Having trouble trying to solve bit level manipulation puzzle in C -

so have use bitwise manipulation solve problem.

should duplicate effect of c expression (x*63), including overflow behavior. examples: multi63(1) = 63    multi63(-1) = -63 available ops: ! ~ & ^ | + << >> 

maybe i'm not understanding being looked im trying different variations , each time result either close needs returned isn't correct.

this playing with. figured if mask x, 1 perimeter, know whether i'm multiplying negative or positive.

int y = x>>31;  return ~(y^x)  

this returns:

 test times63(2147483647[0x7fffffff]) failed... ...gives -2147483648[0x80000000]. should 2147483585[0x7fffffc1] 

and if try return 2147483585[0x7fffffc1] tells me need return -2147483648[0x80000000] i'm confused need return.

as general(!) rule can consider these formulas expression x*n, using shifts , addition/subtraction only:

a: (x << n) + (x << n-1) << + ... + << (x << m)

b: (x << n+1) - (x << m)

n can viewed sequence of 0's , 1's [(0...0)(1...1)(0...0)(1...1)]. have consider runs of 1's bit position n down bit position m, n == m possible.

in case n = 63, 011_1111 in binary. have n = 5 , m = 0. simple example assume x = 2:

using b: (2 << 6) - (2 << 0) == (2*64) - (2*1) == 128 - 2 == 126 (you can try out yourself, works fine.)

just demonstrate process on number, assume n = 55 , x = 2. 55 in binary: 011_0111. time have 2 sequences of 1's. n1 = 5, m1 = 4 , n2 = 2 , m2 = 0.

using b n1/m1: (2 << 6) - (2 << 4) == (2*64) - (2*16) == 128 - 32 == 96

using b n2/m2: (2 << 3) - (2 << 0) == (2*8) - (2*1) == 16 - 2 == 14

adding both results yields 110, desired value.