Why is strlen working fine without '\0'?

Question

As title, I have some question using char* in c. For example, if I write this

char *a = calloc(5, 5);
a[0] = '1';
a[1] = '1';
a[2] = '1';
a[3] = '1';
a[4] = '1';
printf("a = %s, length = %d", a, strlen(a));

and the output is

a = 11111, length = 5

Why is strlen working fine without '\0'? Can someone help me understand?

`char *a = calloc(5, 5)` corresponds to `char a[25] = {0};`. — BLUEPIXY, Apr 26 '17 at 04:38
Moving target is hard to answer. Especially after an [answer](http://stackoverflow.com/a/43625150/2410359). Maybe things will change again in the next five minutes. — chux - Reinstate Monica, Apr 26 '17 at 04:41
"Why is strlen working fine without '\0'?" --> What do you think is specified to happen? Programmer broke the rules, why should the compiler create code that catches the mistake? Welcome to the world of undefined behavior. — chux - Reinstate Monica, Apr 26 '17 at 04:44
@tonylin Never modify a question, whatever the case might be. — , Apr 26 '17 at 04:46
@chux, I'm old school apparently... old compilers, old man, old format specifiers; I'll now remember what `%zu` said though :) Is there a reason for `%zu` over `%lu`? my compiler says use `%lu`, on this end. — l'L'l, Apr 26 '17 at 04:50
If you've got an answer and need to change the question, change the question without completely invalidating the answer — add revised code. Since the revised code invokes undefined behaviour (causing `strlen()` to read outside the bounds of the 5-byte array allocated by `calloc()`), any result is OK — crash, 5, or any other answer, or other behaviour, at all. — Jonathan Leffler, Apr 26 '17 at 04:57
@l'L'l `%lu` for `sie_t` is plain wrong without a cast. For example in 64-bit Windows `long int` is still 32-bits but `size_t` 64-bits, so anything can happen. — Antti Haapala -- Слава Україні, Apr 26 '17 at 05:13
@AnttiHaapala: Interesting that the compiler suggests that: `format specifies type 'int' but the argument has type 'unsigned long'` `%d ~~ %lu`; good to know - thanks. — l'L'l, Apr 26 '17 at 05:57
@l'L'l no, `%zu` is the correct one. `strlen` returns `size_t`. In 32 and 64-bit windows `%zu` corresponds to `%llu` — Antti Haapala -- Слава Україні, Apr 26 '17 at 07:09
@AnttiHaapala: I'm not using windows, but linux, although I presume what you mention would be relative for both? — l'L'l, Apr 26 '17 at 17:20

score 3 · Accepted Answer · answered Apr 26 '17 at 04:37

3

calloc(5, 5) allocates and zeroes 25 bytes. You assign the first five of these, but the sixth is still '\0'.

answered Apr 26 '17 at 04:37

aschepler

70,891
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Sorry, I typed wrong. It's calloc(5,sizeof(char)) – Tony Lin Apr 26 '17 at 04:44
@tonylin if it's `calloc(5,sizeof(char))` then you get undefined behaviour. `calloc(5,sizeof(char))` allocated 5 bytes. It seems to work in your case because the byte immediately following the 5 allocated bytes is 0 by chance, but actually it's value is indetermined, so if you run your program on another computer or if you compile it with another compiler or maybe if you just reboot your computer it may very well not work any more. – Jabberwocky Apr 26 '17 at 07:56

Why is strlen working fine without '\0'?

1 Answers1