I've been working trying to read some C code and I've found some operators that I don't know:

What's the use of &= and |= operators when used for microprocessors programming?


2 Answers 2


These statements are equivalent:

x = x & 0x01;

x &= 0x01;

It means to perform a bitwise operation with the values on the left and right-hand side, and then assign the result to the variable on the left, so a bit of a short form. If you're not familiar with bitwise operations, I suggest you start getting familiar with those first - the & being a bitwise AND and the | being a bitwise OR.

Hope that helps!

  • 5
    \$\begingroup\$ "A compound assignment of the form E1 op= E2 differs from the simple assignment expression E1 = E1 op (E2) only in that the lvalue E1 is evaluated only once." \$\endgroup\$
    – starblue
    Commented Jun 3, 2011 at 6:38

&= is and equals, |= is or equals. These perform bit-wise operations with the left hand and right hand arguments, and assign the result into the left hand side.

  • \$\begingroup\$ This definition may look a bit weird for someone unfamilliar with the language. It could be interpreted as "(foo == 0 |= 1)" would be a valid condition. \$\endgroup\$
    – JulioC
    Commented Jun 3, 2011 at 4:06
  • \$\begingroup\$ @Júlio Souza - how could you interpret it that way? how would you assign (0|1) into 0? assigning a value into a constant is nonsensical, and should be a strong clue that the construct 0 |= 1 is not valid. \$\endgroup\$
    – JustJeff
    Commented Jun 3, 2011 at 10:17
  • \$\begingroup\$ It's a valid interpretation if you say that the = means is equal to (that's what equals means!). == means equals, = means assign ... to. \$\endgroup\$
    – stevenvh
    Commented Jun 3, 2011 at 11:18
  • \$\begingroup\$ @Stevenvh - either way, (foo == 0) |= 1 doesn't make any sense, and neither does foo == (0 |= 1) \$\endgroup\$
    – JustJeff
    Commented Jun 3, 2011 at 21:11

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