Iterating an array of a pointer to a char

[benbalach]

I'm having trouble finding a way of going through each character of a pointer to an array of chars.

#include "std_lib_facilities.h" // Comes with the book I'm reading

int main()
{
	char* input;
	char temp = 0;
	while(cin >> temp && temp != '!')
	{
		input = new char(temp);
	}
	// How do I print the characters in char* input one by one?
	return 0;
}

Thanks in advance to anyone who replies!

[Peter87]

cout << temp;

[Chervil]

Maybe like this - though this could be made shorter, kept it this way to detail each step.

#include <iostream>

using std::cin;
using std::cout;
using std::endl;

int main()
{
    const int size = 100;
    char array[size];           // here the input will be stored
    
    cout << "Please enter some text:" << endl;
    cin.getline(array, size);   // get up to size-1 characters from user

    
    char* input = array;        // input points to the first character of the array
    
    while (*input)              // test current character is not null terminator
    {
        char temp = *input;     // get current character in temp
        cout << temp;           // output it
        ++input;                // increment to next character
    }
    
    cout << endl;
    
    return 0;
}

[benbalach]

Why doesn't this work:

#include "std_lib_facilities.h"

int main()
{
	char* input;
	char temp = 0;
	while(cin >> temp && temp != '!')
	{
		input = new char(temp);
	}
	for(int i = 0; i < 5; ++i)
	{
		cout << input[i] << "\n"; // Doesn't print correct values
	}
	return 0;
}

[vlad from moscow]

In which chapter of the Straustrup's book is there this example of code? It has no any sense.

[benbalach]

@vlad from moscow

Chapter 17 Exercise 7:

Write a program that reads characters from cin into an array that you allocate on the free store. Read individual characters until an exclamation mark (!) is entered. Do not use std::string. Do not worry about memory exhaustion.

I did the exercise, but I want to make sure that what I entered is actually stored. But I can't find I way to print anything from the array.

This is how the book taught me to allocate and initialize an array:

char* a = new char[10];
for(int i = 0; i < 10; ++i)
{
	a[i] = 65 + i;
	cout << a[i] << "\n";  // prints A, B, C, etc..
}

So why doesn't my code print out char* input? I tried it without a for-loop like this:

/* ... */
cout << input[0] << "\n";
cout << input[1] << "\n";
cout << input[2] << "\n";
cout << input[3] << "\n";
/* ... */

But it still doesn't print the right values I entered.

[vlad from moscow]

The purpose of the exercise is to demonstrate the limited nature of arrays and the advantage of using standard class std:;string (exercise 8).

So the code of the exercise 7 can look very simply

#include <iostream>

int main()
{
   const size_t N = 10;

   char *s = new char[N];

   char c;
   size_t i = 0;
   // wait until the program will abort:)
   while ( std::cin >> c && c != '!' ) s[i++] = c;
}  

[vlad from moscow]

Only when you will do exercise 8 you may not write s[i]. Instead you can use either operator += or member function push_back or even append. For example

#include <iostream>
#include <string>

int main()
{
   const size_t N = 10;
   std::string s;
   s.reserve( N );

   char c;
   while ( std::cin >> c && c != '!' ) s += c; // or s.push_back( c ) or s.append( 1, c )
}

[benbalach]

@vlad from moscow

Thank you very much for explaining everything to me, now I understand it.