What does the ">>=" operator do in Solidity?

Question

I've seen this operator used in one of Mikhail Vladimirov's articles:

function mostSignificantBit(uint x) public pure returns (uint r) {
    require (x > 0);
    if (x >= 2**128) { x >>= 128; r += 128; }
    if (x >= 2**64) { x >>= 64; r += 64; }
    if (x >= 2**32) { x >>= 32; r += 32; }
    if (x >= 2**16) { x >>= 16; r += 16; }
    if (x >= 2**8) { x >>= 8; r += 8; }
    if (x >= 2**4) { x >>= 4; r += 4; }
    if (x >= 2**2) { x >>= 2; r += 2; }
    if (x >= 2**1) { x >>= 1; r += 1; }
}

What is the difference between the normal shift operator >> and >>=?

Answer 1

>> and << operators are the shift operators.

From Solidity docs:

The result of a shift operation has the type of the left operand, truncating the result to match the type. Right operand must be unsigned type. Trying to shift by signed type will produce a compilation error.

• For positive and negative x values, x << y is equivalent to x * 2**y.
• For positive x values, x >> y is equivalent to x / 2**y.
• For negative x values, x >> y is equivalent to (x + 1) / 2**y - 1 (which is the same as dividing x by 2**y while rounding down towards negative infinity).

But, >>= and <<= operators work like += or -=.

• x >>= 1 is equivalent to x = x >> 1.
• x <<= 1 is equivalent to x = x << 1.