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C pointer syntax

Name: Anonymous 2007-11-14 12:42

If I declare an array of arrays like this:

const char** p0 = { "##  ", "##  ", "    ", "    " };

I get these warnings:
type_d.h:13: warning: initialization from incompatible pointer type
type_d.h:13: warning: excess elements in scalar initializer
type_d.h:13: warning: (near initialization for ‘p0’)
type_d.h:13: warning: excess elements in scalar initializer
type_d.h:13: warning: (near initialization for ‘p0’)
type_d.h:13: warning: excess elements in scalar initializer
type_d.h:13: warning: (near initialization for ‘p0’)

but if I alter the syntax to

const char* p0[] = { "##  ", "##  ", "    ", "    " };

the warnings are all gone. what's exactly the difference between using * [] and ** ? I really thought they were just the same, just like * and [] or ** and [][].

If I did it like p0[][] = { ... would it be different at all?

Name: Anonymous 2007-11-18 14:17

That these two expressions are equal only in value.
What other contexts are there for those expressions?

type != type *
They still point to the same memory location.

/might/ compile to better code
Explain.

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