Question

I am a little confused by the following C code snippets:

printf('Peter string is %d bytes\n', sizeof('Peter')); // Peter string is 6 bytes 

This tells me that when C compiles a string in double quotes, it will automatically add an extra byte for the null terminator.

printf('Hello '%s'\n', 'Peter'); 

The printf function knows when to stop reading the string ‘Peter’ because it reaches the null terminator, so …

char myString[2][9] = {'123456789', '123456789' }; printf('myString: %s\n', myString[0]); 

Here, printf prints all 18 characters because there’s no null terminators (and they wouldn’t fit without taking out the 9’s). Does C not add the null terminator in a variable definition?

Answer 1

Your string is [2][9]. Those [9] are [‘1’, ‘2’, etc… ‘8’, ‘9’]. Because you only gave it room for 9 chars in the first array dimension, and because you used all 9, it has no room to place a ‘\0’ character. redefine your char array:

char string[2][10] = {'123456789', '123456789'}; 

And it should work.