Thread: (name-age) pair vector program help needed

Here is what I gather the exercise sounds like it wants you to do:

Read 5 names into a vector of strings. Btw, don't use cin.ignore() after reading a string. A string can read anything, so you will end up blocking and ignoring the first newly inputted character.
Read 5 ages into a vector of ints. Btw, there is no point in sorting either vector every iteration. The requirement is that the vector shall be sorted eventually, not kept sorted every time there is a new insertion.
Make a copy of the names vector.
Sort the names vector.
For every element in the names vector, try to find the corresponding index in the old names vector. Use that index to find the right age. Using that information, create a new vector of ages.

Originally Posted by Osman Zakir

But now how do I find the right age?

Since you have found the position of the name in the original name vector, you can thus use that position/index to access the corresponding element in the age vector.

By the way, this is a horrible way of doing things. It would be better to at least have a simple struct to represent a name-age pair:

Code:

struct Person
{
    std::string name;
    double age;
};

Then create a single std::vector<Person> to be sorted. Alternatively, one might use a std::vector<std::pair<std::string, double>> (sacrificing the benefit of the member names for less work), or a std::map<std::string, double> (in which case the sorting is implicit).

If you really need to have parallel arrays, then perhaps sorting pairs of pointers to their elements would be better.

Originally Posted by laserlight

By the way, this is a horrible way of doing things. It would be better to at least have a simple struct to represent a name-age pair:

I agree it's horrible, but the exercise says it should be done this way. Still, it is a good advice for future code.

Originally Posted by Osman Zakir

But the names vector has been sorted already, and the ages may have been entered in a different order. How would I find the right age in that case?

Did I not mention this in my original reply? Read it again.

Btw, what does this do?
vector<string> names_copy(sort(names.begin(), names.end()));
Did you compile your code before posting?

Originally Posted by Osman Zakir

But the names vector has been sorted already, and the ages may have been entered in a different order. How would I find the right age in that case?

Even before Elysia mentioned it, this was in the hint provided by your instructor: "Before sorting name, take a copy and use that to make a copy of age in the right order after sorting name."

Originally Posted by laserlight

Even before Elysia mentioned it, this was in the hint provided by your instructor: "Before sorting name, take a copy and use that to make a copy of age in the right order after sorting name."

Even with this hint instructor has chosen some very convoluted way of doing things

What point of the hints I gave did you not understand?

Originally Posted by Osman Zakir

Anyway, now I'd like some help on how to use the index of the name in the copied names vector to search out the corresponding age and print it for each name. And yeah, even if I do manage to understand the logic, I don't have the syntax down at this point yet. I'm not sure how to do this.

The syntax is incredibly simple. It is a matter of logic, not syntax, unless you don't know the syntax for accessing the elements of a vector by index, but you obviously do know that syntax.

Let's suppose that we have four names and ages in two parallel lists:

Code:

Trent    30
Alice    25
Mallory  15
Bob      20

After we sort the list of names, we get:

Code:

Alice
Bob
Mallory
Trent

Now, we want to print the list of ages corresponding to the sorted list of names. We start with the first element of the sorted list, i.e., Alice, and find that in the original list, Alice is at index 1. Therefore, we print the age at index 1, i.e., we print 25. Then, we go on to the next element of the sorted list, i.e., Bob. We find that in the original list, Bob is at index 3. Therefore, we print the age at index 3, i.e., 20. So on and so forth:

Code:

25
20
15
30

As such, the syntax required is just ages[index], where it is your job to figure out what exactly is index: in your code, is it i or j? If you understand the logic, the answer would be obvious. If you don't, complete mastery of the syntax of all the programming languages in the world will not help you.

EDIT:
By the way, instead of this:

Code:

vector<string> names_copy(names.begin(), names.end());

You might as well write:

Code:

vector<string> names_copy(names);

Remember to sort names after doing this.