HELP: uint16_t and >> operator used in following program

what does "uint8_t" or "uint16_t" mean?
Can you tel me output of following statement if you can then please explain ?

TWI_SLA_CAM=0xC0
uint8_t sla;
uint16_t eeaddr=0x12;
sla = TWI_SLA_CAM |(((eeaddr >> 8) & 0x07) << 1);

uint8_t is unsigned type containing 8 bits (on most compilers unsigned char will do)
uint16_t is unsigned type containing 16 bits - on most compilers unsigned short will do the work

What do YOU think will be the output?

i am not able to get wat
"eeaddr>> 8"
do ,bcoz eeaddr =0x12 even after being uint16_t

http://www.lmgtfy.com/?q=operator+%3E%3E+C

I am unable to figure out how a uint16_t eeaddr be shifted to right by 8 bits, becoz the value stored in eeaddr is 0x12, thus shifting it by 8 bits gives 0.
and finally gives sla=0xc0; that is TWI_SLA_CAM

Quote:

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Originally Posted by pvr_3

I am unable to figure out how a uint16_t eeaddr be shifted to right by 8 bits, becoz the value stored in eeaddr is 0x12, thus shifting it by 8 bits gives 0.
and finally gives sla=0xc0; that is TWI_SLA_CAM

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So? What is the problem with this? Seems to be correct analisys.