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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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    Between consecutive "sequence points", an object's value can be modified only once by an expression. The only "sequence point" in your line of code is the ";". This means x is modified twice. Once by the =, and once by the ++. This code has no well defined meaning. It is not surprising that different implementations gave different answers.

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    Between consecutive "sequence points", an object's value can be modified only once by an expression. The only "sequence point" in your line of code is the ";". This means x is modified twice. Once by the =, and once by the ++. This code has no well defined meaning. It is not surprising that different implementations gave different answers.

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