Difference between string created by array and pointer?

Question

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What is the difference between char s[] and char *s in C?

What is the difference between this:

char arr[] = "Hello, world!";

And this:

char *arr = "Hello, world!";

Where do both strings' memory get allocated? Why am I not able to change content of latter string?

Strictly speaking, the latter should read `const char *arr = "hello world!";` — Nemo, Sep 14 '11 at 04:39
@Nemo: It *should*, but strictly speaking C doesn't require it. C string literals are not `const`, but attempting to modify one is undefined behavior — Keith Thompson, Sep 14 '11 at 05:25
@teacher: This question has been asked many times on SO. A quick search gave these links. (a) http://stackoverflow.com/questions/1704407/what-is-the-difference-between-char-s-and-char-s-in-c (b) http://stackoverflow.com/questions/1880573/c-difference-between-char-var-and-char-var — Bhaskar, Sep 14 '11 at 08:02

Mateen Ulhaq · Answer 1 · 2011-09-14T04:53:56.073

5

The first one is writable memory allocated specifically for arr, which is an array of chars. You can modify it without invoking undefined behavior. This is completely legal:

char arr[] = "Hello, world!";
arr[1] = 'i';

The second one is a pointer to a read-only string. Therefore, this is undefined behavior:

char *parr = "Hello, world!";
parr[1] = 'i'; // Cannot write to read-only memory!

In some compiler implementations:

char *a = "Hello, world!";
char *b = "Hello, world!";

a[1] = 'i';
// b[1] == 'i';

This isn't guaranteed - I'm only including it to give you an 'intuitive' feel of why it's undefined behavior.

edited Sep 14 '11 at 04:53

answered Sep 14 '11 at 04:35

Mateen Ulhaq

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are both strings created on stack? – teacher Sep 14 '11 at 04:36
No, the read-only string is stored in a read-only part of the memory of you're process. – Simon Sep 14 '11 at 04:37
@teacher: depends on where you declare it, and in the second example the actual string is in readonly memory. – Ed S. Sep 14 '11 at 04:37
@Ed S. i am using both inside a function... – teacher Sep 14 '11 at 04:38
1

@teacher - `arr` resides on stack but the value pointed by `arr` resides on read only location whose location is implementation dependent. – Mahesh Sep 14 '11 at 04:42
1

err..No one mentions Second is called a **String Literal**? – Alok Save Sep 14 '11 at 04:50
@Mehash: No, the contents of `arr` resides on the stack. Otherwise it would be not be modifiable. – Ed S. Sep 14 '11 at 04:56
For the char* case you can actually do parr[i] = 'i'; but on some compilers it will print the original string :) – Jesus Ramos Sep 14 '11 at 05:13
@Jesus: For the char* case, `parr[i] = 'i'` has undefined behavior. It can cause your program to crash (if you're lucky). – Keith Thompson Sep 14 '11 at 05:27
@Als: In both cases `"hello, world"` is a string literal. – Keith Thompson Sep 14 '11 at 05:27
@Keith just tried in GCC, doesn't crash and prints the original string http://ideone.com/xtf9r – Jesus Ramos Sep 14 '11 at 05:28
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@Jesus: That's one of the infinitely many possible results of undefined behavior. I just tried this with gcc: `char *s = "hello, world"; s[0] = 'H';` and the program died with a segmentation fault. – Keith Thompson Sep 14 '11 at 05:30
@Keith hence the undefined part, I was just bringing up a point that it may seem like an odd bug if that occurs because of the behavior and the person not understanding how it works. – Jesus Ramos Sep 14 '11 at 05:31

Difference between string created by array and pointer?

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