C# allows to mark function argument as output only:
void func(out int i)
{
i = 44;
}
Is it possible to do something similar in C/C++? This could improve optimization. Additionally is should silence out warnings "error: 'myVar' may be used uninitialized in this function", when variable is not initialized and then passed to function as output argument.
I use gcc/g++ (currently 4.4.7) to compile my code.
Edit: I know about pointers and references, this is not what I am looking for. I need something like this:
void func(int* __attribute__((out)) i)
{
*i = 44;
}
void func2()
{
int myVal; // gcc will print warning: 'myVar' may be used uninitialized in this function
func(&myVal);
//...
}
Edit 2: Some extra code is needed to reproduce warning "'myVar' may be used uninitialized in this function". Additionally you have to pass -Wall -O1 to gcc.
void __attribute__((const)) func(int* i)
{
*i = 44;
}
int func2()
{
int myVal; // warning here
func(&myVal);
return myVal;
}
"Is it possible to do something similar in C/C++?"
Not really. Standard c++ or c doesn't support such thing like a output only parameter.
In c++ you can use a reference parameter to get in/out semantics:
void func(int& i) {
// ^
i = 44;
}
For c you need a pointer, to do the same:
void func(int* i) {
// ^
*i = 44;
// ^
}
Note that there are no distinctions between out and in&out parameters, unless you use a const reference (which means input only):
void func(const int& i) {
// ^^^^^
i = 44;
}
When you want to pass an arg as output in C++ you pass it as a reference :
void func(int &i)
{
i = 44;
}
and you must initialize it before doing anything on it.
Note : You can specify when you want the arg to be just an input with a const ref :
void func(const int &i)
{
i = 44;
}