Hi Doug, you should consult a C-book first, you will definitely find it.
Its the first one. Look: struct foo *bar[10]; So 'bar' is an array of 10 pointers to 'struct foo'. sizeof(bar) will give you 40 on an 32-bit machine, and 80 on an 64-bit machine (assuming that a pointer on an 32-machine is 4 bytes, and on an 64-bit machine its 8 bytes). memset(bar, 0, sizeof(*bar)); If you use the name of an array in an expression, the compiler will automatically convert it into a pointer (to the first element). So 'bar' is now a pointer, and *bar will give you the size of a pointer to 'struct foo' (which is the first element and which will be 4 or 8 bytes). But keep in mind that 'sizeof' is a compile-time operator. If you have a pointer like what-ever-type *foo; then sizeof(*foo); will always give you the size of 'what-ever-type'. That means that sizeof(**bar); will give you the size of 'struct foo'. A last comment: better not use 'memset' here. The reason is that a NULL-pointer isn't something which has a 0-bytes-pattern in memory. Its only the case that most C-implementations do it that way. But a C-implementation is free to use whatever it likes to represent a NULL-pointer internally. So while 'memset' will work on some platforms (and compilers), on other platforms you will have unintialized pointers, and not NULL-pointers. Hope this helps. On Thu, Nov 10, 2011 at 10:39:29AM +0800, Doug Brewer wrote: > Hello, > > I have the following lines in C: > > struct foo *bar[10]; > > If I want to clear out that structure, should I use > > memset(bar, 0, sizeof(bar)); > or > memset(bar, 0, sizeof(*bar)); > > Thank you.
