Re: Initial values of File scoped and Block level variables

In article <twWvf.6437$M%4.1213@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
Joseph Dionne <jdionne@xxxxxxxxxxx> wrote:
>For uninitialized global variables, the loader(LD) allocates data segment
>memory using calloc() which does the "initialization" to NULL for all
>elements ...

This is not in fact how it is implemented on Unix-like systems with
an "ld" linker or equivalent (gld). There is no call to calloc()
involved. The symbols are simply managed in a table until it is
time to assign locations to them; if they are still marked "bss"
at that point, they typically go into a "bss segment" (whose details
are file-format dependent -- a.out, COFF, and ELF are all at least
slightly different here; and some COFF and ELF formats also offer
a "small data segment" area into which small variables are collected
for register-offset addressing in the runtime image).

All of these details are irrelevant to portable C programs, which
need only concern themselves with the fact that static-duration
objects are initialized, as if by an explicit "= { 0 }". (Note
that braces are allowed for scalars; 0 suffices to set a double to
0.0 or a pointer to NULL; and only the first element of any aggregate
need be initialized to cause the rest to be initialized. Hence,
regardless of the object type to which the letter T is mapped, the
line:

T var = { 0 };

is always valid in C. If T is a function type, any initialization
is invalid, of course.)

>So ...
>
>struct {
> int ii;
> long ll;
> char a[32];
>} gvar;
>
>creates an entry in the disk program image that the compiler emits to disk
>that informs the loader to allocate 1 element of sizeof(gvar) memory.

In general, "gvar" in this case will be in a bss segment, along
with all other bss variables. When you actually run the binary,
the OS does not allocate one small chunk of memory at a time.
Instead, one or more entire pages or segments are allocated (or
marked for zero-fill, on a paging OS) based on the *total* size
given in the bss segment. This is quite OS-dependent, of course,
and assumes that the machine's all-bits-zero is the correct way
to initialize "double"s and pointers and so on.

>Memory "allocated" between braces "{}" is actually stack memory
>which retains the data from the last function to use that stack
>memory. This memory can/will change from call to call being
>altered by the "other" functions that are called in between.

It took me quite a while to figure out what you meant by this. As
stated, this is wrong, because it claims that, in:

void f(void) {
static int x;
...
}

x will be "stack memory". In fact, x is yet another "bss segment"
(or, potentially, small-data-segment) variable. The critical thing
is not the braces, but the fact that x has static storage duration.
If x had automatic storage duration -- which is indeed the default
for block-scope variables -- it would, as you say, typically be
stored on a (usually single, often OS-provided) stack.

Note however that IBM S/370 C compilers allocate "stack frames"
using the same, general-purpose "heap" memory that malloc() and
company use. The system does not provide, nor use, a stack: it
simply builds a LIFO data structure that *acts as* a stack (but is
not a single contiguous area, unlike typical OS-provided stacks).
(The S/370 approach has some big advantages when dealing with
languages that use coroutines.)
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.
.

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