Amol Mohite wrote:
So when gcc pads it to 8 bytes, can we say for sure that next allocation
might be at 8byte boundary ?
No. You can be sure that it will satisfy any CPU-imposed alignment
requirements, but that's all.
--
Glynn Clements [EMAIL PROTECTED]
Hi!
So when gcc pads it to 8 bytes, can we say for sure that next allocation
might be at 8byte boundary ?
Thanks
I meant something like :
*i = ((int *)c++);
I guess this will work like doing (int *)++c.
Sorry i meesd the brackets.
will be more clear next time.
Thank anyway.
Amol Mohite wrote:
say I have a char *c pointing to an array of 10 bytes.
When i do (int *)c++ (increment before typecast -- forget the brackets for
nw), and equate it t an int *i then
*i will return 2nd to fifth bytes as integer. Is this correct ?
No. `(int *)++c' or `(int *)(c+1)'
On Fri, 26 Mar 1999, Amol Mohite wrote:
# say I have a char *c pointing to an array of 10 bytes.
you mean:
char myarray[10], *c;
c = myarray;
# When i do (int *)c++ (increment before typecast -- forget the brackets for
# nw), and equate it t an int *i then
please re-read what you type.
