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?

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.