I want to write a wrapper for my C function as I want to call it from python code.So my wrapper function is
static PyObject* my_func_wrapper(PyObject *self, PyObject *args)
{
PyObject* obj_Ptr = my_func();
return Py_BuildValue('O', obj_Ptr);
}
My C my_func() has no params, and return MyClass pointer converted to (void *).
When I run my setup.py script I'm getting this
warning: passing argument 1 of ‘Py_BuildValue’ makes pointer from integer without a cast [-Wint-conversion]
Py_BuildValue('O', obj_Ptr);
and when I call my_func_wrapper from python I'm getting Seg.Fault Error in return line.
How can I fix this?
Thanks.
Here is the prototype of Py_BuildValue:
PyObject *Py_BuildValue(char *format, ...);
So the first argument should be "O" and not 'O'.
I'm not sure about you second arg either : you are using my_func() result instead of getting a pointer to this function.
i'm using an array of pointer (which are addresses of functions).
I'm parsing multiple types of data (char *, int, etc) to them.
To avoid any type error i'm using multiples void *.
That's why i'm confused because the compiler says that they are incompatible type (thus, it work when I compile it).
The array prototype is : void *(*arg_handler[4])(void *arg);
I'm using a function called list to save the different addresses and return the function address with the specific arguments:
void *list(int x, void *arg)
{
arg_handler[0] = &my_putstr;
arg_handler[1] = &my_put_printable;
arg_handler[2] = &my_put_nbr;
arg_handler[3] = &my_put_nbr;
return (arg_handler[x](arg));
}
I'm calling list from the main function through:
list(f_type(s[x + 1]), va_arg(args, void *));
but I can't figure out why I got this error:
warning: assignment from incompatible pointer type [-Wincompatible-pointer-types]
for :
arg_handler[0] = &my_putstr;
arg_handler[1] = &my_put_printable;
arg_handler[2] = &my_put_nbr;
arg_handler[3] = &my_put_nbr;
any Idea ?
As alk said the warnings were appearing because the 4 functions where not typed as void *. So instead of making an array for each type I converted the functions as void * as so :
arg_handler[0] = (void *)my_putstr;
arg_handler[1] = (void *)my_put_printable;
arg_handler[2] = (void *)my_put_nbr;
arg_handler[3] = (void *)my_put_nbr;
So I'm doing an assignment where we need to pass functions we've made ourselves into a library provided for us.
Tree * createBinTree(int (*comparePointer) (TreeDataTypePtr data1,
TreeDataTypePtr data2), void (*destroyPointer) (TreeDataTypePtr));
Is the code I've been provided.
My function pointer for the comparePointer is
int(*compStr)(void *, void *) = &compareName;
Compare is
int compareName(void * rest1, void * rest2);
But when I pass it through like so
Tree * myTree = createBinTree((compstr)(str1,str2),die(str1));
I only get an error on compstr which is "passing argument 1 of createBinTree makes pointer from integer without a cast" and "expected 'int (*)(void *, void *) but argument is of type int.
You need to pass the function, not call the function and pass the return value:
Tree* myTree = createBinTree( compstr , die );
But this is still not correct. The types of compstr and die must be compatible with the parameter types of the function createBinTree:
int(*compStr)( TreeDataTypePtr , TreeDataTypePtr ) = &compareName;
in which case, compareName must be a function of the same type. The same goes for die.
Note that only casting function pointers to the compatible type will cause undefined behavior when they are used to call the function. The original type of the function pointer must be compatible with the type used to call the function. This will cause undefined behavior:
int(*compStr)(void *, void *) = &compareName;
Tree* myTree = createBinTree( ( int(*)(TreeDataTypePtr,TreeDataTypePtr ) )compstr , die );
I am confused with function pointer declaration.
I have an api abc() which takes an argument as so:
void abc(void (*my_func)(void *p), int, int)
If I want to pass my function as an argument to that api, I am declaring it in my .h file:
void (*xyz)(void *p)
and defining as:
void *(xyz)(void *p){
statements;
}
but this throws an error. Please correct me.
you just need to declare it:
void xyz(void *p);
with the implementation the same way.
When you pass it into your api, the type system figures out it out automatically:
abc(xyz,someint,anotherint);
The (*xyz) means that it is a function pointer.
