Re: how the following code works?
On Jun 16, 10:38 am, Ravi <ra.ravi....@xxxxxxxxx> wrote:
main() {
float a = 5.375;
char *p;
int i;
p = (char*)&a;
for(i=0;i<=3;i++)
printf("%02x",(unsigned char)p);
}
oops, you had taken liberty over the size of float!
printf("%02x",(unsigned char)p);
Never know what you are trying to do by this statement.
the above code gives the binary representation of a.
No, it doesn't!
Moreover, even if you correct all the mistakes, I don't think you will
get the binary representation, unless CHAR_BIT==1,in which case your
implementation is non-conforming!
You will get only "byte representation".
how does it work?
It doesn't work at all.
.
Relevant Pages
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