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C Problem?

Jim Leifker 

unsigned char x=1;
x=(x++)<<3;

Using uVision 6.12
Gives answer of 8
It shifts binary 1 left and doesn't increment.
(even when optimization set to level 0)

Using Micorsoft C++ 6.0
Gives answer of 9
It shifts binary 1 left then increments.


I was a little confused with both answers because I was thought that
the increment would happen first because of the ( ) around x++ then
shift left three giving 16 but that's not the case.

This gives me 16 in both Keil and MS.
x=(++x)<<3; works it gives 16


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  • 0 in reply to Michael Fielding

    Welcome to the wonderful club comprised of those bitten by race conditions. We'll get you your membership card tomorrow.

    In point of fact, there doesn't need to be an index in this expression to cause a problem. Essentially, if you have interrupt-driven code, you must imagine that any instruction that is not "atomic" (i.e. completed without interruption) can be stopped in the middle.

    Thus, something as simple as

    a = b;

    where a and b are 16-bit values and you're using, say, an 8051 can cause problems if they are modified in main code and used in an ISR or vice-versa.

Reply
  • 0 in reply to Michael Fielding

    Welcome to the wonderful club comprised of those bitten by race conditions. We'll get you your membership card tomorrow.

    In point of fact, there doesn't need to be an index in this expression to cause a problem. Essentially, if you have interrupt-driven code, you must imagine that any instruction that is not "atomic" (i.e. completed without interruption) can be stopped in the middle.

    Thus, something as simple as

    a = b;

    where a and b are 16-bit values and you're using, say, an 8051 can cause problems if they are modified in main code and used in an ISR or vice-versa.

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