Understanding binary conversion implementation in C++ with bit operations

Question

I am interested in understanding the implementation of converting decimal to binary. Could somebody clarify the purpose of using left-shift and Right-shift in the following code?

void static inline unsignedToBinary(unsigned x, char*& bin)
{
  bin = (char*) malloc(33);

  int p = 0;
  for (unsigned i = (1 << 31); i > 0; i >>= 1)
    bin[p++] = ((x&i) == i) ? '1' : '0';
  bin[p] = '\0';
}

Answer 1

This is a straightforward implementation of binary conversion that uses bit operations.

• Variable i represents the mask - an int containing 2^k value, where k is the position of the bit.
• The initial value is 2³¹, produced with left-shifting 1 by 31.
• for loop uses >>= 1 to right-shift the mask until 1 gets shifted out of it, making i == 0.
• At each iteration x&i is compared to i. The comparison succeeds when x contains 1 in the position where i has its 1; it fails otherwise.

Note: Although using malloc in C++ is certainly allowed, it is not ideal. If you want to stay with C strings, use new char[33] instead. A more C++-like approach would be using std::string.