Digvijoy Chatterjee writes: > In 5/25/06, Andrew Haley <aph@xxxxxxxxxx> wrote: > > > > You declare fp to be type void (*fptr) (int). > > > > Therefore if you call a function through fp, you have to pass an int. > > This is because of the type of variable fp, not because of A. > > > > You can easily demonstrate this by compiling: > > > > typedef void (*fptr) (int); > > > > int main() > > { > > fptr fp; > > fp(); > > } > > > > > I want the name of the function where the compiler while parsing a .c > > > file associates a function pointer ,to the function which will be > > > called at runtime. > > > > This never happens, and therefore I can't answer your question. > > > > Andrew. > > > Right , > Which is the file then which has code , to compare the erroneous call > fp(); to its declaration as in the typedef Look at convert_arguments in c-typeck.c. > ?where does the compiler know fp() is a function call ? case CPP_OPEN_PAREN: /* Function call. */ c_parser_consume_token (parser); if (c_parser_next_token_is (parser, CPP_CLOSE_PAREN)) exprlist = NULL_TREE; else exprlist = c_parser_expr_list (parser, true); c_parser_skip_until_found (parser, CPP_CLOSE_PAREN, "expected %<)%>"); expr.value = build_function_call (expr.value, exprlist); expr.original_code = ERROR_MARK; break; in c-parser.c. Andrew.