Printing character array in C

Question

I know I am supposed to put '/o' at end of character array but When I want to print "printf ("%s\n", kk);" , it gives "abcdepqrst". Why is this happening? This is the program I am executing.

#include<stdio.h>
int main()
{

char kk[]={'a','b','c','d','e'};
char s[]="pqrst";

printf("%s\n",s);

printf("%s\n",kk);

}

Output:

pqrst

abcdepqrst

I tried reversing the order in which I declare the array by declaring array 's' before array 'kk' here, ideone link, but I am still getting the same output. I think it has something do with how ideone allocates memory to variables.

#include<stdio.h>
int main()
{
char s[]="pqrst";
char kk[]={'a','b','c','d','e'};


printf("%s\n",s);

printf("%s\n",kk);

}

Output:

pqrst

abcdepqrst

Also see this: http://stackoverflow.com/questions/10943033/why-are-strings-in-c-usually-terminated-with-0 — 7VoltCrayon, Aug 27 '15 at 20:20

score 4 · Accepted Answer · answered Aug 27 '15 at 20:18

4

The printf() function expects a null terminated string but you are passing a character array with no null terminator. Try changing your array to:

char kk[]={'a','b','c','d','e','\0'};

When you use string literal syntax to initialize your s array, the null terminator is automatically added:

char s[] = "pqrst"; // s is {'p','q','r','s','t','\0'}

answered Aug 27 '15 at 20:18

mattnewport

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Please check the edit. – Aditya Varma Aug 27 '15 at 20:23
What's the point of all of these downvotes? – Eugene Sh. Aug 27 '15 at 20:24
@EugeneSh.; Beware of ghost souls!! – haccks Aug 27 '15 at 20:24
Is the behaviour undefined as said by @juanchopanza. – Aditya Varma Aug 27 '15 at 20:26
@c0de222 yes, it is undefined behaviour, typically you will just see whatever is in memory after your non-terminated string until there happens to be a 0 byte or you cause an access violation. In your case it happens the compiler has chosen to place your other string `s` immediately after `kk` in memory. – mattnewport Aug 27 '15 at 20:28

juanchopanza · Answer 2 · 2015-08-27T20:25:17.067

4

kk is not null-terminated, so printf doesn't know where to stop. Invoking printf on a string that is not null-terminated is undefined behaviour (UB). That means you cannot expect any particular outcome from your program. What you see is one manifestation of UB.

You need to add a null-terminator:

char kk[]={'a','b','c','d','e', '\0'};

edited Aug 27 '15 at 20:25

answered Aug 27 '15 at 20:18

juanchopanza

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Okay. So we can not expect a particular output? – Aditya Varma Aug 27 '15 at 20:25
2

@c0de222 Right, UB means pretty much anything can happen. – juanchopanza Aug 27 '15 at 20:25
But practically, it will just print everything in the memory starting at `kk` until the first zero is encountered. Which in your case at the end of `s`. And as the `s` and `kk` are allocated on stack, it is going "backwards". – Eugene Sh. Aug 27 '15 at 20:28

score 0 · Answer 3 · answered Aug 27 '15 at 20:18

0

You need to add a '\0' character to kk as it's last element like so:

char kk[]={'a','b','c','d','e','\0'};

or printf is just going to run off the end of the array.

answered Aug 27 '15 at 20:18

7VoltCrayon

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Please check the edit. – Aditya Varma Aug 27 '15 at 20:23
I commented on your question with a link to an answer which clears the concept. – 7VoltCrayon Aug 27 '15 at 20:24
Thanks for the link. – Aditya Varma Aug 27 '15 at 20:32

Printing character array in C

3 Answers3