Stroustrup Ex.7, Chap.4 - C++ Syntax - c++

Question

I'm new to C++ and, as many others here, I'm trying to learn it from Bjarne Stroustrup's Programming -- Principles and Practice Using C++.
I'm stuck on Exercise 7, Chap.4., where the idea is to write a calculator that, when the input is either an integer and/or a string followed by a character (+,-,* or /), the output should announce "the sum/diff/prod/ratio of" input 1 and input 2 is the result; so if ("two" 3 *) is the input, the output should be "the product of 2 * 3 = 6".
Here's Stroustrup's solution (I'm leaving Stroustrup's comments):
-- There's no copyright infringement, as this is all from his website --
/*The solution uses two functions (in addition to main():
initialize_numbers() to initialize the vector of number string
representations
get_number() to read a number that is either a string or a sequence of
digits
*/
vector<string> numbers; // representation of numbers as strings
// numbers[i] is the string representation for i
// for numbers[0] to numbers[numbers.size()-1]
void initialize_numbers()
{
numbers.push_back("zero");
numbers.push_back("one");
numbers.push_back("two");
numbers.push_back("three");
numbers.push_back("four");
numbers.push_back("five");
numbers.push_back("six");
numbers.push_back("seven");
numbers.push_back("eight");
numbers.push_back("nine");
numbers.push_back("ten"); // why not? :-)
}
int get_number()
{
const int not_a_symbol = numbers.size(); // not_a_symbol is a value that does not correspond
// to a string in the numbers vector
int val = not_a_symbol;
if (cin>>val) return val; // try to read an integer composed of digits
cin.clear(); // clear string after failed attempt to read an integer
string s;
cin>>s;
for (int i=0; i<numbers.size(); ++i) // see if the string is in numbers
if (numbers[i]==s) val = i;
if (val==not_a_symbol) error("unexpected number string: ",s);
return val;
}
int main()
try
{ initialize_numbers();
cout<< "please enter two floating-point values separated by an operator\n The operator can be + - * / % : ";
while (true) { // "forever"; that is until we give an unacceptable input or make a computations error
int val1 = get_number();
char op = 0;
cin>>op; // get the operator
int val2 = get_number();
string oper; // text appropriate for an operator
double result;
switch (op) {
case '+':
oper = "sum of ";
result = val1+val2;
break;
case '-':
oper = "difference between ";
result = val1-val2;
break;
case '*':
oper = "product of ";
result = val1*val2;
break;
case '/':
oper = "ratio of ";
if (val2==0) error("trying to divide by zero");
result = val1/val2;
break;
case '%':
oper = "remainder of ";
if (val2==0) error("trying to divide by zero (%)");
result = val1%val2;
break;
default:
error("bad operator");
}
cout << oper << val1 << " and " << val2 << " is " << result << '\n';
cout << "Try again: ";
}
}
More specifically, my problem is with the following part:
int get_number()
{
const int not_a_symbol = numbers.size(); // not_a_symbol is a value that does not correspond
// to a string in the numbers vector
int val = not_a_symbol;
if (cin>>val) return val; // try to read an integer composed of digits
cin.clear(); // clear string after failed attempt to read an integer
etc etc etc...
}
I just don't understand what's going on here, in the big context. I'm having trouble understanding this whole get_number() function, and how it relates to the rest of the code.
1 - Why assign the value of number.size() to not_a_symbol? What does this accomplish?
2 - if (cin >> val) - why is that the conditional? val is == size of vector numbers, which is 11, so is the conditional the number 11? How is that helpful?
And what is it returning? Itself?
3 - // try to read an integer composed of digits - how is that accomplished, and why is that helpful?
Thank you, and sorry for the long format of the question.

Answer 1

In for that is in whole function get_number() i goes from 0 to one less than numbers.size() and puts i where input string not containing digits compared with one of strings in numbers vector into val(so you convert name of number into value of number). After that you check if val is same as size of vector numbers because if it is, there was no match(someone entered word that is not name of any number you can handle).
From if (cin >> x) - Why can you use that condition? cin>>val(cin reads from input into variable val) will return false if you enetred atleast one letter.
If you enter number with no letters you can return it (we want either string representing name of number or a normal digit).

Answer 2

Sorry to take away your answer, man, but I figured it out by myself, and the thing is much more simple (and clever).
The for-loop works as it should, comparing input strings to the strings inside the vector and returning the correspondent index number.
But the reason to assign the value of numbers.size() to not_a_symbol, giving val the value of numbers.size() resides in the fact that, if the first IF fails, and only then, the second will become true, since val was already initialized. That's why there's 2 separated IF statements, instead of an IF-ELSE: an ELSE would not make the previously initialized val count, because the input from string "s" would take over, preventing val's initial value to work on (val=not_a_symbol) inside the ELSE.
Forget the functions, put it all inside the main:
int main() {
vector<string> numbers{ "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten" };
int not = numbers.size();
int val = numbers.size(); //val is initialized
string s;
cin >> s;
for (int i = 0; i<numbers.size(); ++i)
if (numbers[i] == s) val = i; // first IF; if this condition is not met, it will return val as it was initialized (val=numbers.size() or 11)
if (val == not) val = 88; // then this condition will be checked. It will be true. It will return val = 88 (a random number);
cout << "val is: " << val << "\n" << "not is: " << not << "\n";
}
So, it's not a matter of comparing val with the vector's number of elements, is a matter of val already being equal to it and that fact acting upon the failing of the first condition.