I want convert an int into 2 bytes representing that int. Probably must use bitwise and bit shifting, but I dont know what to do.
I need to shift an unsigned int to the right more than 32 times and still get a proper answer of zero instead of the random or original number. E.g 8 >> 40 should = 0 but it returns a random number.
I was looking through some of the .net source yesterday and saw several implementations of GetHashcode with something along the lines of this:
I am confused on the Bit Shift Operator开发者_运维知识库s (in c#). Can you shed light on why the value of \"a\" below returns \"1\" instead of \"4294967296\":
Lets say I got this for example (from java obfuscation) with a highly overflowed shift value x = buffer[count + -3] << 0x8f553768 & 0xff00
I am converting some Java code to Javascript to run on node.js, and I ran into something peculiar with bit shifting.
I would like to perform the division of num by 60 which is not power of two using right shift operation. How do I do this?
C++03 standard tells us that the result of applying the bitwise shift operators to signed types can be UB and Impl. defined for ne开发者_JAVA技巧gative values. My question is following: why for operat
Here is one program #include<stdio.h> #include<stdlib.h> int main() { un开发者_运维问答signed char a=0x80;
How does a computer know that (int x, y) x << y means shift over y bits? I don\'t mean the shift part. I mean the y part. Does the computer shift x by one and subtract one from y until y == 0? I
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