Retrieve info about the &&x which keep the address of &x

Question

I'd like to explore the feats of pointer with the following codes:

#include <stdio.h>
int x = 3;
int main(void)
{
    printf("x's value is %d, x's address is %p", x, &x);
    //printf("x's address is stored in", &&x);
}

It works properly and get outputs

$ ./a.out
x's value is 3, x's address is 0x10b1a6018

When I utilize &x, a memory space is set aside for it to keep the address 0x10b1a6018, so an address is printed.

Sequently, I intent to get info about the address which store another address.

#include <stdio.h>
int x = 3;
int main(void)
{
    printf("x's value is %d, x's address is %p", x, &x);
    printf("x's address is stored in", &&x);
}

But it report error as:

$ cc first_c_program.c 
first_c_program.c:14:40: warning: data argument not used by format string [-Wformat-extra-args]
    printf("x's address is stored in", &&x);
           ~~~~~~~~~~~~~~~~~~~~~~~~~~  ^
first_c_program.c:14:42: error: use of undeclared label 'x'
    printf("x's address is stored in", &&x);
                                         ^
1 warning and 1 error generated.

How could I retrieve information about the memory address which store the address of value x.

"When I utilize &x, a memory space is set aside for it " No, there is no memory reserved to store that address. It is only temporarily evaluated and passed to `printf`. This might be in a register. You cannot get the address of this value. — Gerhardh, Oct 19 '18 at 07:53
I love your comment "This might be in a register", cos, it should not be a ghost. @Gerhardh, could you please transmit it to answer. — AbstProcDo, Oct 19 '18 at 08:03
I guess C standard should claim, variable as direct variable, pointer as indirect variable to eliminate the confusions for the newbie. — AbstProcDo, Oct 19 '18 at 08:29

score 5 · Answer 1 · answered Oct 19 '18 at 07:52

5

&x is a temporary value. It is not stored in memory and does not have an address.

Similarly &42 is invalid because 42 is not stored in memory (it's also a temporary value).

The slightly weird error message you get is because gcc implements a unary && operator, which can be used to get the address of a label. This is a non-standard extension of the C language.

To get a better error message, use & &x (with a space) or &(&x).

But what you want simply doesn't exist.

answered Oct 19 '18 at 07:52

melpomene

84,125
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1

much better now , `first_c_program.c:14:43: error: cannot take the address of an rvalue of type 'int *'`, thank you. – AbstProcDo Oct 19 '18 at 08:24

Antti Haapala -- Слава Україні · Accepted Answer · 2018-10-19T17:01:22.397

4

&x is a non-lvalue expression. It does not have a location. & requires an lvalue as its operand (or a function).

&&x is not a valid construct at all in standard C. The address is not necessarily stored anywhere at all - in your program as x is a global variable, it is a compile/link-time constant!

Instead GCC uses unary &&x as an extension to get the address of jump label x for computed goto. This does not break C conformance by itself as && is parsed as a single token and && is not allowed as an unary operator in standard C.

I.e. you can write

    static void *array[] = { &&foo, &&bar, &&hack };

    int i = 1;
    goto *array[i]; // jump to label bar

    ...

foo: ...
bar: ...
baz: ...

edited Oct 19 '18 at 17:01

answered Oct 19 '18 at 07:52

Antti Haapala -- Слава Україні

129,958
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literally, could I understand it as only the lvalue set aside a memory space? – AbstProcDo Oct 19 '18 at 08:00
1

@riderdragon No, &x gives you the address where x is stored. It already has an address whether you use &x or not. – nos Oct 19 '18 at 08:03
1

@nos that's not exactly correct either (or at all), if we consider local variables etc. – Antti Haapala -- Слава Україні Oct 19 '18 at 08:03
1

@riderdragon Yes, but if you declare a variable `register`, it won't have an address either. It's weird that way. – melpomene Oct 19 '18 at 08:04
1

@AnttiHaapala According to C it is, and the compilers that can optimize that away is an implementation detail. – nos Oct 19 '18 at 08:04
1

@nos the C does not say that a variable has an address. It says it has a constant address in that 2 pointers to it compare equal. – Antti Haapala -- Слава Україні Oct 19 '18 at 08:06
"it is a compile/link-time constant!", but &x is a pointer, a pointer has its address itself rather a variable which does not not has an address itself compiled in symbol table. – AbstProcDo Oct 19 '18 at 08:07
1

If you count 2 + 3 mentally, and mark the result on paper, where did you store the 2? – Antti Haapala -- Слава Україні Oct 19 '18 at 08:13
1

If you see that the paper is on a table, where did you write that the paper is on the table? – Antti Haapala -- Слава Україні Oct 19 '18 at 08:14
1

@riderdragon A pointer is a variable that holds some address. `&x` yields a value which is the address of `x`. If you put this value into a pointer, you can take the address of that pointer similar to how you took the address of `x`. – Swordfish Oct 19 '18 at 08:14
I guess I got it now, pointer is a variable in its first place. @Swordfish – AbstProcDo Oct 19 '18 at 08:15
@ 1: i put the `2` in the 485,123th brain cell from the left. 2: i don't have to write down where the paper is as long as i can see it. – Swordfish Oct 19 '18 at 08:16
@Swordfish emphatically **no**. `&x` is an expression that has a pointer type. – Antti Haapala -- Слава Україні Oct 19 '18 at 08:16
1

@AnttiHaapala How does my sentence "`&x` yields a value which is the address of `x`" contradict or deny that this value has a type? – Swordfish Oct 19 '18 at 08:18
@riderdragon `&x` is a non-lvalue expression. It does not have an object whose address you could take. – Antti Haapala -- Слава Україні Oct 19 '18 at 17:00

score 1 · Answer 3 · answered Oct 19 '18 at 07:53

1

To do what you want, you need to store &x in a variable:

int* y = &x;
printf("%p", &y);

answered Oct 19 '18 at 07:53

Code-Apprentice

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Retrieve info about the &&x which keep the address of &x

3 Answers3