Initializing an array of string pointers and print the same through a function call not Working, Why?

Question

I have to initialize an array of string pointers to point to 4 worker names. Then i want to call a function func1() which prints out all the names. And i have to do this using either pass by referebce or values or may be a hybrid of these, anything is fine.

I have already been through following queries of similar type

how to initialize string pointer?

c++ initialization of an array of string pointers

returning array of string from function not working as expected

Also tried to use the Books Deitel and deitel and C++ Primer yet, could not understand or clarify my understanding regarding array of string pointers, to a point that now i am even getting confused with the fundamental basics of this.

Although, having gone through numerous similar questions online i could get only this far, as per the code I have written below.

My Code:-

#include <iostream>
#include <string>
#include <cstdlib>

using namespace std;

void func1(string *str, int *num)
{
    for (int i = 0; i <= (*num); i++)
    {
        cout<<"\n"<<str[i]<<"\n";
    }
    cout<<"reached to the end of calling function\n\n";
}

int main() 
{
    int num; 
    cout<<"\nplease input your prefered size of array\n";
    cin >> num;  // Number of  names elements
    cout<<"\nplz enter the desired names\n";
    string *str = new string[num];
    for (int i=0;i<num;i++)
    {
        getline(cin, str[i]);
    }

    cout<<"now printing all the names through a calling function\nusing appropriate parameters\n";

    func1(str, &num);

    return 0;
}

Present Behaviour:-

Code executes and takes the input from me, but as soon as i give the second-to-the-last value, it gives a prompt error saying "file1.exe has stopped working". However, it does carry on with the remaining task and calls func1() and prints the ending line of the called function.

Output:

please input your prefered size of array

4

plz enter the desired names

Pratik

Pratik Vyas

Pratik Vyas A

now printing all the names through a calling function using appropriate parameters

Pratik

Pratik Vyas

Pratik Vyas A

---

Process exited after 14.24 seconds with return value 255 Press any key to continue . . .

And i have to close the terminal due to the prompt error that i get as soon as i have entered the third name Pratik Vyas A as described above.

I understand that this may be some really basic level issue, but only if my brain could process other numerous ways given online, that i would have saved this question space for some other worthy important questions. any help appriciated highly.

Answer 1

If you add the line

cin.ignore();

before reading input, this won't happen.

The source of the problem is that when you read input with cin like

cin >> num;

cin reads up to the newline character, and stops there (before the newline, leaving the newline in the buffer). When you call getline, the first thing it sees is the \n left in the buffer from when the user entered a number, so you get an empty string in the first element of your array. Notice how your code left a gap of three lines before printing the names, when you only told it to leave one line - it's because the middle of those lines was the (empty) first string.

Here's the main method with that change made, I've tested this code on my computer and it works correctly:

int main()
{
    int num;
    cout<<"\nplease input your prefered size of array\n";
    cin >> num;  // Number of  names elements
    cout<<"\nplz enter the desired names\n";
    string *str = new string[num];
    cin.ignore();
    for (int i=0;i<num;i++)
    {
        getline(cin, str[i]);
    }

    cout<<"now printing all the names through a calling function\nusing appropriate parameters\n";

    func1(str, &num);

    return 0;
}

Answer 2

After you read the number into num the newline that followed it is still in the stream, so the first iteration of the loop reads an empty string.

You could debug this for yourself by changing the loop to show you what it's doing:

for (int i=0;i<num;i++)
{
    if (getline(cin, str[i]))
      cout << "got: " << str[i] << std::endl;
    else
      cerr << "Failed to read!\n";
}

You should learn to debug your own code like this, to find out what it's really doing, instead of just asking here.

Answer 3

Use vector and string:

typedef std::vector<std::string> stringVec;

void foo(const stringVec& names){
    for (int i=0;i<names.size();i++){std::cout << names[i] << std::endl;}
}

int main(){
    stringVec s;
    s.push_back("Peter");
    s.push_back("Klaus");
    foo(s);
}

... and dont use using namespace std; (cant find the link but I will look for it)

Answer 4

From the comments of posted in response to the question, one particular comment identifies a main, fundamental and clear flaw in the code design, which when rectified, the code functions right on target fulfilling the main purpose. Responded by @crashmstr

"func1's for loop goes to <= num, which is one past the end of the array."-crashmster

,hence exhibiting an 'out of bounds' or 'out of range' kind of behaviour in most compilers (generally all), with some compilers, while still giving the final output on screen alongside of the prompt-box error, for eg:- codeblocks, devcpp, codebox, scripter etc. gratitude.

below is small but specific correction in the code

#include <iostream>
#include <string>
#include <cstdlib>

using namespace std;

void func1(string *str, int *num)
{
    for (int i = 0; i < (*num); i++)
    {
        cout<<"\n"<<str[i]<<"\n";
    }
    cout<<"reached to the end of calling function\n\n";
}

int main() 
{
    int num; 
    cout<<"\nplease input your prefered size of array\n";
    cin >> num;  // Number of  names elements
    cout<<"\nplz enter the desired names\n";
    string *str = new string[num];
    cin.ignore();   //important input by @Chris Hengler as it controls
                    // aspects of out of bounds behaviour prevention
    for (int i=0;i < num;i++)
    {
        getline(cin, str[i]);
    }

    cout<<"now printing all the names through a calling function\nusing appropriate parameters\n";

    func1(str, &num);

    return 0;
}

now works fine. grats to Chris and Crash for responses.