C programming strange output, what did I do wrong? - c

Question

#include<stdio.h>
#include<string.h>
int main()
{
int j;
char password[8] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'};
j = strlen(password);
printf("Size = %d\n", j);
return 0;
}
Outputs:
Size = 8
But this code
#include<stdio.h>
#include<string.h>
int main()
{
int j;
char password[8] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'};
char enteredpassword[9];
j = strlen(password);
printf("Size = %d\n", j);
return 0;
}
Outputs:
Size = 14
The difference between the two codes is that unused "enteredpassword[9]" array, is that supposed to change the string length of password[8] from 8 to 14?

Answer 1

strlen expects a null-terminated string. Your character array lacks null terminator
char password[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', '\0'};
// ^^ ^^^^
j = strlen(password);
Calling strlen on a string that is not null-terminated is undefined behavior, meaning that your program could crash or return an unpredictable result. Note how the change removes the hard-coded length of the password array, letting the compiler figure out the correct size.

Answer 2

Your password is not a null-terminated string. If you initialize it conventionally:
char password[] = "abcdefgh";
or
const char *password = "abcdefgh";
then calling strlen on it will give you the expected answer.
(Or, if you're bound and determined to do it the hard way, use
char password[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', '\0'};
or
char password[9] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', '\0'};
.)

Answer 3

Your program has invoked undefined behaviour , therefore , possible explanation is anything can appear as output (if lucky you get seg fault).
char password[8] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'}; //not a C-style string
password is not a null-terminated string and passing it to strlen will cause UB.
j = strlen(password); //will invoke UB
Either write -
char password[9] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h','\0'};
or
char password[]="abcdefgh";

Answer 4

The function strlen() is to be used with strings. In C strings are array of characters terminated by a \0. But your array is not NULL \0 terminated.
Try the following
char password[9] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', '\0'};

Answer 5

Your code has undefined behavior, since you're calling strlen() on arrays of characters that are not properly 0-terminated, i.e. they are not strings.

Answer 6

The normal way to initialize a string in C is something like that:
char mystr[100]="test string";
Because every string ends up with a NULL character ('\0'), so you must initialize your string like the following:
char password[8] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', '\0'};
In this case, your password string was null-terminated and the strlen() will return the correct answer.