How to pass an expression as an argument/parameter? - c++

Question

Say I have a polynomial function f(x), and I want to use this expression into either the expression g(x) or h(x), depending on which the user chooses. Is this possible?
for example
int main() {
float fexpression = /* polynomial function f(x) here */;
/* some code where user picks g(x) or h(x) */
if (/* gexpression == true */)
cout << gfunction(fexpression);
else /* h expression == true */
cout << hfunction(fexpression);
return 0;
}
float gfunction(float j){
float var = /* some function representing g(f(x)) == g(j) */;
return var;
}
float hfunction(float j){
float var = /* some function representing h(f(x)) == h(j) */;
return var;
}
it just doesnt seem right to me that I can pass a variable that is an expression, such as
float fexpression = ....

Answer 1

You want a function pointer.
Make the expression a function.
float fexpression( float x )
{
return (5*(pow(x,4))) + (3*(pow(x, 3))) + (10 * x) - 5
}
This code, as you wrote it, will then pass the functionality itself.
if (/* gexpression == true */)
cout << gfunction(fexpression);
else /* h expression == true */
cout << hfunction(fexpression);

Answer 2

I tried to explain details of this in the comment section of Drew's answer, but it got lost in the formatting... so had to start another answer.
#DrewDorman's is right in saying you need to just define your expression as a function:
float fexpression( float x )
{
return (5*(pow(x,4))) + (3*(pow(x, 3))) + (10 * x) - 5
}
What is missing (for you) is the following (which you would get from reading Drew's link...):
define your gfunction as follows (for instance)
float gfunction(float (*funcp)(float)) {
// lots of stuff to determine the value of x
return (*funcp)(x);
}
This is how gfunction can evaluate your polynomial. Do something similar for hfunction
When you call gfunction, you have to reference fexpression as follows:
cout << gfunction(&fexpression);
the & matters... it was also missing from Drew's answer.