bitwise-operators

Parsing webp file header in Kotlin to get its height and width, but getting unexpected results

有些话、适合烂在心里 提交于 2020-11-29 19:27:49
问题 I am trying to read the WebP image header, according to the WebP Container Specification of Extended File Format. fun get24bit(data: ByteArray, index: Int): Int { return ((data[0 + index].toInt()) or (data[1 + index].toInt() shl 8) or (data[2 + index].toInt() shl 16)) } fun get32bit(data: ByteArray, index: Int): Int { return get24bit(data, index) or (data[3 + index].toInt() shl 24) } // data -> File(fileName).readBytes() for testing purpose fun webpExtract(data: ByteArray) { println(String

NAND logical bitwise operation in ARM

天涯浪子 提交于 2020-11-29 08:33:24
问题 Is there a way to perform a bitwise NAND operation on the bits in two registers in ARM7, either with the existing AND, OR and EOR operations or other instructions? 回答1: Sure; AND the two registers and then EOR the result with all 1's (for the negation). 回答2: and then mvn (move not). From GCC explorer int nand(int a, int b) { return ~(a & b); } nand(int, int): and r0, r0, r1 mvn r0, r0 bx lr 来源: https://stackoverflow.com/questions/21207561/nand-logical-bitwise-operation-in-arm

Difference between & and && in C?

半世苍凉 提交于 2020-11-25 02:38:29
问题 What is the difference between & and && in C? My teacher gave me this example: int a = 8; int b = 4; printf("a & b = %d\n", a & b); printf("a && b = %d\n", a && b); Output: a & b = 0; a && b = 1; I'm not sure why this would return true in one scenario and false in another. 回答1: & is bitwise and and && is logical and . The expression x && y will return 1 if both x and y is non-zero, and 0 otherwise. Note that if x is zero, then y will not be evaluated at all. The expression x & y will perform

Difference between & and && in C?

时光怂恿深爱的人放手 提交于 2020-11-25 02:38:24
问题 What is the difference between & and && in C? My teacher gave me this example: int a = 8; int b = 4; printf("a & b = %d\n", a & b); printf("a && b = %d\n", a && b); Output: a & b = 0; a && b = 1; I'm not sure why this would return true in one scenario and false in another. 回答1: & is bitwise and and && is logical and . The expression x && y will return 1 if both x and y is non-zero, and 0 otherwise. Note that if x is zero, then y will not be evaluated at all. The expression x & y will perform