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?
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17\$\begingroup\$ I don't believe those operators are any different between micros and other platforms. \$\endgroup\$– KellenjbJun 2, 2011 at 22:44
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4\$\begingroup\$ For the complete masterclass graphics.stanford.edu/~seander/bithacks.html \$\endgroup\$– Toby JaffeyJun 2, 2011 at 22:47
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\$\begingroup\$ Related: stackoverflow.com/questions/1354138/… \$\endgroup\$– Steve MelnikoffJun 3, 2011 at 8:43
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\$\begingroup\$ this type of question belongs on stackoverflow \$\endgroup\$– Jason SMar 5, 2012 at 13:58
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2 Answers
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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!
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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\$– starblueJun 3, 2011 at 6:38
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&= 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.
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\$\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\$– JulioCJun 3, 2011 at 4:06
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\$\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\$– JustJeffJun 3, 2011 at 10:17
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\$\begingroup\$ It's a valid interpretation if you say that the
=meansis equal to(that's whatequalsmeans!).==meansequals,=meansassign ... to. \$\endgroup\$– stevenvhJun 3, 2011 at 11:18 -
\$\begingroup\$ @Stevenvh - either way, (foo == 0) |= 1 doesn't make any sense, and neither does foo == (0 |= 1) \$\endgroup\$– JustJeffJun 3, 2011 at 21:11