Re: segfault w/ block, but not file scope

Chris Torek wrote:

[..]

Mr Dionne appears to believe that C's arrays are an exception, and
are passed by reference.  While C's arrays *are* exceptional, this
is not where the exception occurs.  C remains pass-by-value.  The
gimmick is that the "value" of an array is a pointer to the array's
first element.  In:

    void f(void) {
        char buf[100];
        char *p;

        p = buf;

we attempt to copy the "value" of buf -- an array -- to the pointer
variable p.  The "value" of the array is a pointer to the array's
first element, so this sets p to point to &buf[0].  Likewise, if we
go on to call a function g() and pass the "value" of buf:

        g(buf);
        ...
    }

then g() receives, as its value, a pointer to the array's first
element -- a pointer to buf[0].  Within g(), as for any simulation
of by-reference in C, we have to use the unary "*" operator in
order to write to buf[0]:

    void g(char * ptr) {
        * ptr = 42;
        ...
    }

and if we fail to prefix "ptr" with the unary "*" operator when
assigning, we will "re-bind" it so that it no longer points to
"buf" at all:

        ptr = "oops";

Now *ptr is no longer 42 (in any current character set anyway):
ptr now points to the first element of an anonymous (unnamed) array
of 5 "char"s that are set to {'o','o','p','s','\0'} and must not
be changed (they may or may not actually be read-only, but the
effect of attempting to change them is undefined).


Look who is being humorous. Sir, in you example, "buf" will never change is memory value, even *if* you pass it by reference, "& buf". You might cause a program error, but buf will forever point to the same memory address.


[..]

I think we all can agree the pascal and c++ support pass by reference by different syntax. I assert the c uses yet another syntax to implement pass by reference, nothing more or less.
.

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