Question

 char 
 char* c[30];
    c = (char*) malloc(30*sizeof(char) );

How does this give an incompatible declaration in built in function warning and and incompatible types in assignment error in the line where i have declared malloc . According to the syntax of malloc , i shouldnt have any error

Answer 1

You have declared c as an array. You can’t change the value of an array, and an array is not a pointer.

c has the type (char*)[30](i.e. an array of 30 char pointers) , not char* as your cast suggests.

If you’re trying to create a dynamically allocated char “array” for 30 chars, use

char *c = malloc(30);

If you really want an array of 30 char pointers, and e.g. allocate space for 30 chars in each of its elements, use

int i;
for(i = 0; i < sizeof c/sizeof c[0]; i++) {
  c[i] = malloc(30);
  if(c[i] == NULL) {
    //handle error
   }
}