Go Classes C programming Course

Answer 1

Assuming short = 2 bytes, and int = 4 bytes.
-9 is stored in memory in it’s 2’s complement form: 

-9

1111 1111

1111 0111

 

Now for assigning a to iy, unsigned (zero) extension will occur (since the source, ie. a, is of unsigned type.

So iy will look something like this in memory:

this equals 65527

0000 0000

0000 0000

1111 1111

1111 0111

 

Now you should note that any time a char or short type is used in an expression it is first implicitly promoted to int type.

So for first printf:- first a is promoted to int following unsigned extension. And then we have to print it as a decimal integer because of the format specifier. Thus, it will print 65527.

For the second printf, we have to print the same promoted a as an unsigned integer which yields similar results as above: 65527.

And similarly you can reason for the next two printf.

Comments

Yes got it, now I understood what mistake I had done.

Thanks a lot.

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No worries.
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