What does “>>=” mean in Linux kernel source code?

问题

in Linux kernel source code, I find below code:

    h++;
    pending >>= 1;

It is part of __do_softirq(void). But what does ">>=" mean? Why isn't it ">>" as I remember? Thanks!

回答1:

It simply does

pending = pending >>1;

In short it divides by 2 an unsigned int.

That's the same construct than +=, /=, etc.

It's not just pending >>1 as you remember because that wouldn't store the result of the shift operation in the variable.

回答2:

It's equivalent to

pending = pending >> 1;

Which bitshifts right the bits in pending. This would have the effect of dividing an unsigned int by 2. >> and << are the bitshift operators, and the combination with = behaves the same way += and /= do.

来源：https://stackoverflow.com/questions/11246466/what-does-mean-in-linux-kernel-source-code