encountered strange type in c++

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What does this mean?

C#

char const *  pWhiteList[] =
{
    "str1",
    "str2",
};

I know what char * means.

I also know what char someVal[50] means.

But I have never seen them defined together, like above.

Any explanations welcome.

PS. I think similar construct one encounters in main in C++, e.g.,

C++

int _tmain(int argc, _TCHAR* argv[])

Again above we have this combination of char* and []:

C++

_TCHAR* argv[]

.
So what does this mean?

Posted 9-Oct-13 4:40am

_coder

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4 solutions

Solution 3

Adding one more...

I think you may be confused by the layout, rewrite it like this and see if it makes more sense:

C++

char const *  pWhiteList[] = { "str1", "str2" };

For some reason they have it laid out like they would if declaring a struct or typedef, but its just a const pointer array to a couple strings.

Posted 9-Oct-13 5:26am

Ron Beyer

Comments

_coder 17-Oct-13 4:13am

so pWhiteList[0] is actually a pointer right?

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Richard MacCutchan · Accepted Answer · 2013-10-09T05:23:00

Solution 2

C++

char const *  pWhiteList[] =

Reading right to left:

pWhiteList[] : is an array
*            : of pointers
const        : to constant
char         : characters

Posted 9-Oct-13 5:23am

Richard MacCutchan

OriginalGriff · Accepted Answer · 2013-10-09T05:03:00

Solution 1

It declares an array of constant pointer-to-characters and initialises it.
In other words it creates an array of two strings, and assigns it to pWhiteList.

Posted 9-Oct-13 5:03am

OriginalGriff

Comments

Philippe Mori 11-Oct-13 20:42pm

In fact, a pointer to constant char. To be a const pointer, const would have to be after the *.

Andreas Gieriet · Accepted Answer · 2013-10-09T07:07:00

Reading C/C++ declaration may be hard.
I had my troubles too until lately I learned to read these reliably.

Start at the entity name (e.g. x) and do the following:

"declare x as "
for all the directly following (...) and [...] say "function (...) returning " and "array [...] of " respectively
for all directly prefixing pointer operators (e.g. *) say "pointer to " (add const, volatile and other pointer operators)
if the above was nested (e.g. (*x)) continue a 2.
what remains is the type, so you say "type"

E.g.

`int x`	declare x as int
`int *x`	declare x as pointer to int
`const char *x[4]`	declare x as array [4] of pointer to const char
`const char (*x)[4]`	declare x as pointer to array [4] of const char
`long *f(double)`	declare x as function (double) returning pointer to long
`long (*f)(double)`	declare x as pointer to function (double) returning long

Cheers
Andi