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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


Parents
  • 0 in reply to Sergey Dymchishin 

    unsigned char x = 1;
x = (x++) << 3;
    "MSVC result '9' ... can be explained only if suppose that 'x++' evaluated after assigning '8' to x."

    Which is what I'd expect from a post increment:

    "x++" means "take the value of x and then, when you've done with it, increment x by one"

    In this case, "when you've done with it" means, "when you've shifted it"

    However, I agree with the consensus that this is ambiguous and best avoided!

Reply
  • 0 in reply to Sergey Dymchishin 

    unsigned char x = 1;
x = (x++) << 3;
    "MSVC result '9' ... can be explained only if suppose that 'x++' evaluated after assigning '8' to x."

    Which is what I'd expect from a post increment:

    "x++" means "take the value of x and then, when you've done with it, increment x by one"

    In this case, "when you've done with it" means, "when you've shifted it"

    However, I agree with the consensus that this is ambiguous and best avoided!

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