CODEWITHSUNDEEP
carrayspointers
2013Asker
2013Asker

Reputation: 2058

C - Why pointer to pointer isn't the same thing as pointer to array?

If I write:

char string[] = "some string";
char **ptr = &string;
printf("%s\n",*ptr);

It prints nothing and gives a warning: warning: initialization from incompatible pointer type [enabled by default]

Now, if I write the following:

char *string = "another string";
char **ptr = &string;
printf("%s\n",*ptr);

It works all right.

Shouldn't string[] decay to a pointer similar to *string and work? Why doesn't it?

Upvotes: 3

Views: 337

Answers (4)

glglgl
glglgl

Reputation: 91017

The first string is an array. Its address points to this array:

+-+-+-+-+-+-+-+-+-+-+-+
|s|o|m|e| |s|t|r|i|n|g|
+-+-+-+-+-+-+-+-+-+-+-+

The address of this array is a pointer to, well, an array, so a *char[].

The second string is a pointer pointing to the same string which lies (probably, but depending on the implementation) in read only space:

+------+    +-+-+-+-+-+-+-+-+-+-+-+
| addr | -> |s|o|m|e| |s|t|r|i|n|g|
+------+    +-+-+-+-+-+-+-+-+-+-+-+

The address of this pointer is a **char.

Upvotes: 0

Nobilis
Nobilis

Reputation: 7448

Actually if I can add to Timothy's answer, the char (*ptr)[12] = &string; type of pointer is fairly unusual as it is a pointer to an array of particular type and size.

Let's see what happens when we print its address before and after we've incremented it by one:

char string[] = "some string";
char (*ptr)[12] = &string;
printf("Before: %p\n", (void*)(ptr));
printf("After:  %p\n", (void*)(ptr+1));

Which prints:

Before: 0xbfec44e0
After:  0xbfec44ec

Notice how in the second case the pointer has been moved 12 blocks of memory. Since the pointer knows the size of the type it points to this applies in the case of an array as well, not just for any plain old type. Which in our case is 12.

There's an indepth explanation of this here, that you migh want to read.

Upvotes: 0

Timothy Jones
Timothy Jones

Reputation: 22125

That's not how the decay works in in this case. See the C faq:

6.12 Q: Since array references decay into pointers, if arr is an array, what's the difference between arr and &arr?

A: In Standard C, &arr yields a pointer, of type pointer-to-array-of-T, to the entire array.

So, when you do:

char string[] = "some string";
char **ptr = &string;

The assignment fails, because &string is of type "pointer to char array of length 12". You could instead write:

char (*ptr)[12] = &string;

Which (while almost certainly not what you're trying to do) reads "declare ptr as a pointer to an array of 12 chars"

If you really want to get a pointer-to-a-pointer, then you could always use an intermediary variable:

char string[] = "some string";
char *ptr = string;
char **doublepointer = &ptr;
printf("%s",*doublepointer);

Upvotes: 9

Santhosh Pai
Santhosh Pai

Reputation: 2625

There's an concept by name pointer to an array.You are using an pointer to point to an array and that's not the way to do it . If you want to point an pointer to an array try the below way you would get the required output .

char string[] = "some string";
char (*ptr)[]=&string;
printf("%s",*ptr);

Upvotes: 4

Related Questions