array of pointers arithmetic

Question

I was experimenting with array of pointers address calculation arithmetic.I got confused with the output of the above code.Could anyone explain whats happening?

void foo()
{
    int i=10,k=3,l=20,m=30;
    int *ary[2];
    ary[0]=&i;
    int b=20;
    ary[1]=&k;

    printf("%d\n",ary[0][1]);
}

Output is 3

Second program

void foo()
{
    int i=10,k=3,l=20,m=30;
    int *ary[2];
    ary[0]=&i;
    int b=20;
    ary[1]=&b;

    printf("%d\n",ary[0][1]);
}

Output is 20.

How is address calculation done in these above codes?

What @Jon Lin said, and how are you calling foo? There's a chance for undefined behavior... — K-ballo
– K-ballo, Commented Jun 24, 2012 at 6:44
The difference is that ari[1] is assigned &k in the first program and &b in the second. — ObscureRobot
– ObscureRobot, Commented Jun 24, 2012 at 6:49
sorry....changed the code now...now it has no undefined behaviour.. — Vindhya G
– Vindhya G, Commented Jun 24, 2012 at 6:50

sepp2k · Accepted Answer · 2012-06-24 06:52:13Z

3

In both code samples ary[0] is a pointer to a single integer. So when you do array[0][1] you're accessing that pointer out of bounds. So the behavior of both of your code samples is undefined.

The reason that you're getting the behavior you're seeing on your particular compiler is probably, that all the variables whose address you don't take are stored in registers and not in memory (or possibly they aren't stored at all because you never use them).

So in example 1 the only variables in memory are i and k. And in example 2 the only variables in memory are i and b. So in example 1 k is the variables that comes directly after i in memory and example 2 that variable is b.

answered Jun 24, 2012 at 6:52

sepp2k

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K-ballo · Accepted Answer · 2012-06-24 06:52:03Z

0

Your argument is

int *ary[]

and in function arguments, array decay to pointers. So you basically have

int** ary

You then set the pointer at index 0, presumably a valid index, to the value of a single variable.

ary[0]=&i;

Then your code prints

ary[0][1]

which is basically

( &i )[1]

triggering undefined behavior, hence whatever result you get is a valid result.

answered Jun 24, 2012 at 6:52

K-ballo

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2 Comments

Vindhya G Over a year ago

my speculation was about how compiler does this.shouldnt it give value of variable immediately next to it.why does it gives value of k first time and then value of b next.shouldnt it give same variable each time?

K-ballo Over a year ago

@vindhya: When you trigger undefined behavior, the compiler has free pass to do whatever it wants to do. It could just crash, it could just return some random value, it could just return the variable immediately next to it... It could even decide that under no defined behavior will you call foo, and hence generate no code for it.

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