Question

Very simple questions guys, but maybe I’m just forgetting something.
In 64bit linux, a long is 8bytes correct?
If that’s the case, and I want to set the 64th bit, I can do the following:

unsigned long num = 1<<63;

Whenever I compile this, however, it gives me an error saying that I’m left shifting by more than the width.
Also, if I wanted to take the first 32bits of a long type (without sign extension), can I do:

num = num&0xFFFFFFFF;

or what about:

num = (int)(num);

Thank you.

Answer 1

In 64bit linux, a long is 8bytes correct?

Need not be. Depends on the compiler than on the underlying OS. Check this for a nice discussion.
What decides the sizeof an integer?

Whenever I compile this, however, it gives me an error saying that I’m
left shifting by more than the width

Everyone have already answered this. Use 1UL

Also, if I wanted to take the first 32bits of a long type (without
sign extension), can I do:

num = num&0xFFFFFFFF;
or what about:

num = (int)(num);

num = num&0xFFFFFFFF. This will give you the lower 32-bits. But note that if long is just 4 bytes on your system then you are getting the entire number. Coming to the sign extension part, if you’ve used a long and not unsigned long then you cannot do away with the sign extended bits. For example, -1 is represented as all ones, right from the 0th bit. How will you avoid these ones by masking?

num = (int)(num) will give you the lower 32-bits but compiler might through a Overflow Exception warning if num does not fit into an int