Why doesn't fgets take a char pointer as the first argument?

Question

Why does fgets segfault when I pass a char pointer as the first argument? For example:

char *p;
fgets(p, 10, stdin); // segmentation fault

What's with the char pointer p, when we just declare it as char *p (as in the above example)? thanks!

using the exact code as described p point to NULL and fgets() need memory to save the 10 characters — pascal sautot, Oct 18 '21 at 09:15
You always have to answer the question "To what valid block of memory does my pointer point?", e.g. What address does my pointer hold as its value? A few links that provide basic discussions of pointers may help. [Difference between char *pp and (char*) p?](https://stackoverflow.com/a/60519053/3422102) and [Pointer to pointer of structs indexing out of bounds(?)...](https://stackoverflow.com/a/60639540/3422102) (ignore the titles, the answers discuss pointer basics) — David C. Rankin, Oct 18 '21 at 09:16
@pascalsautot, it doesn't point to `NULL`, it's uninitialized. — frippe, Oct 18 '21 at 09:22
I understand why new users post answers to this. But if you are a 10k+ rep and/or C gold badge user coming across a FAQ question like this, the correct action is to open the [C tag wiki](https://stackoverflow.com/tags/c/info), scroll down to "FAQ" to find a list of canonical duplicates, then close vote for the appropriate dupe. In this case one was found below FAQ -> Pointers. — Lundin, Oct 18 '21 at 09:27
@frippe what difference do you make between uninitialiated (value is 0) and assigned by NULL (value is 0) both end up with the same problem a pointer that does not point on a memory space — pascal sautot, Oct 19 '21 at 14:52
@pascalsautot Uninitialized does not imply 0, it means it contains a garbage value and could point _anywhere_ — frippe, Oct 19 '21 at 15:01

frippe · Answer 1 · 2021-10-18T09:23:02.207

1

p contains a garbage value. You need to initialize it first.

char *p = (char*)malloc(10);
if (p == NULL) {
    // handle error
}
//use p

// Once you're done with p
free(p);

or

char p[10];  // Automatic storage

edited Oct 18 '21 at 09:23

answered Oct 18 '21 at 09:11

frippe

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@XMan no need for "thank you comments". If you find an answer useful, then upvote it. – Jabberwocky Oct 18 '21 at 09:27

tstanisl · Answer 2 · 2021-10-18T09:14:12.007

0

Pointer p must point to a buffer.

char buf[10];
char *p = buf;
fgets(p, 10, stdin);

should work now. Note that array expressions decays to pointers so the code could be simpified to:

char p[10];
fgets(p, 10, stdin);

edited Oct 18 '21 at 09:14

answered Oct 18 '21 at 09:12

tstanisl

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thanks, if we just declare "char *p", where the pointer p point to? – XMan Oct 18 '21 at 09:13
_"if we just declare "char *p", where the pointer p point to?"_: it points nowhere, it contains an undetermined value. – Jabberwocky Oct 18 '21 at 09:20

score 0 · Answer 3 · answered Oct 18 '21 at 09:15

fgets can and does take a valid pointer to a character buffer as the first argument. In your case, your char *p is not valid. It has not been initialised and probably points to a junk value.

You can either initialise it to a static string or a character array which contains some sensible value.

score 0 · Answer 4 · answered Oct 18 '21 at 09:20

0

Here p is just a pointer and it doesn't point to anything since it has not been initialized yet. You can initialize it via allocating memory to it.

char *p = (char*)malloc(10);

answered Oct 18 '21 at 09:20

Akasam Ramalakshmi

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Why doesn't fgets take a char pointer as the first argument?

4 Answers4