0

can anyone help me with following bit-wise manipulation.thanks!

```
/*
* copyLSB - set all bits of result to least significant bit of x
* Example: copyLSB(5) = 0xFFFFFFFF, copyLSB(6) = 0x00000000
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 5
* Rating: 2
*/
int copyLSB(int x) {
}
*
* logicalShift - shift x to the right by n, using a logical shift
* Can assume that 1 <= n <= 31
* Examples: logicalShift(0x87654321,4) = 0x08765432
* Legal ops: ~ & ^ | + << >>
* Max ops: 16
* Rating: 3
*/
int logicalShift(int x, int n) {
return 2;
}
/*
* bitCount - returns count of number of 1's in word
* Examples: bitCount(5) = 2, bitCount(7) = 3
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 40
* Rating: 4
*/
int bitCount(int x) {
return(x>>1);
}
/*
* bang - Compute !x without using !
* Examples: bang(3) = 0, bang(0) = 1
* Legal ops: ~ & ^ | + << >>
* Max ops: 12
* Rating: 4
*/
int bang(int x) {
return 2;
}
/*
* leastBitPos - return a mask that marks the position of the
* least significant 1 bit. If x == 0, return 0
* Example: leastBitPos(96) = 0x20
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 6
* Rating: 4
*/
int leastBitPos(int x) {
return 2;
}
/*
* TMax - return maximum two's complement integer
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 4
* Rating: 1
*/
int tmax(void) {
return 2;
}
/*
* isNonNegative - return 1 if x >= 0, return 0 otherwise
* Example: isNonNegative(-1) = 0. isNonNegative(0) = 1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 6
* Rating: 3
*/
int isNonNegative(int x) {
return 2;
}
/*
* isGreater - if x > y then return 1, else return 0
* Example: isGreater(4,5) = 0, isGreater(5,4) = 1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 24
* Rating: 3
*/
int isGreater(int x, int y) {
return 2;
}
/*
* divpwr2 - Compute x/(2^n), for 0 <= n <= 30
* Round toward zero
* Examples: divpwr2(15,1) = 7, divpwr2(-33,4) = -2
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 15
* Rating: 2
*/
int divpwr2(int x, int n) {
return 2;
}
/*
* abs - absolute value of x (except returns TMin for TMin)
* Example: abs(-1) = 1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 10
* Rating: 4
*/
int abs(int x) {
return 2;
}
/*
* addOK - Determine if can compute x+y without overflow
* Example: addOK(0x80000000,0x80000000) = 0,
* addOK(0x80000000,0x70000000) = 1,
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 20
* Rating: 3
*/
int addOK(int x, int y) {
return 2;
}
```

*Edited 3 Years Ago by mike_2000_17*: Fixed formatting