On Thursday, August 14, 2008 at 12:56:45 (-0500) Maurice LeBrun writes:
> On Thursday, August 14, 2008 at 10:18:32 (-0700) Alan W. Irwin writes:
> > On 2008-08-14 09:58-0500 Maurice LeBrun wrote:
> >
> > > For C & C++, a const pointer means the memory it points to cannot be
> altered.
> > > The pointer can. So for example
> > >
> > > const char *foo = "bar";
> > >
> > > (or later reassignment) is perfectly legit.
> >
> > After years of dabbling at C, this pointer stuff still makes my head spin
> so
> > could you expand a bit more, Maurice? What actually happens with the
> above
> > statement? After the declaration of foo as a pointer to a const char is
> foo
> > initialized as a pointer to the first character in the const "bar" string?
>
> Correct.
>
> > Or would a better way to look at it be that all character strings like
> "bar"
> > can be interpreted as pointers so the above is simply a pointer
> assignment?
>
> This too is correct. All string literals in a program are stored in the
> executable, as a `strings <program>` will demonstrate. The proper way to
> refer to these is via a const char* since they are considered fixed.
>
> If you want a char* string that's modifiable, you need to define a character
> buffer and copy a string into it. Whenever unsure, just construct a simple
> test and try it (I keep several handy for that purpose). For example, what
> if you assign a literal to a char* and then try to modify it?
>
> char *foo = "0";
> foo[0] = "1";
>
> then gcc complains:
> tst.cc: In function ‘int main(int, char**)’:
> tst.cc:15: error: invalid conversion from ‘const char*’ to ‘char’
Whoa, tripped up in my own example. This is a syntax error because
"1" is a const char *, and foo[0] is a char. This is what I meant
to write:
char *foo = "0123";
foo[0] = '0';
Now gcc doesn't complain, since you're assigning char to char. But what
happens when I run the program?
$ ./tst
Segmentation fault (core dumped)
since I was trying to write into unmodifiable memory.
> because apparently it's smart enough to know that even tho foo is defined as
> char*, it's actually pointing to a const char*. I'm no language lawyer, but
> it would seem that this syntax is supported for legacy reasons.
Forget this part, I was mistaken.
--
Maurice LeBrun
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