0 
#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.

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4
  • 1
    With e.g. char tno[9] you have a "string" for eight characters. Remember that char strings in C++ are really called null-terminated byte strings. Don't forget space for the terminator, or you will go out of bounds and have undefined behavior. Commented Nov 2, 2017 at 7:25
  • And why use p? The expression (p + i)->name is equal to p[i].name which of course is the same as a[i].name. Commented Nov 2, 2017 at 7:27
  • Please consider switching to std::string instead of c-strings, and cout instead of puts. Smart pointers instead of raw pointers. And also consider switching to some new IDE. Commented Nov 2, 2017 at 7:33
  • include<iostream.h> use a compiler from this century. Commented Nov 2, 2017 at 7:34

2 Answers 2

Reset to default
2

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

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1 Comment

"strcpy does is that it also adds \n to the end" No, strcpy does not add a newline at the end. However, strcpy does copy the null terminator i.e. '\0' from the string literal | "it tries to add it to tno [10]" No, the null terminator is at index 9 (which is out of bounds of the output array).
0

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.

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5 Comments

@Vedant why would there be padding?
Refer to:geeksforgeeks.org/…
I think structs containing char and char [] can be effectively unaligned, and don't require padding. The only way I could see the observed behavior of the OP, was by a structure with no padding.
@mksteve the OP said he doesnot get output for the names, but according to the image you have created in the answer, only the first and last chacter has \0, all the other have valid characters, so why dont these character print...? Why is OP getting completly blank output...?
@Vedant '\0' marks the end of the string, everything afterwards is not considered

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