What does a && operator do when there is no left side in C?

Question

I saw a program in C that had code like the following:

static void *arr[1]  = {&& varOne,&& varTwo,&& varThree};

varOne: printf("One") ;
varTwo: printf("Two") ;
varThree: printf("Three") ;

I am confused about what the && does because there is nothing to the left of it. Does it evaluate as null by default? Or is this a special case?

Edit: Added some more information to make the question/code more clear for my question. Thank you all for the help. This was a case of the gcc specific extension.

Answer 1

It's a gcc-specific extension, a unary && operator that can be applied to a label name, yielding its address as a void* value.

As part of the extension, goto *ptr; is allowed where ptr is an expression of type void*.

It's documented here in the gcc manual.

You can get the address of a label defined in the current function (or a containing function) with the unary operator &&. The value has type void *. This value is a constant and can be used wherever a constant of that type is valid. For example:

void *ptr;
/* ... */
ptr = &&foo;

To use these values, you need to be able to jump to one. This is done with the computed goto statement, goto *exp;. For example,

goto *ptr;

Any expression of type void * is allowed.

As zwol points out in a comment, gcc uses && rather than the more obvious & because a label and an object with the same name can be visible simultaneously, making &foo potentially ambiguous if & means "address of label". Label names occupy their own namespace (not in the C++ sense), and can appear only in specific contexts: defined by a labeled-statement, as the target of a goto statement, or, for gcc, as the operand of unary &&.

Answer 2

This is a gcc extension, known as "Labels as Values". Link to gcc documentation.

In this extension, && is a unary operator that can be applied to a label. The result is a value of type void *. This value may later be dereferenced in a goto statement to cause execution to jump to that label. Also, pointer arithmetic is permitted on this value.

The label must be in the same function; or in an enclosing function in case the code is also using the gcc extension of "nested functions".

Here is a sample program where the feature is used to implement a state machine:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main(void)
{
    void *tab[] = { &&foo, &&bar, &&qux };

    // Alternative method
    //ptrdiff_t otab[] = { &&foo - &&foo, &&bar - &&foo, &&qux - &&foo };

    int i, state = 0;

    srand(time(NULL));

    for (i = 0; i < 10; ++i)
    {
        goto *tab[state];

        //goto *(&&foo + otab[state]);

    foo:
        printf("Foo\n");
        state = 2;
        continue;
    bar:
        printf("Bar\n");
        state = 0;
        continue;
    qux:
        printf("Qux\n");
        state = rand() % 3;
        continue;
    }
}

Compiling and execution:

$ gcc -o x x.c && ./x
Foo
Qux
Foo
Qux
Bar
Foo
Qux
Qux
Bar
Foo

Answer 3

I'm not aware of any operator that works this way in C. Depending on the context, the ampersand in C can mean many different things.

Address-Of operator

Right before an lvalue, e.g.

int j;
int* ptr = &j;

In the code above, ptr stores the address of j, & in this context is taking the address of any lvalue. The code below, would have made more sense to me if it was written that way.

static int varOne;
static int varTwo;
static int varThree;

static void *arr[1][8432] = { { &varOne,&varTwo, &varThree } };

Logical AND

The logical AND operator is more simple, unlike the operator above, it's a binary operator, meaning it requires a left and right operand. The way it works is by evaluating the left and right operand and returning true, iff both are true, or greater than 0 if they are not bool.

bool flag = true;
bool flag2 = false;
if (flag && flag2) {
    // Not evaluated
}
flag2 = true;
if (flag && flag2) {
   // Evaluated
}

Bitwise AND

Another use of the ampersand in C, is performing a bitwise AND. It's similar as the logical AND operator, except it uses only one ampersand, and performs an AND operation at the bit level.

Let's assume we have a number and that it maps to the binary representation shown below, the AND operation works like so:

0 0 0 0 0 0 1 0
1 0 0 1 0 1 1 0
---------------
0 0 0 0 0 0 1 0

In C++ land, things get more complicated. The ampersand can be placed after a type as to denote a reference type (you can think of it as a less powerful but safe kind of pointer), then things get even more complicated with 1) r-value reference when two ampersands are placed after a type. 2) Universal references when two ampersands are placed after a template type or auto deducted type.

I think your code probably compiles only in your compiler due to an extension of some sorts. I was thinking of this https://en.wikipedia.org/wiki/Digraphs_and_trigraphs#C but I doubt that's the case.