bitwise-or

How do I use Java's bitwise operators in Kotlin?

由 霸气de小男生 提交于 2020-05-25 09:01:48
问题 Java has binary-or | and binary-and & operators: int a = 5 | 10; int b = 5 & 10; They do not seem to work in Kotlin: val a = 5 | 10; val b = 5 & 10; How do I use Java's bitwise operators in Kotlin? 回答1: You have named functions for them. Directly from Kotlin docs As of bitwise operations, there're no special characters for them, but just named functions that can be called in infix form. for example: val x = (1 shl 2) and 0x000FF000 Here is the complete list of bitwise operations (available

How do I use Java's bitwise operators in Kotlin?

由 百般思念 提交于 2020-05-25 08:59:50
问题 Java has binary-or | and binary-and & operators: int a = 5 | 10; int b = 5 & 10; They do not seem to work in Kotlin: val a = 5 | 10; val b = 5 & 10; How do I use Java's bitwise operators in Kotlin? 回答1: You have named functions for them. Directly from Kotlin docs As of bitwise operations, there're no special characters for them, but just named functions that can be called in infix form. for example: val x = (1 shl 2) and 0x000FF000 Here is the complete list of bitwise operations (available

Bitwise operation in C language (0x80, 0xFF, << )

由 一个人想着一个人 提交于 2020-05-15 10:45:33
问题 I have a problem understanding this code. What I know is that we have passed a code into a assembler that has converted code into "byte code". Now I have a Virtual machine that is supposed to read this code. This function is supposed to read the first byte code instruction. I don't understand what is happening in this code. I guess we are trying to read this byte code but don't understand how it is done. static int32_t bytecode_to_int32(const uint8_t *bytecode, size_t size) { int32_t result;

Bitwise operation in C language (0x80, 0xFF, << )

由 ╄→尐↘猪︶ㄣ 提交于 2020-05-15 10:45:12
问题 I have a problem understanding this code. What I know is that we have passed a code into a assembler that has converted code into "byte code". Now I have a Virtual machine that is supposed to read this code. This function is supposed to read the first byte code instruction. I don't understand what is happening in this code. I guess we are trying to read this byte code but don't understand how it is done. static int32_t bytecode_to_int32(const uint8_t *bytecode, size_t size) { int32_t result;

How to read result of bitwise operator OR (|)?

由 Deadly 提交于 2020-01-03 05:52:05
问题 I would like the confirmation on bitwise operators inside Android XML files. For example this line android:layout_gravity="center_horizontal|bottom" How should I read it? Are the rules inherited from Java or Android does its own interpretation of bitwise operators? Are all bitwise operators supported? If yes, could you show me the example of other operators (link is fine as well)? Thanks 回答1: The value of android_layout_gravity is parsed by Android code. As documented here, the only

Purpose of bitwise OR of an integer with its negative

由 瘦欲@ 提交于 2019-12-23 12:25:31
问题 I was curious about the implementation and representation of NaN in both IEEE single- and double-precision floating points, and I found this implementation of an "is NaN" function. Namely: int isnan(double x) { int32_t hx,lx; // Move lower 32 bits of double to lx, higher 32 to hx. EXTRACT_WORDS(hx,lx,x); // Remove sign bit, since -NaN and NaN are both NaN. hx &= 0x7fffffff; // Equivalent to hx |= (lx != 0). hx |= (u_int32_t)(lx|(-lx))>>31; // Difference is negative iff (hx & 0x7ff00000) ==

Does bitwise-or guarantee an evaluation ordering?

由 五迷三道 提交于 2019-12-19 05:11:32
问题 Say I have this code: unsigned int func1(); unsigned int func2(); unsigned int func3(); unsigned int x = func1() | func2() | func3(); Does C++ guarantee that func1() will be called first, then func2(), and then func3()? Or is the compiler allowed to call the functions in any order it feels like? Also, is the compiler allowed to implement a short-circuit optimization here if it wants to? (e.g. if func1() returned ~0, could the compiler decide not to bother calling func2() or func3(), because

bitwise and logical AND/OR in terms of hex result

由 让人想犯罪 __ 提交于 2019-12-13 03:06:12
问题 so if I have x = 0x01 and y = 0xff and I do x & y , I would get 0x01 . If I do x && y do I still get 0x01 , but the computer just says its true if its anything than 0x00 ? My question is are the bits the same after the operation regardless of bit-wise or logical AND/OR, but the computer just interprets the final result differently? In other words are the hex values of the result of & and && (likewise | and ||) operations the same? edit: talking about C here 回答1: The answer depends on the

How does C++ do bitwise “or” operations on negative numbers?

由 放肆的年华 提交于 2019-12-01 17:47:54
问题 When I give to a variable such value: e = 17|-15; , I get -15 as an answer after compiling.I can't understand what arithmetic c++ uses. How does it perform a bit-wise OR operation on negative decimals? 回答1: It's just doing the operation on the binary representations of your numbers. In your case, that appears to be two's complement. 17 -> 00010001 -15 -> 11110001 As you can see, the bitwise OR of those two numbers is still -15 . In your comments above, you indicated that you tried this with

How does C++ do bitwise “or” operations on negative numbers?

由 一曲冷凌霜 提交于 2019-12-01 17:35:50
When I give to a variable such value: e = 17|-15; , I get -15 as an answer after compiling.I can't understand what arithmetic c++ uses. How does it perform a bit-wise OR operation on negative decimals? Carl Norum It's just doing the operation on the binary representations of your numbers. In your case, that appears to be two's complement . 17 -> 00010001 -15 -> 11110001 As you can see, the bitwise OR of those two numbers is still -15 . In your comments above, you indicated that you tried this with the two's complement representations, but you must have done something wrong. Here's the step by
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