Function pointers are best handled with typedefs. So it is guaranteed that there is nothing wrong.
I would do the following:
// define a type for the function (not its pointers, as you can often read)
typedef void my_func_t(void *p);
void abc(my_func_t*, int, int);
// declaration in order to be type-safe - impossible if only the pointer would be typedef'd
my_func_t my_func_impl;
// definition:
void my_func_impl(void *p)
{
do_something_with(p);
}
and then you can call your abc() with abc(my_func_impl, 47, 11). You can put a & before my_func_impl there in order to point out that it is the function address you wish to obtain, but it is optional.
An alternative would be to write
typedef void (*my_func_p)(void *p);
and use my_func_p instead of my_func_t *, but this has the disadvantage that you cannot write my_func_t my_func_impl;.
Why would you want to do that?
Well, if, by any coincidence or accident, the function definition or the typedef is changed, they won't match any longer, but the collision is not declared as error, but only as warning (Mismatch pointer). OTOH, my_func_t my_func_impl; serves as a kind of prototype, which causes a function header mismatch, which is an error.
Simply declare and define your function as you would any other:
void xyz(void *p);
void xyz(void *p){
// ...
}
and pass a pointer to the API:
abc(xyz, 42, 7);
Function names are automatically interpreted as function pointers where appropriate. You can also explicitly take the address of the function, if brevity isn't your thing:
abc(&xyz, 42, 7);
If I understood correctly, you want xyz to be the function that is passed to abc, right?
As the argument my_func indicates, you have a pointer to a function that takes void * as argument and returns void
type (*func_pointer)(type, type, type, .......)
^ ^ ^ ^ ^ ^
| | | | | |
| | | argument types
| | pointer name
| This is a function pointer
return type
Therefore, you need to declare xyz as:
void xyz(void *p);
And the same in implementation:
void xyz(void *p){
statements;
}
What you are doing wrong is that, the line you wrote in the .h file defines a function pointer, named xyz. The function pointer has no value, because you never wrote xyz = some_function;
What you have written in the source file is a function, also with name xyz that takes a void * as input and returns a void *, instead of void which was your intention.
Maybe this helps you get less confused:
When you write int *x;, x is a pointer to int. Then you can have int y; that doesn't have an extra * and write x = &y;.
It's the same with functions. If you have void (*funcptr)(void *p);, then all you need is a function that says void some_func(void *p){} (again without the extra *) and write funcptr = some_func;. You don't need the & since function names are in fact pointer to the function. You could put it to be more explicit though.
The first argument of 'abc' is the pointer of function returning 'void' and having 'void *' as an argument...
So your code should look like:
void
myFunc (void *)
{
// ... my statements
}
...
abc (myFunc, 10, 20);
This works
void abc(void* (*my_func)(void*), int a, int b) {
my_func(0);
}
void *(xyz)(void *p) {}
int main() {
abc(xyz, 0, 0);
return 0;
}
When you write void (*my_func)(void *p) it means pointer to function, that returns void
And void (*my_func)(void *p) it means pointer to function, that returns pointer
Hallo All,
I have this method:
void *readFileLocal(char filename[]){
.....
}
Now i want to start this method a a thread:
char input[strlen(argv[1])];
strcpy(input,argv[1]);
pthread_t read,write;
pthread_create(&read, NULL, &readFileLocal, &input);
But during compilation it gives this warning:
file.c:29: warning: passing argument 3 of ‘pthread_create’ from incompatible pointer type
/usr/include/pthread.h:227: note: expected ‘void * (*)(void *)’ but argument is of type ‘void * (*)(char *)’
How can I parse an char array to my funcation over pthread_create without this warning ?
Thanks for helpt
Just use this:
pthread_create(&read, NULL, readFileLocal, input);
And consider changing your function's signature to:
void *readFileLocal(void *fileName) { }
When you are passing a pointer to function (like the one you are using in readFileLocal parameter) you don't need to put &.
Also, when you have an array (like input in your case) you don't need & since in C arrays can be used as pointers already.
Functions for threads need to be prototyped:
void *func(void *argv);
As with all void pointers you then need to interpret ("cast") the pointer to a meaningful entity. You readFileLocal functions then becomes:
void *readFileLocal(void *argv)
{
char *fname = argv; // Cast to string
// Rest of func
}