C++ pointer array of structure

Question

#include<iostream.h>
#include<conio.h>
#include<string.h>
#include<stdio.h>

struct telephone
{
    char name[10];
    char tno[9];
};

void main()
{ 
    telephone a[5];
    clrscr();
    telephone* p;
    p = a;
    strcpy(a[0].name, "Aditya"); // initializing the array
    strcpy(a[1].name, "Harsh");
    strcpy(a[2].name, "Kartik");
    strcpy(a[3].name, "Ayush");
    strcpy(a[4].name, "Shrey");
    strcpy(a[0].tno, "873629595");
    strcpy(a[1].tno, "834683565");
    strcpy(a[2].tno, "474835595");
    strcpy(a[3].tno, "143362465");
    strcpy(a[4].tno, "532453665");

    for (int i = 0; i < 5; i++)
    {  
        puts((p+i)->name);cout<< " ";   //command for output
        puts((p+i)->tno );cout<<endl;
    }
    getch();
}

In this code, while taking output, I am not getting output of names. I only get output for (p+0)->name and not for anything else, but if I don't initialize the telephone number then I get output of names.

Answer 1

A struct is stored in continuous memory locations , so when you assign the tno variable, data higher than its bound size, is attempted to be stored in it, the leftover bits get added to the next memory location .

In your code tno [9] , so it can store max 9 chars , though you give it 9 chars only, but what the strcpy does is that it also adds \\0 to the end, it tries to add it to tno [10] , which does not exist and goes out of bounds and stores it in some other memory location , probably that of the next array, this leads to undefined behaviour.

You just need to change the struct definition as follows:

struct telephone
{ 
   char name[10];
   char tno[10]; // if you intend to store 9 digit number
 }; 

Just remember if you intend to store x chars in a character array , then your array must be of size x + 1 . If using character arrays seems to be difficult, maybe you can use std::string

Answer 2

Telephone number needs to be at least one bigger. The telephone number is overunning by one byte into the next array, and changing the name to '\\0'

strcpy(a[0].tno ,"873629595" );

turns a[1].name from

"Harsh\0"

into

"\0arsh\0"

Structure layout

+---+---+---+---+---+---+---+---+---+---+
| A | d | i | t | y | a | \0|   |   |   |
+---+---+---+---+---+---+---+---+---+---+
+---+---+---+---+---+---+---+---+---+
| 8 | 7 | 3 | 6 | 2 | 9 | 5 | 9 | 5 | 
+---+---+---+---+---+---+---+---+---+
+---+---+---+---+---+---+---+---+---+---+
| \0| a | r | s | h | \0|   |   |   |   |
+---+---+---+---+---+---+---+---+---+---+
+---+---+---+---+---+---+---+---+---+
| 8 | 3 | 4 | 6 | 8 | 3 | 5 | 6 | 5 | 
+---+---+---+---+---+---+---+---+---+

None of the telephone numbers has space for the null terminator, and is overwriting beyond its space. This is in fact the next array elements name.