Why does that works in c++ char arrays

Question

I have the next code:

    char arr[6] = "Hello";
    strcpy(arr, "Hello mellow");
    cout << strlen(arr) << ", " << arr << endl; // 12, Hello mellow


    char arr1[] = "Hello";
    strcpy(arr1, "Hello mellow");
    cout << strlen(arr1) << ", " << arr1 << endl; // 12, Hello mellow

So, why does that work? Why doesn't it get limited somehow? Whatever I put instead of "Hello mellow", it works and prints it out.

it sometimes works, but also may not or even crash the program. — SHR, Dec 01 '14 at 15:31
`Why doesn't it get limited somehow?` That question you asked is the answer to your question. What do you mean by "limited somehow"? Crashing? A message bos popping up saying "it's a bad thing you did"? Nothing happening? That's what is meant by undefined behavior. None of us, even yourself knows what "limited somehow" means -- it could mean anything to anyone. — PaulMcKenzie, Dec 01 '14 at 15:44

Mathieu Rodic · Answer 1 · 2014-12-01T15:49:19.880

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In general, C and C++ don't check for boundaries (unlike other higher-level languages: Java, PHP, Python, Javascript, etc).

This means that if you try to strcopy, say, a 13-bytes string such as "Hello mellow" to a character array, it won't check whether or not the given array has been instantiated with enough memory to contain the string. It will just copy the given string, character by character, to the given memory pointer.

What happens here, is that you write at some places in memory you are not supposed to access; once in a while, this program might just crash, with no other indication than: segmentation fault.

If you happen to try this...

char arr1[8];
char arr2[8];
strcpy(arr1,"Hello mellow");
printf("%s\n", arr1);
printf("%s\n", arr2);

...it is very likely (but not 100% sure, see comments) you would get the following output:

Hello mellow
llow

Why? Because the second char[] would have been overwritten by the data you tried to put in the first one, without it having enough reserved space for it.

See: http://en.wikipedia.org/wiki/Stack_buffer_overflow

edited Dec 01 '14 at 15:49

answered Dec 01 '14 at 15:30

Mathieu Rodic

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Why do you mention stack overflow? – doctorlove Dec 01 '14 at 15:33
or any other output because it is UB – No Idea For Name Dec 01 '14 at 15:34
http://accu.org/index.php/journals/1857 - "Demons May Fly Out Of Your Nose" – doctorlove Dec 01 '14 at 15:36
@NoIdeaForName which is why I wrote "very likely" :) However, most compilers reserve memory blocks next to each other. – Mathieu Rodic Dec 01 '14 at 15:36
@doctorlove : very good article :) – Mathieu Rodic Dec 01 '14 at 15:42

score 4 · Accepted Answer · answered Dec 01 '14 at 15:31

4

It works because strcpy doesn't check that the destination array is at least as large as the source one. Your code invokes undefined behavior as you call strcpy with invalid arguments, and because the behavior is undefined, anything can happen; In your case, the memory is silently overwritten. Your program could crash as well.

answered Dec 01 '14 at 15:31

Columbo

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Well, most modern (all?) OSs won't crash on this. – Bart Friederichs Dec 01 '14 at 15:32
@BartFriederichs Probably not. I'm not very familiar with the internal working. – Columbo Dec 01 '14 at 15:32
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@BartFriederichs still UB – No Idea For Name Dec 01 '14 at 15:33
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@Columbo, because they use something called "protected memory" where each program runs in its own part of memory. The program may crash, but the PC probably won't. – Bart Friederichs Dec 01 '14 at 15:33
Definitely UB, but certainly no OS crash. If it did, that would be a very easy way to DoS machines. – Bart Friederichs Dec 01 '14 at 15:34

score 1 · Answer 3 · answered Dec 01 '14 at 15:35

Native arrays in C/C++ are very low-level abstractions that are treated as pointers to memory locations in many use cases. So, when passing arr to strcpy, all strcpy knows is the address of arr[0]. As a result, there is no possibility of bounds checking. This is a very good thing for performance reasons. It is up to the programmer to ensure that he/she uses these low-level constructs safely, for instance by using strncpy and giving an appropriate bound, or using std::vector and checking for bounds explicitly or using std::vector::at to check bounds when accessing a location.

score 0 · Answer 4 · answered Dec 01 '14 at 15:32

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I think that's because there's no check runtime nor compile time and if you're Lucky, you won't get a segmentation fault;)

answered Dec 01 '14 at 15:32

tmnd91

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**Unlucky** to not get a segmentation fault in my opinion. – eerorika Dec 01 '14 at 15:32

Why does that works in c++ char arrays

4 Answers4