Basic command line REPL calculator

Question

I'm pretty confident in this code but I want to be sure I haven't missed anything to do with optimization or error handling. The program takes input in the form of {number} {operator} {number} and prints the result. If the input is invalid it prints nothing.

Example input: 1.0 + 2.0 Output: 3

#include <iostream>
#include <string>
#include <stdexcept>
#include <regex>
double calculate(const double a, const char op, const double b)
{
 switch (op)
 {
 case '+':
 return a + b;
 case '-':
 return a - b;
 case '*':
 return a * b;
 case '/':
 if (b == 0.0)
 throw std::invalid_argument {"Divide by zero"};
 return a / b;
 default:
 throw std::invalid_argument {{op}};
 }
}
int main()
{
 const std::regex valid_input {
 "([0-9]*\\.?[0-9]*) ([\\+\\-\\*\\/]) ([0-9]*\\.?[0-9]*)"};
 std::string line;
 std::smatch terms;
 double a;
 double b;
 char op;
 while (std::getline(std::cin, line))
 {
 if (!std::regex_match(line, terms, valid_input))
 continue;
 a = std::stod(terms[1]);
 b = std::stod(terms[3]);
 op = terms[2].str().front();
 std::cout << calculate(a, op, b) << '\n';
 }
}

Answer 1

Escaping escape characters becomes confusing.
Use the appropriate string literal.

 // Your String
 "([0-9]*\\.?[0-9]*) ([\\+\\-\\*\\/]) ([0-9]*\\.?[0-9]*)"
 // A Raw String (no escape characters)
 R"RAW(([0-9]*\.?[0-9]*) ([\+\-\*\/]) ([0-9]*\.?[0-9]*))RAW"

Don't think your regular expressions are good enough.

([0-9]*\.?[0-9]*)
// Can map
.
<blank>

I would do:

// If you have the decimal point then you require at least on digit after it.
// If there is no decimal point then you require at least one digit.
(([0-9]*(\.[0-9]+))|[0-9]+)

Don't think regular expressions are the best way to capture number operator number. The stream operators support this already so prefer to use those.

while (std::getline(std::cin, line))
{
 std::stringstream lineStream(line);
 double a;
 double b;
 char op;
 char emptyTest;
 if (lineStream >> a >> op >> b && !(lineStream >> emptyTest)) {
 // Enter here if:
 // 1: We correctly read a op and b
 // 2: There are zero extra non-space characters on the line
 // We find this because the read to emptyTest fails.
 }

Declare your variables as close to the location that you use them. This also allows you to initialize them on declaration (which is always nice).

// Don't do this.
double a;
double b;
char op;
 // Now you can see the type and know that they are initialized.
 double a = std::stod(terms[1]);
 double b = std::stod(terms[3]);
 char op = terms[2].str().front();

Sure you can use a switch that is totally fine. But you can practice using the command pattern in this situation.

static std::map<char, std::function<double(double, doubel>> const actionMap = {
 {'+', [](double lhs, double rhs){return lhs + rhs;}},
 {'-', [](double lhs, double rhs){return lhs - rhs;}},
 {'*', [](double lhs, double rhs){return lhs * rhs;}},
 {'/', [](double lhs, double rhs){
 if (b == 0.0) {
 throw std::invalid_argument {"Divide by zero"};
 }
 return lhs / rhs;
 }
};
// Now use the action map to find the function you want to call.
auto find = actionMap.find(op);
if (find == actionMap.end()) {
 throw std::invalid_argument {{op}};
}
return find->second(a, b);

• Brady Dean: Thanks for the regex advice. After submitting the question I rewrote main to use stringstream and it works really good

Answer 2

• What happens if I enter only + surrounded by spaces?
• The program should not abort if I enter 0 / 0, but continue with the next line.
• I'd like to omit the spaces around the operator.
• In the regular expression, the operator part can be written as [+\\-*/].