? Is it at the OS level of processing? 回答1: The computer will already be expecting the data to be in (or not in) two's complement form (otherwise there wouldn't be a way of telling if it is - 2 or 254). And yes, that would probably be decided at the OS-level. You