Allocate memory for a pointer to an array in C

Question

I have a pointer array defined declared as

char (*c)[20]

When allocating memory using malloc

c=malloc(sizeof(char)*20);

or

c=(char*)malloc(sizeof(char)*20);

I get a warning as "Suspicious pointer conversion"

Why?

• Richard J. Ross III: Why would you be using malloc there? c is already initialized because it is an array.
• Jerry Coffin: If you just want a pointer to 20 chars, chances are you just want char *c=malloc(20);.
• Akash: no, using size of (c) gets the pointer size
• Kerrek SB: @RichardJ.RossIII: No, c is just a pointer, and uninitialized.
• pmr: @Alex K. Please try to avoid small typo edits and fix the whole post. I lost my changes and had to redo them. That could have saved everybody some work.

Answer 1

In this declaration

char (*c)[20];

c object has type char (*)[20].

We know that in C malloc return type is void * and that there is an implicit conversion between void * to any object pointer types.

So c = malloc(whatever_integer_expression) is valid in C. If you get a warning, you are probably using a C++ compiler or you are using a C compiler but forgot to include the stdlib.h standard header.

But c = (char*) malloc(whatever_integer_expression) is not valid C because there is no implicit conversion between char * type and char (*)[20] type. The compiler has to (at least) warn.

Answer 2

First of all, make sure you have stdlib.h included.

Secondly, try rewriting it as

c = malloc(sizeof *c);

I suspect you're getting the diagnostic on the second case because char * and char (*)[20] are not compatible types. Don't know why the first case would complain (at compile-time, anyway) unless you don't have stdlib.h included.

edit

Remember that you will have to dereference the pointer before applying the subscript; that is, your expressions will have to be

(*c)[i] = val;
printf("%c", (*c)[j]);

etc.

Alternately you could write c[0][i] in place of (*c)[i], but that's probably more confusing if c isn't supposed to act like a 2-d array.

Answer 3

Because you are defining C as a pointer to a static array of chars, not as a pointer to an array of chars, that is, a pointer to the first char.

Change

char (*c)[20];

for

char * c;

a

• Daniel Fischer: No, c is defined as a pointer to an array of 20 char. Array of pointers to char would be without the parens.

Answer 4

For an array of 20 characters, counting the NUL terminator, you don't need a pointer

char array[20];

for a pointer to char, you don't need an array

char *pointer;

A pointer to char can point to an array

pointer = array;

to part of the array (assuming no 'funny' business with the NUL terminator)

pointer = &array[10];

or to a bunch of bytes allocated with malloc

pointer = malloc(20);

Answer 5

just use as

c=(char(*)[20])malloc(sizeof(char)*20);

Reason:

1, as @ouah said.

2, in K&R style compiler, the above one is valid and expected.