What does x->a = x->b, ++x++ ->b; do in the given program?

Question

int main() 
{
    struct data{
        int a;
        int b;
    }y[4] = {1,10,3,30,2,20,4,40};

    struct data *x = y;
    int i;

    for(i = 0; i < 4; i++)
    {
         x->a = x->b, ++x++ ->b;
         printf("%d%d\t", y[i].a, y[i].b);
    }


    return 0;
}

How does this code work?

Don't change the question after getting answers and comments. And don't make the question to broad. `->` is the structure reference operator ( Google it for more information on that ) and `= { ... }` initializes all the elements of the array with the values in the initializer list. Play with it for a while and you'll get the hang of it :-) — Spikatrix, Aug 22 '15 at 16:03
No professional or enthusiast programmer would care because all such code is garbage, as is this question. — Martin James, Aug 22 '15 at 17:37
I'm voting to close this question as off-topic because the code in the question is execrable and is of zero, or negative, use to future visitors. — Martin James, Aug 22 '15 at 17:39

haccks · Answer 1 · 2015-08-22T15:48:57.787

4

The statement

x->a = x->b, ++x++ ->b;  //note the comma operator after x->a = x->b

is equivalent to

x->a = x->b;     // Assign member `a` to first member of element of array y 
++( (x++) ->b);  // Fetch the member b of element of array y and then increment 
                 // it by 1. Increment `x` by 1 after `b` is fetched.

edited Aug 22 '15 at 15:48

answered Aug 22 '15 at 15:42

haccks

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Grzegorz Szpetkowski · Answer 2 · 2015-08-22T16:03:56.380

The ++x++ is well defined in this context:

++x++->b

that is equivalent to:

++((x++)->b)

This means that two different x and x->b lvalues are incremented, not single one (which would be an UB). You may rewrite it into two statements:

++(x->b);
x++;

The full expression statement:

x->a = x->b, ++x++ ->b;

is valid, since there is sequence point (because of comma operator) after x->a = x->b is evaluated ~~(there were no side effects in this one, so result of this subexpression was discarded)~~.

score 1 · Answer 3 · answered Aug 22 '15 at 15:50

This expression

x->a = x->b, ++x++ ->b;

uses the comma operator and in fact for clarity can be split in two statements

x->a = x->b;
++x++ ->b;

The second subexpression can be imagine the following way

struct data *temp = x;
++x;
++temp->b;

that is it increases data member b of the structure pointed to by pointer x and increases the pointer itself.

So data member b of the current structure (object of type struct data) is increased and the pointer will point to the next structure.

What does x->a = x->b, ++x++ ->b; do in the given program?

3 Answers3