Charles Mills wrote:
I wonder how C passes a potentially 16-byte string by value.
Not sure what you mean by that question... a C string constant
is an array of characters; otherwise a C 'string' is typically
a pointer to a character or a declared array of characters.
When you pass an array as a parameter; it is converted to
a pointer to the first element and the pointer is passed.
When you pass a pointer - a pointer is passed.
For example:
myfunc("A string");
invokes the function 'myfunc' with the address of the array of
nine characters defined as { 'A', ' ', 's', 't', 'r', 'i', 'n', 'g', '\0' }.
And - myfunc would typically be declared/defined as
void myfunc(char *);
Note that if you declare myfunc this way:
void myfunc(char [9]);
because you are declaring an array as the parameter, the parameter
value is treated as a pointer. That is, even with that declaration, C would
pass a pointer to the first element of the array.
Why is this?
Because of the C rule that an array name is converted to the address of
the first element of the array, and that subscripting operations then
devolve
into pointer arithmetic.
For example:
int array[10];
array[5] = 5;
The expression 'array[5]' is completely the same as
(&array[0] + 5)
and that's what the C language actually does...
Note that since [] is actually just an operator in C, you can
code the equivalent:
5[array];
which says "add 5 to the address of the first element of 'array'".
It's a little confusing to write it that way - but that's the same semantics
as "add the address of the first element of 'array' and 5" since pointer
addition is commutative.
(Try it :-) )
The only way to pass a "string by value" would be to
declare an array inside a structure, and pass the structure
by value... which is well-defined. That is, at that point you
are passing a structure, not an array (the structure happens to
contain an array...)
- Dave Rivers -
--
riv...@dignus.com Work: (919) 676-0847
Get your mainframe programming tools at http://www.dignus.com
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