CODEWITHSUNDEEP
cstringmemory-addressc-strings
Casey Kuball
Casey Kuball

Reputation: 7955

Is &"string" the same address as "string"?

I have some generated C code that looks like this:

char *example[] = {&" ",&"\n", &"\t"};

If I were to write this myself, I would drop the ampersands (&), as string literals are already of type char *. To me, it looks like example should have been generated declared as a char**[].

Can I be sure that the pointer is the same with or without the &s? Is this taking the address of an address well defined in C?

Edit: I am investigating a warning on some software currently in a lifecycle that doesn't accept any changes. Is printf("%p", &"hello world"); always the same as printf("%p", "hello world");, or is this dependent on the compiler?

Upvotes: 2

Views: 159

Answers (1)

ouah
ouah

Reputation: 145829

Is &“string” the same address as “string”?

No, their type is different.

 "hello world"

object has type char [12]. As for other arrays in an expression context, it is converted to char *.

But:

 &"hello world"

has type char (*)[12].

So in your example, it also means that:

char *example[] = {&" ",&"\n", &"\t"};

is an invalid declaration as the array elements in the initializer list are not of type char *.

This is the correct declaration:

char *example[] = {" ", "\n", "\t"};

for the sake of this question, if you had to use the &, here is the correct declaration:

char (*example[3])[] = {&" ",&"\n", &"\t"};

Upvotes: 7

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