Converting bit position to mask value

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I want to calculate the value of the mask dynamically based on bit position.

For Example: The mask value for the 17th bit in a 32-bit value is 0x00020000 and that of the 18th bit is 0x00040000. So if I know the bit positions like 17, 18 etc., how can this be converted dynamically to mask values in C? Of course left shifting is one method (1<<17 or 1<<18). But I am thinking left shifting may consume too many instructions! Or is left shifting itself the best and efficient method?

c
embedded
asked on Stack Overflow Mar 29, 2015 by Mahesha Padyana • edited Mar 29, 2015 by emlai

2 Answers

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Basically, it won't consume too many instructions. What you are doing may be fine in one operation. Here is an answer that explains this on SO: Which is faster: x<<1 or x<<10? Basically there is no difference in speed.

Please see this answer on SO too for implementations: How do you set, clear, and toggle a single bit?

answered on Stack Overflow Mar 29, 2015 by Drakes • edited May 23, 2017 by Community
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In most architectures support a single atomic instruction for logical-shift-left. For example:

  • MIPS: SSL
  • ARM: LSL
  • x86: SHL
  • AVR: LSL

PIC microcontrollers are however a more complex story. Different versions of PIC have different instruction sets, and not all include a logical-shift-left. Low-end (prior to PIC24/sdPIC) PIC instruction sets are not particularly suited to C code generation in any case since they lack direct support for many apparently insignificant C operations.

You should check your target's instruction set and the actual assembler/machine code generated by your compiler - most compilers have support for outputting assembler listings, or you can disassemble the output in your debugger or simulator, or a standalone disassembler.

In the unlikely absence of an logical-shift-left or a compiler optimisation that makes best use of the existing instruction set, you could (at no insignificant memory cost) use a look-up table to at least optimise for time, but in most cases a part with no logical-shift-left is likely to also have very small memory resources, so it is perhaps an impractical solution.

answered on Stack Overflow Mar 29, 2015 by Clifford

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