Oh I see. You can't. The way you've done it - taking numeric input for the operation the user wants to do - means you can only accept numerical commands. But the assignment explicitly says you should get + for add, - for subtract and so on. So instead of int which_one; cin >> which_one; you need to declare which_one to be a char so that the user can enter a character command.
Also note the assignment states "Use a separate function enter_code to prompt the user for instruction code" and mentions another function as well - if you don't do this then after all the work you've done you'll still fail it.
cin >> option;
cout<<"Invalid operation; please re-enter"<<endl;
The assignment doesn't ask you to display a menu.
One thing I'm not clear on is why left_operand *and* right_operand are needed. On a handheld calculator typically you use the current accumulator and a single new number, so you would enter + to add, then enter a new number and press =, and that would add the new number to the accumulator. Perhaps you could ask what should happen if the user enters +: does the program ask for two numbers, or does it ask for one and add the number to the accumulator? If it asks for two numbers and + adds those, what does it do with the number in the accumulator - does the new result overwrite that?
void enter_operand(float *left, float *right, char opt)
float compute(float left, float right, char opt)
case '+': return left+right;
case '-': return left-right;
case '*': return left*right;
case '/': return left/right;
case 'p': return pow(left,right);
case 'c': return 0;
case 's': return sqrt(left);
case 'l': return log(left);
case 'n': return ln(left);
Well there you go. Surprised myself; I don't normally do homework, but watching you struggle with this is too painful. This is the first pass - I haven't compiled this code so I don't know if there are any silly errors (there probably are). You'll have to figure out what to #include to make the function calls strchr, log, pow etc work OK but I imagine those will be in your course notes somewhere.
I've resolved the question about the accumulator by having the result overwrite the previous value. If you want to add up a few numbers then you'll need to keep re-entering the current result, e.g. to add 2,3,5,2 you need to do 2+3=5, 5+5=10, 10+2=12. This is not like a calculator where you would usually do 2+3+5+2= and that would display 12 with the interim results displayed as you enter each option.