uint16_t and uint32_t question!

Hi,
I am getting confused between uint16 and uint32.
uint16 has to 2 bytes[16 bits] and uint32 has 4 bytes[32 bits].

In my problem, uint16t value= 1112 , when i convert 1112 into binary i get 10001011000 [only 12 bits are present].

So how does the structure of 16 bits of 1112 look?

Second question
uint16t value= 1112

How does (uint32_t)value look? [32 bits representation of 1112]

Any pointers would be helpful

Regadrs
Niran

> So how does the structure of 16 bits of 1112 look?
It has 4 leading zero bits, then 10001011000

> How does (uint32_t)value look? [32 bits representation of 1112]
It has 20 leading zero bits, then 10001011000

Quote:

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Originally Posted by NiranjanRavi199

uint16 has to 2 bytes[16 bits] and uint32 has 4 bytes[32 bits].

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You should be aware that while it is conventional to say that a byte is 8 bits, in C the number of bits in a byte is denoted by CHAR_BIT, and while it must be at least 8, it could be larger.

Quote:

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Originally Posted by NiranjanRavi199

So how does the structure of 16 bits of 1112 look?

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You're dealing with unsigned integers so that's trivial: fill in the remaining leading bits with 0s.

Quote:

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Any pointers would be helpful

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Code:

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int *foo;

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I couldn't help myself...

Quote:

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Originally Posted by click_here

Code:

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int *foo;

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i couldn't help myself...

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bwahahahahaha! :)

Thank you for your help