Address of pointers

Question

Looking at the following Code:

char* c = (char*) malloc(sizeof(char)*BUF);

I know that doing

printf("%c\n", *c);

gives me the first character in the string. Also,

printf("%s\n", c);

prints the whole string, where c is the address of the first character. However, what is the meaning of &c? Is it the address of the address? Thanks in advance!

There are 1000s post on StackOverflow related to this. Please do some search — P0W, Jan 27 '17 at 10:02
"*the whole string*" There is no (C-)string in the code you show. — alk, Jan 27 '17 at 10:03
It's much more **faster** to do a search than writing a post. — Meninx - メネンックス, Jan 27 '17 at 10:04
`printf("%s\n", c);` Will invoke undefined behaviour if c is not null terminated. — George, Jan 27 '17 at 10:04
@George: even more (or better "less" ;-), the memory is not initialised at all. — alk, Jan 27 '17 at 10:05
Yup, also while it probably won't cause an issue here, [casting the result of _malloc_](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc), can mask errors/warnings. — George, Jan 27 '17 at 10:06

score 2 · Answer 1 · answered Jan 27 '17 at 10:02

2

&c is the adress of your variable c (which itself contains the adress of your allocated array).

answered Jan 27 '17 at 10:02

rom1v

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Address of pointers

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