character pointer

Question

Why this code does not work?

int main(){
  char *str ="abcde";
   scanf("%s",str);
  printf("%s",str);
}

but this works?

int main(){
  char str[] ="abcde";
   scanf("%s",str);
  printf("%s",str);
}`

This question is answered by a [SOFAQ entry](http://loungecpp.wikidot.com/faq#stringliterals) — Seth Carnegie, Feb 19 '12 at 01:20

score 5 · Accepted Answer · answered Feb 19 '12 at 00:34

5

In the first code, you declare a pointer, which points to a string literal: "abcde".
This might be a constant, and you will not be able to change it.

The second code is declaring an array and filling it with ['a','b',c','d','e','\0'], and is not a constant - so you can change it.

answered Feb 19 '12 at 00:34

amit

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can the input for str[] case take more values than the size with which it initialized(6 in this case). – code4fun Feb 19 '12 at 00:38
@gaurav: I did not understand the question [in comment], can you clarify it? – amit Feb 19 '12 at 00:39
I think he's asking if the array is of size 6 can the input take more than 6 characters. Which the answer is yes and you must carefully check for that. – Timeout Feb 19 '12 at 00:41
yeah.i was asking that coz we are not specifying the size here..it seems it accepts few more characters unitl there is no more free continuous space – code4fun Feb 19 '12 at 00:48
@gaurav: You are guaranteed to have array of the size of the string [including the '\0']. Everything other then that - is undefined behavior - it might work and it might fail. – amit Feb 19 '12 at 00:52

score 4 · Answer 2 · edited May 23 '17 at 11:48

4

Because char *str ="abcde"; is a pointer to string literal which is most likely stored in read-only memory.

char str[] ="abcde"; is an array initialized with "abcde".

You should also check out Difference between char* and char[]

edited May 23 '17 at 11:48

Community

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answered Feb 19 '12 at 00:35

LihO

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String literals are not `const`. They just (may be) in read-only. – asaelr Feb 19 '12 at 00:48
*"An ordinary string literal has type "array of n const char" and static storage duration."* – LihO Feb 19 '12 at 00:59
2

That's C++ standard. We are talking about C. See http://stackoverflow.com/a/2245983/1055952 – asaelr Feb 19 '12 at 01:05
@asaelr: Nice one. I thought it's same in C. I have edited my answer, it should be OK now. – LihO Feb 19 '12 at 01:08
1

Note that "read-only memory" in this context probably doesn't mean hardware ROM; it's more likely just a segment of memory that's write-protected by the OS. (Though it could be actual ROM on an embedded system.) – Keith Thompson Feb 19 '12 at 01:22

score 1 · Answer 3 · answered Feb 19 '12 at 00:36

When string value is directly assigned to a pointer, it’s stored in a read only block(generally in data segment) that is shared among functions
char *str  =  "GfG";
...
char str[] = "GfG";  /* Stored in stack segment like other auto
variables */  *(str+1) = 'n';   /* No problem: String is now GnG */

http://www.geeksforgeeks.org/archives/5328

character pointer

3 Answers3