C++ declarations (arrays and pointers)

Re: C++ declarations (arrays and pointers)


"joeschmoe9 8" <joeschmoe9876@ gmail.comwrote in message
news:0e3b22c7-b4dd-4b04-9398-5933d293e16c@n5 8g2000hsf.googl egroups.com...

>i know that:
>
double *a[n]; is an array of n pointers
>
double (*b)[n]; is pointer to array of size n
>
but what is:
>
double ( *c[n] )( ); and
>
double ( *d( ) )[n];

http://ieng9.ucsd.edu/~cs30x/rt_lt.rule.html

-Mike

Re: C++ declarations (arrays and pointers)

joeschmoe98 wrote:

still confused, at least point me in the right direction. please

When you post something new to a thread, please do not remove the
existing conversation. If re-post your question, please include your
best guess at what the confusing expressions mean.

Re: C++ declarations (arrays and pointers)

still confused, at least point me in the right direction. please

Re: C++ declarations (arrays and pointers)

joeschmoe98 wrote:

i just dont understand what the empty set of parenthesis are for

Please don't snip relevant context.

Empty parentheses in a declaration mean that the thing to their left is
a function.

Re: C++ declarations (arrays and pointers)

i just dont understand what the empty set of parenthesis are for

Re: C++ declarations (arrays and pointers)

joeschmoe98 writes:

i know that:
>
double *a[n]; is an array of n pointers
>
double (*b)[n]; is pointer to array of size n
>
but what is:
>
double ( *c[n] )( ); and
>
double ( *d( ) )[n];

That looks like a question on your homework, which you should try to figure
out on your own.


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i know that:

double *a[n]; is an array of n pointers

double (*b)[n]; is pointer to array of size n

but what is:

double ( *c[n] )( ); and

double ( *d( ) )[n];