should a right-shift on a signed int16 do sign extension?

[truevenik]

Hi the following code doesn't do sign extension. Am I missing something or is the compiler missing something?

signed int16 result = -25152;

result >>= 4;

When I replace result >>=4; with result = result/16; it starts doing what I want...


Thanks
-Ben

[Eric Minbiole]

According to ISO/ANSI C, when shifting signed numbers it's up to the compiler to decide whether to do sign extension, or to just shift in 0's. CCS likely chose the simpler '0' approach to keep code size to a minimum.

You might be able to do something like

[rnielsen]

When you do a shift >> ALL of the bits will be moved and zero's will be put in their place. This includes the sign bit. It will be moved four places and a zero will be in it's original place. This will have a drastic effect on your variable's value. Hence, -25152 (1001110111000000) if shifted four places will be 2524 (0000100111011100). You can't simply shift bits to get a divide-by result. You need to remember that each bit doubles the last bit's value (1, 2, 4, 8, 16, 32, 64....).

Ronald

[Guest]

thanks guys - that clears it up.