Getting "conflicting types for function" in C, why?

Question

I'm using the below code:

char dest[5]; char src[5] = "test";  printf("String: %s\n", do_something(dest, src));  char *do_something(char *dest, const char *src) {     return dest; } 

The implementation of do_something is not important here. When I try to compile the above I get these two exception:

error: conflicting types for 'do_something' (at the printf call)
error: previous implicit declaration of 'do_something' was here (at the prototype line)

Why?

Answer 1

You are trying to call do_something before you declare it. You need to add a function prototype before your printf line:

char* do_something(char*, const char*); 

Or you need to move the function definition above the printf line. You can't use a function before it is declared.

Answer 2

In "classic" C language (C89/90) when you call an undeclared function, C assumes that it returns an int and also attempts to derive the types of its parameters from the types of the actual arguments (no, it doesn't assume that it has no parameters, as someone suggested before).

In your specific example the compiler would look at do_something(dest, src) call and implicitly derive a declaration for do_something. The latter would look as follows

int do_something(char *, char *) 

However, later in the code you explicitly declare do_something as

char *do_something(char *, const char *) 

As you can see, these declarations are different from each other. This is what the compiler doesn't like.