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What is an “internal address” in Java?

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问题 In the Javadoc for Object.hashCode() it states As much as is reasonably practical, the hashCode method defined by class Object does return distinct integers for distinct objects. (This is typically implemented by converting the internal address of the object into an integer, but this implementation technique is not required by the Java™ programming language.) It\'s a common miconception this has something to do with the memory address but it doesn\'t as that can change without notice and the

Divide by 10 using bit shifts?

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问题 Is it possible to divide an unsigned integer by 10 by using pure bit shifts, addition, subtraction and maybe multiply? Using a processor with very limited resources and slow divide. 回答1: Editor's note: this is not actually what compilers do, and gives the wrong answer for large positive integers ending with 9, starting with div10(1073741829) = 107374183 not 107374182. It is exact for smaller inputs, though, which may be sufficient for some uses. Compilers (including MSVC) do use fixed-point

Why bit endianness is an issue in bitfields?

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问题 Any portable code that uses bitfields seems to distinguish between little- and big-endian platforms. See the declaration of struct iphdr in linux kernel for an example of such code. I fail to understand why bit endianness is an issue at all. As far as I understand, bitfields are purely compiler constructs, used to facilitate bit level manipulations. For instance, consider the following bitfield: struct ParsedInt { unsigned int f1:1; unsigned int f2:3; unsigned int f3:4; }; uint8_t i; struct

Divide by 10 using bit shifts?

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Is it possible to divide an unsigned integer by 10 by using pure bit shifts, addition, subtraction and maybe multiply? Using a processor with very limited resources and slow divide. Here's what the Microsoft compiler does when compiling divisions by small integral constants. Assume a 32-bit machine (code can be adjusted accordingly): int32_t div10(int32_t dividend) { int64_t invDivisor = 0x1999999A; return (int32_t) ((invDivisor * dividend) >> 32); } What's going here is that we're multiplying by a close approximation of 1/10 * 2^32 and then removing the 2^32. This approach can be adapted to