C programming: Array prints two characters on its own

Question

I'm trying to print a 2 character string. This is the part of that code.

char arraytwo[3]; 
// 2 characters 
for (i = 'a'; i <= 'z'; i++) 
{
    arraytwo[0] = i; 
    for (j = 'a'; j <= 'z'; j++) 
    {
        arraytwo[1] = j; 
        printf("%s\n", arraytwo);
    }
}

The output I am getting is this. For some reason it keeps adding "AZ" at the end of each iteration. What am I missing?

aaAZ
abAZ
acAZ
adAZ
aeAZ
afAZ
agAZ
ahAZ
aiAZ
ajAZ
akAZ

%s means to print a null-terminated string ... not just two characters — M.M
– M.M, Commented May 2, 2017 at 6:01
How are you expecting printf to know that you want it to print two characters? How do you think this is going to work? — David Schwartz
– David Schwartz, Commented May 2, 2017 at 6:02

Sourav Ghosh · Accepted Answer · 2017-05-02 06:11:05Z

5

What you're missing is the definition of a string, it has to be null-terminated, by definition.

Quoting C11, chapter §7.1.1, (emphasis mine)

A string is a contiguous sequence of characters terminated by and including the first null character. [....]

In your case, for arraytwo,

it's automatic storage, and not initialized explicitly.
you did not null-terminate it manually.

So, technically, arraytwo is not a string.

In this usage, as an argument to %s format specifier, out of bound access happens in search of the null-terminator, which causes undefined behavior.

Also quoting chapter §7.21.6.1

s

If no l length modifier is present, the argument shall be a pointer to the initial element of an array of character type.²⁸⁰⁾ Characters from the array are written up to (but not including) the terminating null character. [....]

Solution:

Either initialize the array elements to 0, something like char arraytwo[3] = {0};
Or, manually null terminate your array, like arraytwo[2] = '\0';

before using the array as string.

edited May 2, 2017 at 6:11

answered May 2, 2017 at 6:01

Sourav Ghosh

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5 Comments

sd10 Over a year ago

Thank you for pointing to the right direction. Terminated the string with arraytwo[2] = '\0'; and it now printed the characters correctly.

Sourav Ghosh Over a year ago

@zohan Well, that's good, however, I've just added one more alternative, please refer to the updated answer. :)

sd10 Over a year ago

@Sourabh Thanks. Which book these chapters are from? I need to read it detail. :)

Sourav Ghosh Over a year ago

@zohan well, they are not from any book, the quotes are from C standard, C11.

sd10 Over a year ago

@Sourabh Thanks :)

Martin B. · Accepted Answer · 2017-05-02 06:08:24Z

2

You were on a good way making the array with a length of 3. But you didn't initialize it to be all 0s. So at the beginning of your snipped, the array contains some random garbage and after the array also comes some random stuff.

When you pass a pointer to a printf("%s") function call, printf will output the memory starting with the pointed-to value and incrementing the pointer until it hits a '\0'.

In your case this already happend after 2 random characters. But it does not have to. And if the bell-ringing character (it was '\b' I believe) is there too in the random part of your printf call, your computer might even start to beep.

answered May 2, 2017 at 6:08

Martin B.

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1 Comment

sd10 Over a year ago

Thank you Mike. Understood.

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