How can I create a function pointer to a function where some parameters are set to be fixed upon the definition.
Here is an example what I mean:
Let's say I have the function
int add (int n, int m) {
return n+m;
}
and the function pointer type
typedef int (*increaser)(int);
What I want is a pointer to the function add which fixes the first parameter to 1 and leaves the second paramenter open. Something along the lines of
increaser f = &add(1,x);
How can I accomplish this?
What I want is a pointer to the function add which fixes the first parameter to 1 and leaves the second paramenter open.
There is no such thing in C. The closest you can come is to create a wrapper function and make a pointer to that:
int add1(int x) {
return add(1, x);
}
increaser f = &add1;
If you don't need a pointer to the function then you can use a macro:
#define increaser(x) add(1, (x))
C does not support doing this directly. In C++ there is std::function and bind which can achieve this however.
None the less for a pure C solution the closest you can reasonably get is defining a new function that calls add like:
int increment(int input) {
return add(1, input);
}
Then you can do:
increaser f = &increment;
increaser. To me, it is rather anint_unary_operator, as functions implementing this signature can very well do something completely different.