Weird behavior with stack and heap while using strncpy

Question

I found a very interesting question.

When I'm using following code:

int main() { char * in = "hi, "; char str[10]; strncpy(str, in, 2); printf("output = %s", str); return 0; }

My result is nothing, the printf didn't work.

But if I use this:

int main() { char * in = "hi, "; char * str = malloc(sizeof(char) * 10) ; strncpy(str, in, 2); printf("output = %s", str); return 0; }

I can get what I expect.

Why does this happen? Is it because of stack and heap? How exactly does that makes this huge difference?

This code compiles just fine. Are you sure that you didn't make some other typo? — templatetypedef
– templatetypedef, Commented Jun 11, 2012 at 20:51
Im using ideone.com to do the compiling which makes it says can't be compilied. I am still little bit confused and my computer did not have a gcc complier. Anyone can help me compile or give me a hand. Thanks so much. — windsound
– windsound, Commented Jun 11, 2012 at 20:55
oh, my bad. The way Im using stack will result a nothing comes out. The printf statement didnt work — windsound
– windsound, Commented Jun 11, 2012 at 20:57
man 3 strncpy ... BTW: threre are few case where you actually need strncpy(). This is not one of them. — wildplasser
– wildplasser, Commented Jun 11, 2012 at 21:01

Community · Accepted Answer · 2017-05-23 12:21:14Z

The problem - in both cases - is that your string won't be properly terminated after your call to strncpy.

You specify that 2 characters are copied, and your source string in has a length of 4. So strncpy will copy 2 characters, and since that is less than the length of the source string no null terminator will be added - to understand why this is so, review the docs for strncpy:

If count is reached before the entire string src was copied, the resulting character array is not null-terminated.

In this case you'll want:

str[2] = '\0';

after the 'strncpy'.

The second case may seem to work because the buffer that you get from malloc happens to be initialized to all zeros, but you shouldn't rely on this.

Please review the docs for strncpy, noting the exception for null termination, and in general, be careful with string termination!

For even more detail see: Why are strings in C++ usually terminated with '\0'?

Well, it still like before, it complies but nothing comes out.

P.P · Accepted Answer · 2012-06-11 20:55:10Z

The code compiles fine. The run-time error could be because, you haven't terminated the str with null.

From the man page:

The strncpy() function is similar, except that at most n bytes of src are copied. Warning: If there is no null byte among the first n bytes of src, the string placed in dest will not be null terminated.

Add str[2]=0; after the strncpy().

the result I got is: ideone.com/2V0ZE#view_edit_box it is working but can't print anything out. How so though???
It should be either str[2]=0; or str[2]='\0'; double quote is for string literal.
Thanks a lot. I made so many mistakes. Should I delete this post? I fixed out.

Nathan · Accepted Answer · 2012-06-12 04:57:00Z

You could initializing the memory with zeros to be on the "safe" side.

Have a look at the following code,

int main() { char * in = "hi, "; char str[10]={0}; strncpy(str, in, 2); printf("output = %s", str); return 0; } int main() { char * in = "hi, "; char * str = calloc(10,sizeof(char)); strncpy(str, in, 2); printf("output = %s", str); free(str); //<<Important step return 0; }

str[2] = '\0'; is also a valid solution.

Thanks for helping, actually, my biggest mistake is that I did double qouted instead of single quote on second edition. Stupid mistake I made causes this XD

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