在此C程序示例中，如何在不使用指针的情况下获取数组输出？

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.

Answer 1

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]);
    }
}

Answer 2

You have a few errors in your code

1. main() must return int

2. 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]; 

3. Arrays in c are indexed from 0 to n - 1 and not from 1 to n .

4. 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; 

5. 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' .

6. Your printf() statement is using the wrong specifier, because you are passing a char and the "%s" is specting ac 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]);