invalid converstion from void* to char**

Question

it has been a while since I messed with C code.

I am getting the following error when compiling C code under ubuntu using gcc.

command i am using to compile code is(if these errors are because of comiler I am using , please let me know how to make that go away):

gcc -o runnable mycode.C 

error: invalid conversion from ‘void*’ to ‘char**’

line 39 is :

sequence=malloc(sizeof(char *)*seqNum);

sequence is declared as:

char **sequence;

seqNum is declared as:

int seqNum

Answer 1

You need to cast the result of malloc to be the type you want.

So:

 char **sequence;
 ...
 sequence = (char **)malloc(sizeof(char *) * seqNum);

Also remember that if you're going to use sequence you'll need to allocate a list of "char *"s as you have done but then you don't have the pointed-to memory allocated, it's only allocated the space for the list of pointers themselves.

Part of the reason for this being an error is that assigning between different pointer types can change the alignment required. Malloc is guaranteed to return a pointer to space with alignment suitable for any type.

Answer 2

Added: The fasted solution to Arron's actual problem is provided by sgm in a comment. The text below is all accurate, and hopefully helpful, but a second rate solution to the problem at hand.

---

Your compiler is being very stiffnecked about pointer casts (are you using a c++ compiler?), adding an explicit cast like

sequence=(char**)malloc(sizeof(char *)*seqNum);

should make the error go away. Alternately you might be able to convince the compiler to go easy on you with some kind of option like

$(CC) --lighten-up-baby code.c

which might be preferable if this is in some third party code that you don't really want to hack. Read your compiler documentation to find the option you want. Since all the gccs I have on hand (versions 4.0 and 4.2) are happy with that code, I'm not in a good place to offer advice on switches to turn this behavior off.