c[-1]?

[blurx]

Say if a function accessed this element in an array called c. What does it mean?

[matsp]

It is technically undefined, if c is an array. If c is a pointer INTO an array, then I think it is still undefined by the standard, but it's possible that it "works" correctly - getting the element BEFORE c.


--
Mats

[vart]

It is technically undefined, if c is an array. If c is a pointer INTO an array, then I think it is still undefined by the standard, but it's possible that it "works" correctly - getting the element BEFORE c.


--
Mats

Why would it be undefined?
It should be equivalent to *(c-1)

if c-1 is valide pointer (still pointing to the inside of the array) - you can dereference it

[blurx]

c was a pointer so it did mean one element before. Thanks.

[Snafuist]

It is technically undefined, if c is an array. If c is a pointer INTO an array, then I think it is still undefined by the standard, but it's possible that it "works" correctly - getting the element BEFORE c.

Citing from my copy of the standard:

J.2 Undefined behavior

The behavior is undefined in the following circumstances:

[...]

- Addition or subtraction of a pointer into, or just beyond, an array object and an integer type produces a result that does not point into, or just beyond, the same array object (6.5.6).

I read this as "if c[-1] is an object of the same array as the one that contains c[0], then it's ok". Besides, the subscript operator in E1[E2] is defined as (*((E1) + (E2))) where E1 is a pointer and E2 is an integer.

Hence, the following should be ok:

int a[8];
int *c = a + 4;
if(c[-1] == a[3])
        puts("works");

Greets,
Philip

[Bladactania]

In the case you site there then c[-1] is technically a valid statement. But if you have

int c[8];

then c[-1] would return an unknown value since that address isn't part of the array c. I think this is more in the nature of the original question.

So while c[-1] isn't undefined, except in the case you refer to, it is an unknown value.

[EVOEx]

I agree with snafuist. It should be fine and equal to *(a - 1)...

[grumpy]

In the case you site there then c[-1] is technically a valid statement. But if you have

int c[8];

then c[-1] would return an unknown value since that address isn't part of the array c.

The expressions c[-1] doesn't just yield an "unknown value". In this case (since the first element of c is c[0]) the expression c[-1] or, equivalently, *(c-1) yields undefined behaviour.

[Bladactania]

I don't think that is considered undefined behavior. c[-1] returns the "value" stored in the address that is one before c[0].

Although, I guess that if c[0] is stored on address "0" or the lowest possible address, then c[-1] would generate an error. Or of c[0] is one address after the end of system protected memory.

I append my original statement. c[-1] yields undefined behavior.

[Tanuj_Tanmay]

why will C compiler give error if you try to access a memory.....
the only differnece here is you haven't stored any value there...
it will definitely print some value...but it will be a GARBAGE.

Python....whereas prints the value of last element of array
...suppose i have array of 5 elements then a[-1] is equivalent to a[4]
...but thats in PYTHON

[Tanuj_Tanmay]

But if store some value there...it will print it
example

#include<stdio.h>

int main()
{
	int c[] = {2,5,6,8,3,6};
	c[-1] = 57;
	printf("%d",c[-1]);
	return 0;
}

OUTPUT:
57

[Bladactania]

Yes you can do that, but you are likely to get a seg fault or other unexpected behavior if you do something like that.

Just because something works SOMETIMES doesn't mean it will work ALL times.

[Ideswa]

The memory you're writing to isn't allocated, so it's undefined.

[Tanuj_Tanmay]

Compiler doesn't play dice.....
Can u make out when does it give segmentation fault and when the garbage value...

[anon]

You could be overwriting something important that is located in front of the array in memory?