C: Difference between the two array initializations

Question

Why is there a difference between these two array initializations?

char *message = "HELLO";

if(message[0] == 'H'){
    // true
}

Ok, this works. But this doesn't:

char message[6] = "HELLO";

if(message[0] == 'H'){
    // false
}

Aren't they expressing the same thing?

They are. I am getting the expected behavior with gcc 4.8.2. — R Sahu, Jun 08 '14 at 04:27
It appears that perhaps you gave an incorrect example. Both of those should and do evaluate to true (as well as `if ("HELLO"[0] == 'H')`). — Jason C, Jun 08 '14 at 04:29
it seems like an issue with avr-gcc (on embedded devices), sorry — apscience, Jun 08 '14 at 04:41

score 0 · Answer 1 · answered Jun 08 '14 at 04:27

0

"HELLO" has the type const char *. But due to historical reasons this can be considered as a char *.

char messagd[6] is an array of characters on the stack. A different beast.

You need to use strcpy to populate that array

answered Jun 08 '14 at 04:27

Ed Heal

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score 0 · Accepted Answer · answered Jun 08 '14 at 04:28

In the first code fragment, message is a pointer to a read-only array of characters. You can modify the pointer (e.g. ++message is fine), but you cannot modify what it points to (message[0] = 'X' invokes undefined behavior).

In the second code fragment, message is an array of characters with an initializer. You cannot modify the variable (++message is invalid), but you can modify the contents of the array (message[0] = 'X' is fine).

Also sizeof(message) is likely to be different.

They behave similarly when you simply access message[0]; that is, either you are wrong about your claimed behavior or your compiler has a very surprising bug.

But they are different things and behave differently in several ways, some of which I just enumerated.

C: Difference between the two array initializations

2 Answers2