CODEWITHSUNDEEP
c++
user297850
user297850

Reputation: 8005

the functionalities of two lines of code

When I was trying to learn from an existing program, I could not understand what the following two lines of code try to do?

for(i=0;0==(x&1);++i)x>>=1;


if(0==(x-=y)) return y<<i;

Any explanations would be appreciated.

Upvotes: 5

Views: 193

Answers (4)

David Rodr&#237;guez - dribeas
David Rodr&#237;guez - dribeas

Reputation: 208323

Will return x if x is y left shifted an indeterminate i number of positions.

That is, if x = 01010000 and y = 00000101 it will return x. There is no info in the question to guess what it will return if the condition is not met.

Upvotes: 0

Uri
Uri

Reputation: 89729

Is this an interview question?

The << and >> operators and & as well are all bitwise operations.

Superficially, the first one seems to shift right until it finds a 1 bit, but is destructive.

The other one is quite convulted.

However, without more context it is not clear what the program is trying to do.

Upvotes: 0

Eclipse
Eclipse

Reputation: 45493

for(i=0;0==(x&1);++i)x>>=1

Finds the least significant bit set to 1 in an integer

if(0==(x-=y)) return y<<i;

Subtracts y from x, and if the result is 0, returns y shifted over (toward the more significant bits) by i bits.

Upvotes: 9

Justin Ethier
Justin Ethier

Reputation: 134167

for(i=0;0==(x&1);++i)x>>=1;

This code x>>=1 is shifting the bits of x to the right one place. This will continue as long as 0==(x&1) is true, which means that the right-most bit of x is a 0. i is the number of bits shifted.

if(0==(x-=y)) return y<<i;

This code subtracts y from x. Then, if x is 0 the code returns y shifted to the left by i bits.

Upvotes: 4

Related Questions