How to get array output without using pointers in this C program example?

Question

Here is the C code.

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

main()
{
    int i,n;
    char a[100];

    printf("Enter number of teams");
    scanf("%d",&n);

    for(i=1;i<=n;i++)
    {
        printf("\nenter team %d",i);
        scanf("%s",a);
    }

    for(i=1;i<=n;i++)
    {
        printf("%s",a[i]);
    }
}

My program is simple, but its crashing. i just want to load in a few names into array and then print them back. I am able to do that but at last line, its crashing.

`printf("%s", a[i]);` is wrong because type of `a[i]` is `char`, you must enable compiler warnings. — Iharob Al Asimi, Feb 08 '15 at 17:26
array index start from 0 and goes till n-1. and what are you expecting with `scanf("%s",a);` in a loop? — SMA, Feb 08 '15 at 17:26
You have some serious misunderstandings of very basic stuff. You should read a [good book](http://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list) to get that sorted out. — Baum mit Augen, Feb 08 '15 at 17:28
I'm voting to close this question as off-topic because OP doesn't have basic understanding of strings/arrays in C — P0W, Feb 08 '15 at 17:30
@downvoters(whoever they maybe, I am not addressing any particular person, I am just saying in general) .My account has been banned from asking questions and the stackoverflow has asked me to review my questions. I don't see how this question is wrong by any means. So if there is any improvement, please suggest so, or else please consider upvoting this question, so I get back my ability to ask questions on this site. I like stackoverflow and would like to be part of this community, so I want your help either in terms of suggestions for re-formatting the question or in form of upvotes. Thanks — ThisSiteSucks, Mar 16 '16 at 08:15

shauryachats · Answer 1 · 2015-02-08T17:37:34.623

3

A char[] is equivalent to a string in C when it is terminated with a NUL character (\0). You need to create a char[][], or an array of char[]s.

So, an array of strings in C can be defined as:

char arr[5][100];
//The 5 is the number of strings, and 100 is the length of a single string.

Also, in C, array indexing starts from 0, not 1.

Code:

#include<stdio.h>
int main()
{
    int i,n;
    printf("Enter number of teams");
    scanf("%d",&n);
    char a[n][100];
    for(i=0;i<n;i++)
    {
        printf("\nenter team %d",i);
        scanf("%s",a[i]);
    }
    for(i=0;i<n;i++)
    {
        printf("%s\n",a[i]);
    }
}

edited Feb 08 '15 at 17:37

answered Feb 08 '15 at 17:29

shauryachats

9,975
4
35
48

`char arr[5][100]` is not an array of strings, it's an array of `char` arrays, for the contents to be strings, they must be `'\0'` terminated. And though it's not harmful, c++ is not relevant to this question. – Iharob Al Asimi Feb 08 '15 at 17:35
Sorry, I'll add that. Thanks. @iharob the OP tagged it with `c++` before. – shauryachats Feb 08 '15 at 17:36
Can be defined makes it a much better post. – Iharob Al Asimi Feb 08 '15 at 17:42

Iharob Al Asimi · Accepted Answer · 2015-02-08T17:41:40.230

You have a few errors in your code

main() must return int
a should be an array of strings, and it's an array of char, you can do it with an array of char arrays, declare it as
```
char a[100][100];
```
Arrays in c are indexed from 0 to n - 1 and not from 1 to n.
You must check if the first scanf() worked, furthermore, you must check that n is not too big so this part does that
```
if (scanf("%d", &n) != 1)
    return 1;
if (n > 100)
    n = 100;
```
You should prevent a buffer overflow, for that you can use the length specifier for scanf(), if your array can store 100 char's then you need to use
```
scanf("%99s", a[i]);
```
because you also need the terminating '\0'.
Your printf() statement is using the wrong specifier, because you are passing a char and the "%s" is specting a c string.

Try this

#include <stdio.h>

int
main()
{
    int i, n;
    char a[100][100];

    printf("Enter number of teams");
    if (scanf("%d", &n) != 1)
        return 1;
    if (n > 100)
        n = 100;

    for (i = 0 ; i < n ; i++)
    {
        printf("\nenter team %d",i);
        scanf("%99s", a[i]);
    }

    for (i = 0 ; i < n ; i++)
    {
        printf("%s", a[i]);
    }

    return 0;
}

also, even though you are not explicitly manipulating pointers, you can't avoid that completely, because for example printf() takes a pointer parameter for the "%s" specifier, it's just that in this case you don't explicitly pass a pointer, but the i-th array of the array of char arrays will automatically decay to a pointer in

printf("%s", a[i]);

Minor: Consider stating the most important issues first (#2 & #3). Also use differ sizes in array rather than 100,100, maybe `char a[20][50];` or `#define M,N`. It helps in understanding code like `if (n > 20) .... scanf("%49s", a[i]);` — chux - Reinstate Monica, Feb 08 '15 at 19:58

How to get array output without using pointers in this C program example?

2 Answers2