1. What are the two arithmetic operations? Can the input values, 0 or 1, be incorperated into the arithmetic function?

2. As laserlight pointed out, it is not at a beginner level -> But I'm sure that the OP can appreciate that it can be done.

However, if you are only adding two numbers, you could use arrays to input the two different numbers and the input to determine the offset from the start of the array.

Code:
```int main(int argc, const char *argv[])
{
int user_in=0;
int lookup_arr[6] = {0, 0, 1, 2, 3, 4};

printf("Enter 1 for 1 + 2, or \nEnter 2 for 3 + 4\n");
scanf("%d", &user_in);

arr_offset = 2 * user_in * ((unsigned int)user_in <= 2);

printf("Output: %d + %d = %d", lookup_arr[arr_offset], lookup_arr[arr_offset+1], answer);

return EXIT_SUCCESS;
}```
You couldn't alert the user to invalid inputs without using a conditional operator though...

3. If its just arithmetic operations then no arrays or function pointers are needed.
Just perfrom both calculations in one large expression, multiplying one part by the user_input, and the other part by (1-user_input).
Although in all liklihood, it can be simplified even further from there.

The user gets prompted for a choice and will enter in a 0 or 1....entering a 0 will do an arithmetic operation and give an output and entering a 1 will do a different arithmetic operation giving a different output...the catch is no if, if else, switch, while, do while, or anything like that can be used.
Code:
```double addition(double a, double b) { return a + b; }
double subtraction(double a, double b) { return a - b; }

int main(void)
{
int choice;
printf("Enter 0 for addition or 1 for subtraction: ");
scanf("%d", &choice);

double a, b;
printf("Enter two numbers a b: ");
scanf("%lf %lf", &a, &b);

double result = 0.0;
result += !choice * addition(a, b);
result += choice * subtraction(a, b);
printf("Result is %g\n", result);
}```