hi,
i hope someone can explain how that would be done. i want to wrap a call to int _cprintf(const char * restrict format, ...); in to a dll that calls it.
ex.:
_declspec(dllexport) int __stdcall conprint(const char * restrict format, ...) {
return _cprintf(format, ...); <------ ?????????? how do i call this function???
}
any help would be greatly appreciated.
thanks
henry
In MyDLL.c ...
_declspec(dllexport) int _stdcall function(...
In MyDLL.h...
#pragma lib "MyDLL.lib" // autocreated library header for mydll
_declspec(dllimport) int _stdcall function(...
In your code...
#include "MyDLL.h"
x = function(...
thanx for you reply.
but what i ment, was how does the call inside the function look like.
return _cprintf(format, ...); <------ ?????????? how do i call this function???
the function header gives
_declspec(dllexport) int __stdcall conprint(const char * restrict format, ...) {
and how would i call return _cprintf(format, ...); <------ ?????????? how do i call this function??? inside the function?
maybe you can give me another helpful hint.
henry
The inner one it's just a normal function call that is linked into the DLL when you compile it, just like in an EXE.
Where you might be having a problem is that you have a function with a variable number of parameters inside a function with a variable number of parameters... I've never had to do that one, so maybe one of the others can offer a bit of advice on how to do it.
yes, but how do i call it.
i don't know the parmeters that are given to _declspec(dllexport) int __stdcall conprint(const char * restrict format, ...).
ex.: conprint("%s %d %s %s", a,b,c,d). how do i give the paramets to return _cprintf(format, ...)?
We were writing at the same time...
As I said, I've never had to do that one, so maybe one of the others can help...
thank you very much anyway. i hope someone can help me.
Two thoughts...
1) Just us _cprintf() directly without the DLL wrapper.
2) You can use the standard printf() with no muss or fuss.
Is there some reason to write such a thinly wrapped function?
Quote
1) Just us _cprintf() directly without the DLL wrapper.
2) You can use the standard printf() with no muss or fuss.
Is there some reason to write such a thinly wrapped function?
unfortunately i need to export this function to another compiler/language and i need the console-output, that the other compiler doesn't have. the library (dll) is a cbtree/isam that i want to test with that compiler. testing it with pelles c has certain problems, such as somefunction(char *x, char *y). on return, x and y are not modified as they should, they retain the old values.
You can't do that. C doesn't allow it. You can use vcprintf though, which lets you forward the arguments using a va_list.
#include <stdio.h>
#include <stdarg.h>
int __cdecl conprint(const char *fmt, ...)
{
va_list args;
int ret = 0;
va_start(args, fmt);
ret = _vcprintf(fmt, args);
va_end(args);
return ret;
}
Note that your function has to be __cdecl instead of stdcall because your code doesn't know how many arguments its going to be called with (the number and size of arguments is required for stdcall)
Quote from: rkorko on August 14, 2012, 10:10:40 AM
testing it with pelles c has certain problems, such as somefunction(char *x, char *y). on return, x and y are not modified as they should, they retain the old values.
I can see where it might be a problem in your case, but generally this is a good thing... In C variables have a "scope" marked by { } in which you can affect them, outside of their scope variables with the same names are not affected. For example:
#include <stdio.h>
int func(int x)
{
x += 3;
printf("%d\n",x); //prints 13
return x;
}
int main (void)
{
int x = 10;
func(x);
printf("%d\n",x); //prints 10
x = func(x);
printf("%d\n",x); // prints 13
return 0;
}
Passing in pointers such as *x will let you modify the variable in the calling scope, but not the pointer. For example...
#include <stdio.h>
void func(int *x)
{
*x += 3;
printf("%d\n",x); //prints 13
}
int main (void)
{
int x = 10;
func(&x);
printf("%d\n",x); //prints 13
return 0;
}
The reason this happens is that C does not know Pass By Reference. Everything is Pass By Value. When you call a C function you are passing in a copy of the variable, not the variable itself. Even with pointers you are passing in a copy of the pointer... If you want to change the value of the pointer itself you need to pass in the address of the pointer; a pointer to a pointer... **x ...
In C this is generally a good thing since it prevents same-name variables inside a function from accidentally affecting the overall program...
(Thanks aardvajk :D )
yes - i understand now. thank you.
do you have a solution for the problem with calling a variable parameter function?