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I want to make a LED chaser from right to left in PIC18F4550 microcontroller. Here is a section of the code from a program which I am unable to understand:

LATB=1<<i;  
delay();
LATB=0x00;
delay();
  1. How this work in chasing from right to left?
  2. Why is the delay required?
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  • \$\begingroup\$ Try replacing "less than" with "<". Two of them together make a shift operator. \$\endgroup\$
    – user16324
    Commented Oct 15, 2016 at 10:58
  • \$\begingroup\$ I did some rework on the question and title, not even sure that it's still what OP asks. Rajat, can you check that it's still what you want to know? \$\endgroup\$
    – pipe
    Commented Oct 15, 2016 at 11:28

1 Answer 1

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The operator << is a left-shift. It shifts each bit a number of positions to the left, and fills the missing pieces with zero:

1<<0 equals 0b00000001
1<<1 equals 0b00000010
1<<2 equals 0b00000100
... etc ...

If i increments from 0 to 7 in a loop, you will get a pattern with a bit that appears to move through every position. The actual order of the LEDs will depend on your physical circuit. If you want it to move in a different direction, make i go from 7 to 0 instead.

The delay is simply required to let you see anything at all. Without the delay, it would scroll through every position too fast to see anything at all. A delay in between makes it turn on the LED, wait a little so that you have time to react, and then turn it off again and leave it off for a while.

The last delay is optional. If you remove it, the effect will be slightly different.

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