Traversing Two Dimensional Array

Question

How come Line 1 and Line 2 have more than ten x(s) when the loop should end at ten given the condition of ii < 10?

I am trying to understand how these two dimensional arrays work, any advice is appreciated.

#include <stdio.h>
int main()
{
    char name[3][10];

    for(int i = 0; i < 3; i++){
        for(int ii = 0; ii < 10; ii++){
            name[i][ii] = 'x';
        }
    }


    printf("Line 1: %s.\n", name[0]);

    printf("Line 2: %s.\n", name[1]);

    printf("Line 3: %s.\n", name[2]);


    return 0;
}

FIX:

Thanks to everyone's comments I managed to fix the code. The problem was that I was not reserving one space for the NULL terminator at the end of the array. Since I wanted ten x(s), the solution is to increase the size of the two dimensional array so that it has space for the NULL terminators.

#include <stdio.h>
int main()
{
    char name[4][11];

    for(int i = 0; i < 3; i++){
        for(int ii = 0; ii < 10; ii++){
            name[i][ii] = 'x';
        }
    }


    printf("Line 1: %s.\n", name[0]);

    printf("Line 2: %s.\n", name[1]);

    printf("Line 3: %s.\n", name[2]);


    return 0;
}

You **must** reserve one character for your NUL terminator, taht is `name[i][9]` **must** be 0. — tadman, Apr 07 '21 at 22:07
@tadman ah, you are completely right. I do not know how I missed that. Thanks. Should I take down the question? — Rivf, Apr 07 '21 at 22:10
Possible dupe target : [When does printf(“%s”, char*) stop printing?](https://stackoverflow.com/q/2726301/327083) — J..., Apr 07 '21 at 22:12
You might want to fix it, then post your fix as an answer so you can learn from this mistake. Happens to the best of us! — tadman, Apr 07 '21 at 22:14
`for(int ii = 0; ii < 10; ii++){ name[i][ii] = 'x'; }` --> `for(int ii = 0; ii < 9; ii++){name[i][ii] = 'x'; } name[i][ii] = '\0';` — Support Ukraine, Apr 07 '21 at 22:17
You didn't need to increase the size from 3 to 4. Only the size from 10 to 11 needed to be increased for the null character — risingStark, Apr 07 '21 at 22:30
For your fixed code, you still need to follow 4386427's fix to actually store the nul char. — Craig Estey, Apr 08 '21 at 00:35

Support Ukraine · Accepted Answer · 2021-04-08T06:29:19.237

As it has been discussed in comments, the problem with your code was that you didn't add a NUL to terminate the strings.

However, I'll post an answer as your fix isn't really a fix

Two things to mention:

You don't need to extend the dimension of the array to [4][11] but only to [3][11]. It's only the number of chars in each string that need to be increased to make room for the termination. Changing [3] will give you room for an extra string and that is not what you are looking for. So use [3][11]
Your fix still doesn't add the NUL terminator. Remember that a local variable isn't automatic initialized so you need to do it yourself. In other words - you need some code that explicit sets name[i][10] to NUL.

This is how memory is looking in your original code, your fix and how you really want it.

Below is two ways to get what you want.

Option 1:

char name[3][11];

for(int i = 0; i < 3; i++){
    for(int ii = 0; ii < 10; ii++){
        name[i][ii] = 'x';
    }

    // Add NUL to the end to get a correct C style string
    name[i][ii] = '\0';
}

Option 2:

// Explicit initialize the whole array to zero
char name[3][11] = { 0 };

for(int i = 0; i < 3; i++){
    for(int ii = 0; ii < 10; ii++){
        name[i][ii] = 'x';
    }
}

I prefer option 1 but option 2 will also do the job.

Traversing Two Dimensional Array

1 Answers1