Thread: what does this pointer type conversion do?

Hi All,
I am trying to understand code snippet from some open source compiler project. But I did not get this part when it calls a function lookup(_, _):

function lookup() is expecting a character array for its first argument, and uses this character array to find a match in some linked data structure (e.g symbol table)

Code:

struct symtab *
lookup(char* key, int stype)
{
	struct symtab *sym;

          ..........
		for (sym = tmpsyms[type]; sym; sym = sym->snext)
			if (sym->sname == key)  // just compare address
				return sym;
}

struct	symtab {
	struct	symtab *snext;	/* link to other symbols in the same scope */
	int	soffset;	/* offset or value */
	char	sclass;		/* storage class */
	char	slevel;		/* scope level */
	unsigned int	sflags;		/* flags, see below */
	char	*sname;		/* Symbol name */
	char	*soname;	
	unsigned int	stype;		
	unsigned int	squal;		
        char*   def_file;
	int     lineno;			
};

However, this lookup() function was invoked sometimes in a confusing way:

Code:

struct symtab* x = ..;
struct symtab* sym=lookup((char*) x, 10);

and this invocation of lookup() seems to be equivalent and interchangeable to the following:

Code:

sym = lookup(x->sname, 10);

which makes more sense to me. How and why the former one works?
Any ideas?

Thanks.

tony

You should look into how they insert into symbol table.

Originally Posted by MK27

Why do you think it would not?

"key" will be the first address stored in the submitted struct in this case.

shouldn't it be 'snext' rather than 'sname', given the definition of struct symtab?