How can I transform a user-defined string into an integer - c++

Question

I basically ask a user to input a string, let's say they input "ABC" or"DEF"
Then, I want to set a certain integer = 1 if ABC is entered and 2 if DEF is entered.
If anything else is entered, then I want it to say invalid value.
So at the end, I'll have the integer i assigned to 1 or 2 if a valid value was entered.
#include<iostream>
#include<string>
#include<sstream>
using namespace std;
int main()
{
string input = "";
// How to get a string/sentence with spaces
cout << "Please enter a valid sentence (with spaces):\n>";
getline(cin, input);
cout << ".\n" << "You entered: " << input << endl << endl;
int m
if(input = ABC)
return 0;
}

Answer 1

Very easy:
#include <iostream>
#include <map>
#include <string>
int main()
{
std::map<std::string, int> const v { { "ABC", 1 }, { "DEF", 2 } };
for (std::string line; std::getline(std::cin, line); )
{
auto it = v.find(line);
if (it == v.end())
{
std::cout << "Input '" << line << "' is invalid.\n";
}
else
{
std::cout << "Your input value: " << it->second << "\n";
}
}
}

Answer 2

Your English spec:
” set a certain integer = 1 if ABC is entered and 2 if DEF is entered […] if anything else is entered, then I want it to say invalid value.
Expressed in C++:
if( s == "ABC" ) { x = 1; }
else if( s == "DEF" ) { x = 2; }
else { cout << "Invalid value." << endl; }
If neither of the specified possibilites is true, i.e. in the case of outputting "Invalid value.", then x keeps the value it had, which is how I interpret the word “set”.
At a higher level of abstraction, it seems like you want the user to specify one of two strings, namely “ABC” or “DEF”, which you want to have available as an integer id after the input operation.
Then there are two main possibilities:
you want the id to indicate also the case where the user specified some other string, e.g. a common value for this, or
you want the input operation to only return control when the user has inputted one of the two valid strings.
The last case effectively means that input operation must loop or logically fail. Logical failure can be expressed as an exception, or e.g. by calling std::terminate.
So as you see there is a range of possibilities here, i.e. above the simple “how do I translate this English to C++” the desired functionality is a bit under-specified.

Answer 3

#include <iostream>
#include <map>
#include <string>
#include <initializer_list>
using namespace std;
int main()
{
const std::map<std::string, int> lookupTable = { {"ABC", 1}, {"DEF", 2} };
string input = "";
int m;
while(true) {
cout << "Please enter a valid sentence (with spaces):\n>";
getline(cin, input);
std::map<std::string, int>::const_iterator it = lookupTable.find(input);
if (it != lookupTable.end()) {
m = it->second;
break;
}
}
cout << "You entered: " << input << endl << endl;
cout << "m = " << m << endl;
}
Second solution if you want your m const. This assumes your compiler supports lambda functions:
#include <iostream>
#include <map>
#include <string>
#include <initializer_list>
using namespace std;
int main()
{
const std::map<std::string, int> lookupTable = { {"ABC", 1}, {"DEF", 2} };
string input = "";
const int m = [&]() {
while(true) {
cout << "Please enter a valid sentence (with spaces):\n>";
getline(cin, input);
std::map<std::string, int>::const_iterator it = lookupTable.find(input);
if (it != lookupTable.end())
return it->second;
}
}();
cout << "You entered: " << input << endl << endl;
cout << "m = " << m << endl;
}
If your compiler does not support the above map initialization then you can use instead:
#include <iostream>
#include <map>
#include <string>
using namespace std;
std::map<std::string, int> initializeMap() {
std::map<std::string, int> map;
map["ABC"] = 1;
map["DEF"] = 2;
// add whatever else here
return map;
}
int main()
{
const std::map<std::string, int> lookupTable = initializeMap();
string input = "";
int m;
while(true) {
cout << "Please enter a valid sentence (with spaces):\n>";
getline(cin, input);
std::map<std::string, int>::const_iterator it = lookupTable.find(input);
if (it != lookupTable.end()) {
m = it->second;
break;
}
}
cout << "You entered: " << input << endl << endl;
cout << "m = " << m << endl;
}