Warior4356 - 7 months ago 30

C Question

So I am trying to make a program that does math operations based on user input, however I am running into an issue with trying to set the math operator based on what they give.

Function:

`const operator(int val)`

{

if (val == 1)

{

return +;

}

if (val == 2)

{

return -;

}

}

With a main code looking something like this

`scanf("%d", val)`

output = 4 operator(val) 2

printf("%d", output)

Is there a variable type that I can use in place of an operator? If not is there a way to make a variable/function reference a defined macro?

For example:

`#define plus +`

then reference the macro in the code?

Finally I am aware I could have if cases for each input, however this scales poorly for something like,

`output = 2 operator(val) 5 operator(val) 7 operator(val) 3`

which would require 64 if statements I think to make it work.

Thank you for reading, I am at my wits end on this.

Answer

What you can do is using function pointers.

Lets say you are restricting yourself to integers and add / subtract for this example:

```
int add(int a, int b)
{
return a + b;
}
int subtract(int a, int b)
{
return a - b;
}
// without typedef, the signature of get_operator would be really messy
typedef int (*arithmetic_func)(int,int);
arithmetic_func get_operator(int val)
{
if (val == 1)
{
return &add;
}
if (val == 2)
{
return &subtract;
}
return NULL; // what to do if no function matches?
}
```

Instead of `output = 4 operator(val) 2`

you can write:

```
output = get_operator(val)(4, 2)
```

What happens there is, that the `get_operator(val)`

function call is returning a function pointer to an actual arithmetic function. Then this actual arithmetic function is called with the `(4, 2)`

as parameters.

Source (Stackoverflow)