CODEWITHSUNDEEP
c++arrayspointerschar
dimayak
dimayak

Reputation: 143

C++ arrays as pointers question

Why this is possible: 

char buf[10], *pbuf = buf, **ppbuf = &pbuf;

and this isn't:

char buf[10], **ppbuf = &buf;

As I understand, the second line is just a shorthand of the first one.

Upvotes: 5

Views: 232

Answers (4)

CB Bailey
CB Bailey

Reputation: 793109

&buf has the wrong type. &buf is a pointer to an array of 10 char. ppbuf has to be initialized with a pointer to a pointer to a char.

In the first initialization, while buf denotes an array, it decays to a char* pointing to its first element so the initialization is valid.

You also can't do: char buf[10], **ppbuf = &(&buf[0]); because &buf[0] is not an lvalue so you can't take its address.

Upvotes: 4

13Tazer31
13Tazer31

Reputation: 181

It's easy to see it's not the same when you start from:

*pbuf = buf

Since the indirection of pbuf is buf, pbuf contains the address of buf, thus:

pbuf = &buf

So:

**ppbuf != pbuf
**ppbuf != &buf

Upvotes: 1

Dan Breslau
Dan Breslau

Reputation: 11522

They're not equivalent.

*pbuf = buf

That means, "pbuf is a pointer to type char, whose value is the address of buf." Since buf is an array of chars, this works.

**ppbuf = &pbuf

That means, "ppbuf is a pointer to a pointer to type char, whose value is the address of pbuf." Since pbuf is a pointer to type char, this works.

**ppbuf = &buf

This means, "ppbuf is a pointer to a pointer to type char, whose value is the address of buf." Since buf is an array of chars, this fails.

Upvotes: 5

Marcelo Cantos
Marcelo Cantos

Reputation: 186098

buf is an array, not a pointer, so you can't assign its address to a pointer pointer. pbuf is a pointer (to the first element of the array), so you can assign its address to a pointer pointer.

Think of the code this way:

typedef char* char_p;

char buf[10];
char_p *ppbuf = &buf;  // If this was legal...
*ppbuf = NULL;         // what would this mean?

Upvotes: 2

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