How does strlen(&string[]) work? - c

Question

This might be a bit long question. I was testing some character arrays in C and so came along this code.
char t[10];
strcpy(t, "abcd");
printf("%d\n", strlen(&t[5]));
printf("Length: %d\n", strlen(t));
Now apparently strlen(&t[5]) yields 3 while strlen(t) returns 4.
I know that string length is 4, this is obvious from inserting four characters. But why does strlen(&t[5]) return 3?
My guess is that
String: a | b | c | d | 0 | 0 | 0 | 0 | 0 | \0
Position: 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
strlen(&t[5]) looks at the length of a string composed of positions 6, 7 and 8 (because the 10th character is a NULL terminating character, right)?
OK, then I did some experimentation and modified a code a bit.
char t[10];
strcpy(t, "abcdefghij");
printf("%d\n", strlen(&t[5]));
printf("Length: %d\n", strlen(t));
Now this time strlen(&t[5]) yields 5 while strlen(t) is 10, as expected. If I understand character arrays correctly, the state should now be
String: a | b | c | d | e | f | g | h | i | j | '\0'
Position: 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
so why does strlen(&t[5]) return 5 this time? I've declared a character array of length 10, should then, by the same logic applied above, the result be 4?
Also shouldn't I be running into some compiler errors since the NULL terminating character is actually in the 11th spot? I'm new into C and would very much appreciate anyone's help.

Answer 1

First let me tell you, your "assumption"
String: a | b | c | d | 0 | 0 | 0 | 0 | 0 | \0
Position: 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
is not correct. Based on your code, The values are only "guaranteed" up to index 4, not beyond that.
For the first case, in your code
printf("%d\n", strlen(&t[5]));
is wrong for various reasons,
you ought to use %zu for a size_t type.
&t[5] does not point to a valid string.
Any (or both) of the above causes undefined behavior and any output cannot be justified.
To elaborate, with a defintion like
char t[10];
strcpy(t, "abcd");
you have index 0 to 3 populated for t, and index 4 holds the null-terminator. The content of t[5] onward, is indeterminate.
Thus, &t[5] is not a pointer to the first element of a string, so cannot be used argument to strlen().
It may run out of bound in search of the null-terminator and experience invalid memory access and, as a side-effect, produce a segmentation fault,
It may find a null-terminator (just another garbage value) within the bound and report a "seemingly" valid length.
Both are equally likely and unlikely, really. UB is UB, there's not justifying it.
Then, for the second case, where you say
char t[10];
strcpy(t, "abcdefghij");
is once again, accessing memory out of bound.
You have all together 10 array elements to store a string, so you can have 9 other char elements, plus one null-terminator (to qualify the char array as a string).
However, you're attempting to put 10 char elements, plus a null character (in strcpy()), so you're off-by-one, accessing out of bound memory, invoking UB.

Answer 2

char t[10]; is not initialized so it just contains garbage values 1). strcpy(t, "abcd"); overwrites the first 5 characters with the string "abcd" and a null terminator.
However, &t[5] points at the first character after the null termination, which remains garbage. If you invoke strlen from there, anything can happen, since the pointer passed is not likely pointing at a null terminated string.
1) Garbage = indeterminate values. Assuming a sane 2's complement system, the address of the buffer t is taken, so the code does not invoke undefined behavior until the point where strlen starts reading outside the bounds of the array t. Reference.

Answer 3

Problem 1:
My guess is that
String: a | b | c | d | 0 | 0 | 0 | 0 | 0 | \0
Position: 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
This assumption is wrong.
The array is not initialized to hold 0 values but contains some "random" garbage.
After copying "abcd" the upper half of the array (t[5] etc.) is still untouched resulting in a "random" length of the string due to undefined behaviour.
Problem 2:
If I understand character arrays correctly, the state should now be
String: a | b | c | d | e | f | g | h | i | j | '\0'
Position: 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
Again wrong.
Your array only holds 10 characters. Theyare at index 0..9. Index 10 is out of bounds.
Your copy operation might result in this layout or it might as well just crash while writing out of bounds.
But this is not checked by the compiler. If you run into problems then it will be during runtime.