Assignment after initialization to specific index in an array - c

Question

After assigning 26th element, when printed, still "Computer" is printed out in spite I assigned a character to 26th index. I expect something like this: "Computer K "
What is the reason?
#include <stdio.h>
int main()
{
char m1[40] = "Computer";
printf("%s\n", m1); /*prints out "Computer"*/
m1[26] = 'K';
printf("%s\n", m1); /*prints out "Computer"*/
printf("%c", m1[26]); /*prints "K"*/
}

Answer 1

At 8th index of that string the \0 character is found and %s prints only till it finds a \0 (the end of string, marked by \0) - at 26th the character k is there but it will not be printed as \0 is found before that.

Answer 2

char s[100] = "Computer";
is basically the same as
char s[100] = { 'C', 'o', 'm', 'p', 'u','t','e','r', '\0'};
Since printf stops when the string is 0-terminated it won't print character 26

Answer 3

Whenever you partially initialize an array, the remaining elements are filled with zeroes. (This is a rule in the C standard, C17 6.7.9 §19.)
Therefore char m1[40] = "Computer"; ends up in memory like this:
[0] = 'C'
[1] = 'o'
...
[7] = 'r'
[8] = '\0' // the null terminator you automatically get by using the " " syntax
[9] = 0 // everything to zero from here on
...
[39] = 0
Now of course \0 and 0 mean the same thing, the value 0. Either will be interpreted as a null terminator.
If you go ahead and overwrite index 26 and then print the array as a string, it will still only print until it encounters the first null terminator at index 8.
If you do like this however:
#include <stdio.h>
int main()
{
char m1[40] = "Computer";
printf("%s\n", m1); // prints out "Computer"
m1[8] = 'K';
printf("%s\n", m1); // prints out "ComputerK"
}
You overwrite the null terminator, and the next zero that happened to be in the array is treated as null terminator instead. This code only works because we partially initialized the array, so we know there are more zeroes trailing.
Had you instead written
int main()
{
char m1[40];
strcpy(m1, "Computer");
This is not initialization but run-time assignment. strcpy would only set index 0 to 8 ("Computer" with null term at index 8). Remaining elements would be left uninitialized to garbage values, and writing m1[8] = 'K' would destroy the string, as it would then no longer be reliably null terminated. You would get undefined behavior when trying to print it: something like garbage output or a program crash.

Answer 4

In C strings are 0-terminated.
Your initialization fills all array elements after the 'r' with 0.
If you place a non-0 character in any random field of the array, this does not change anything in the fields before or after that element.
This means your string is still 0-terminated right after the 'r'.
How should any function know that after that string some other string might follow?

Answer 5

That's because after "Computer" there's a null terminator (\0) in your array. If you add a character after this \0, it won't be printed because printf() stops printing when it encounters a null terminator.

Answer 6

Just as an addition to the other users answers - you should try to answer your question by being more proactive in your learning. It is enough to write a simple program to understand what is happening.
int main()
{
char m1[40] = "Computer";
printf("%s\n", m1); /*prints out "Computer"*/
m1[26] = 'K';
for(size_t index = 0; index < 40; index++)
{
printf("m1[%zu] = 0x%hhx ('%c')\n", index, (unsigned char)m1[index], (m1[index] >=32) ? m1[index] : ' ');
}
}