Question about: name.substr(), and name.find()

[Rafael Olarte]

Dear friends,

I appreciate any feedback on this problem.

I am trying to be able to produce an output with first, middle, and last name using a const string value. Therefore, no matter what name is use on the const string value, it should always display the full name.

So far, I have been able to get the first name correctly. However, when I try to get the middle name, it wraps the last name as well...


Thanks for your any valuable advise.

#include <iostream>
#include <string>
 
using namespace std;
 
const string name = "John Paul Jones";
 
int main() {
 
 
int spacePos = name.find(' ');
// int len = name.length();
string firstName = name.substr(0, spacePos);
string middleName = name.substr(spacePos + 1);
 
 
 
cout << firstName + ' ' << middleName << endl;
 
return 0;
 
}

[r035198x]

Dear friends,

I appreciate any feedback on this problem.

I am trying to be able to produce an output with first, middle, and last name using a const string value. Therefore, no matter what name is use on the const string value, it should always display the full name.

So far, I have been able to get the first name correctly. However, when I try to get the middle name, it wraps the last name as well...


Thanks for your any valuable advise.

#include <iostream>
#include <string>

using namespace std;

const string name = "John Paul Jones";

int main() {


int spacePos = name.find(' ');
// int len = name.length();
string firstName = name.substr(0, spacePos);
string middleName = name.substr(spacePos + 1);



cout << firstName + ' ' << middleName << endl;

return 0;

}

Like this?

 
#include <iostream>
#include <string>
using namespace std;
const string name = "John Paul Jones";
int main() {
 int spacePos = name.find(' ');
 string firstName = name.substr(0, spacePos);
 string temp = name.substr((spacePos + 1), name.length());
 int nextSpacePos = temp.find(' ');
 string lastName = temp.substr(nextSpacePos + 1, temp.length()-1);
 string middleName = name.substr(spacePos + 1, nextSpacePos);
 cout << firstName<<endl;
 cout<<middleName<<endl;
 cout<<lastName<< endl;
 return 0;
}
 

[Motoma]

Dear friends,

I appreciate any feedback on this problem.

I am trying to be able to produce an output with first, middle, and last name using a const string value. Therefore, no matter what name is use on the const string value, it should always display the full name.

So far, I have been able to get the first name correctly. However, when I try to get the middle name, it wraps the last name as well...


Thanks for your any valuable advise.

#include <iostream>
#include <string>

using namespace std;

const string name = "John Paul Jones";

int main() {


int spacePos = name.find(' ');
// int len = name.length();
string firstName = name.substr(0, spacePos);
string middleName = name.substr(spacePos + 1);



cout << firstName + ' ' << middleName << endl;

return 0;

}

The problem is you never searched for the second space. Right now middleName is everything from the first space till the end of the line.

[Rafael Olarte]

THANK YOU VERY MUCH. OWESOME!!! Now, I really understand how substitution works in these type of algorithm.

By any change is there any book, or any source where I could get some of these problems for me to practise?

Have a great day!!

Like this?

 
#include <iostream>
#include <string>
using namespace std;
const string name = "John Paul Jones";
int main() {
 int spacePos = name.find(' ');
 string firstName = name.substr(0, spacePos);
 string temp = name.substr((spacePos + 1), name.length());
 int nextSpacePos = temp.find(' ');
 string lastName = temp.substr(nextSpacePos + 1, temp.length()-1);
 string middleName = name.substr(spacePos + 1, nextSpacePos);
 cout << firstName<<endl;
 cout<<middleName<<endl;
 cout<<lastName<< endl;
 return 0;
}
 

[Rafael Olarte]

Thank you! :)

The problem is you never searched for the second space. Right now middleName is everything from the first space till the end of the line.

[Motoma]

Thank you! :)

Come back anytime you have a problem, we'll be happy to help out.