Why address(&) give random variable with array?-CodePudding

char a[] = {'A'};
printf(This is random value %c", &a[0] );

CodePudding user response：

This program invokes undefined behavior, as %c is not the correct conversion specifier for an address. Use %p, and cast the argument to (void*).

Note: In case the argument is of type char*, casting is optional, but for any other type, the casting is necessary.

CodePudding user response：

You are printing the address incorrectly.

%c is for printing characters. To print an address use %p and cast the pointer argument (i.e. the address) to void-pointer. Like

printf(This is random value %p", (void*)&a[0] );

The C standard does not define what is supposed to happen for your program. So in principal anything may happen and random values could be the result. No one can tell for sure what your code wil do (without having expert level knowledge of the specific system you are using).

However, on most systems your code will take 1 byte from the pointer and print it as-if it was a character. So the (likely) reason for your "random characters" is that the address of the array is different every time you start the program. And that is exactly what many systems do...

They use "Address Space Layout Randomization". This basically means that the address of things in your program (here the array) are randomly determined during program start up. The idea is to make it harder for hackers to exploit your program.