bit-shift

I need to extract specific part (no of bits) of a short data type in C. For Example I have a binary of 52504 as 11001101000 11000 and I want First 6 ( FROM LSB --> MSB i.e 011000 decimal 24) bits and rest of 10 bits ( 11001101000 decimal 820). Similarly I want this function to be too generalized to extract specific no of bits given "start" and "end" (i.e chunks of bits equivalent with some decimal value). I checked other posts, but those were not helpful, as given functions are not too much generalized. I need something that can work for short data type of C. Edit I am having the short array

Basically the questions in the title. I'm looking at the MVC 2 source code: [Flags] public enum HttpVerbs { Get = 1 << 0, Post = 1 << 1, Put = 1 << 2, Delete = 1 << 3, Head = 1 << 4 } and I'm just curious as to what the double left angle brackers << does. That would be the bitwise left shift operator. For each shift left, the value is effectively multiplied by 2. So, for example, writing value << 3 will multiply the value by 8. What it really does internally is move all of the actual bits of the value left one place. So if you have the value 12 (decimal), in binary that is 00001100 ; shifting

This question already has an answer here: Unexpected C/C++ bitwise shift operators outcome 6 answers On Visual Studio compiling following C code , the result is 4 . void main() { int c = 1; c = c<<34;} The assembly code as seen from on Visual Studio disassembly window is shl eax,22h From assembly , its easy to see that we are shifting 34. Since the integer here is 4 bytes , from the results it is obvious that modulo operation was carried out on machine level to make this work as shift by 2. I was wondering if this behavior is standardized across platforms or varies across platforms? Undefined

As the question title reads, assigning 2^31 to a signed and unsigned 32-bit integer variable gives an unexpected result. Here is the short program (in C++ ), which I made to see what's going on: #include <cstdio> using namespace std; int main() { unsigned long long n = 1<<31; long long n2 = 1<<31; // this works as expected printf("%llu\n",n); printf("%lld\n",n2); printf("size of ULL: %d, size of LL: %d\n", sizeof(unsigned long long), sizeof(long long) ); return 0; } Here's the output: MyPC / # c++ test.cpp -o test MyPC / # ./test 18446744071562067968 <- Should be 2^31 right? -2147483648 <-

Lets say that I have an array of 4 32-bit integers which I use to store the 128-bit number How can I perform left and right shift on this 128-bit number? Thanks! Remus Rusanu void shiftl128 ( unsigned int& a, unsigned int& b, unsigned int& c, unsigned int& d, size_t k) { assert (k <= 128); if (k >= 32) // shifting a 32-bit integer by more than 31 bits is "undefined" { a=b; b=c; c=d; d=0; shiftl128(a,b,c,d,k-32); } else { a = (a << k) | (b >> (32-k)); b = (b << k) | (c >> (32-k)); c = (c << k) | (d >> (32-k)); d = (d << k); } } void shiftr128 ( unsigned int& a, unsigned int& b, unsigned int& c,

I have this behavior using Java: int b=16; System.out.println(b<<30); System.out.println(b<<31); System.out.println(b<<32); System.out.println(b<<33); output: 0 0 16 32 Is java bit shift circular? IF not, why I get 0 when b<<30 and 16 when b<<32? Bit shifting is not circular; for bit-shifting int s, Java only uses the 5 least-significant bits, so that (b << 0) is equivalent to (b << 32) (is equivalent to (b << 64) , etc.). You can simply take the bit-shifting amount and take the remainder when dividing by 32. Something similar occurs for bit-shifting long s, where Java only uses the 6 least