Re: pointer as argument and parameter

Eric Sosman wrote:

.... snip ...

If you tell the compiler that `b' is an array of `char*',
the same sort of thing goes on, except that the choice of
which rules to apply is a little different. Most of the
time, when you mention the name of an array you get a pointer
to the array's initial element -- hence the equivalence of
`b' and `&b[0]' in the function call. Since `b' is an
array of `char*', `b[0]' is one of those `char*' elements,
and `&b[0]' is therefore a `char**'.

I disagree. b[0] _is_ a char. It happens to be of the same type
as what is pointed at by a char* pointer. However *b is not a
pointer to a char. Thus "&b[0]" is not a char**.

--
Chuck F (cbfalconer at maineline dot net)
<http://cbfalconer.home.att.net>
Try the download section.



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