C programming; lvalue required

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Lvalue required error

I am having an error in my C program

main () {
    int arr[] = {1, 2, 3, 4};
    printf("%d", *arr);
    arr++;
    printf("%d", *arr);
}

When I compile this code I get lvalue required error. for the line with arr++. Any Help!

Show the minimal entire code please. And an array is a pointer constant, you cannot increment that pointer. (you could assign it to some other pointer and increment that other pointer). — Basile Starynkevitch, Feb 27 '12 at 17:13
Is this homework? Because it doesn't look like something you would ever actually want to do. — trutheality, Feb 27 '12 at 17:14
I just wanted to know the basic difference between a pointer and the array name(i.e the first location pointed by the array). — me_digvijay, Feb 27 '12 at 18:07

score 5 · Answer 1 · edited Feb 27 '12 at 17:16

5

arr is a constant, you can't change its value. You can add a

int *p = arr;

And then do a

p++;

edited Feb 27 '12 at 17:16

Bart

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answered Feb 27 '12 at 17:13

zmbq

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2

You should explain here that "arr" isn't merely a constant -- it is an array, not a pointer. Although C makes arrays and pointers look similar, they are not the same sort of thing -- you cannot increment an array. – Perry Feb 27 '12 at 17:47

score 3 · Answer 2 · answered Feb 27 '12 at 17:14

3

Your problem is that arr is an array and arrays are not lvalues. You need a pointer.

int arr[] = {1, 2, 3, 4};
int *p = &arr;
printf("%d", *p);
p++;
printf("%d", *p);

answered Feb 27 '12 at 17:14

David Heffernan

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Isn't &arr an int ** (with some a const in there, of course)? – zmbq Feb 27 '12 at 17:26
1

@ambq No it is not. In your code you are just taking advantage of the fact that `arr` decays to `&arr`. – David Heffernan Feb 27 '12 at 17:36
Thanks, it's been a decade and a half since I last programmed in straight C... – zmbq Feb 27 '12 at 17:56

Jonathan Wood · Answer 3 · 2012-02-27T18:05:52.640

2

lvalue generally refers to the value on the left of an assignment (=) operator. Since arr++ is shorthand for arr = arr + 1, that's what it's referring to.

Basically, arr as an array, and there is no support for changing the value of an array. I think you want a pointer to the array. Pointers can be incremented in the way your code attempts to.

edited Feb 27 '12 at 18:05

answered Feb 27 '12 at 17:15

Jonathan Wood

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Cacho Santa · Answer 4 · 2012-02-27T17:24:04.640

arr is constant, you can not change it's value.

Keep it simple, access the array this way:

int main (int argc, char *argv[]) {
    int arr[] = {1, 2, 3, 4};
    printf("arr[0] == %d\n", arr[0]);
    printf("arr[1] == %d\n", arr[1]);
    printf("arr[2] == %d\n", arr[2]);
    printf("arr[3] == %d", arr[3]);

    return 0;
}

I recommend this documentation for lvalues and rvalues.

And also, this c-faq about "Arrays are not pointers".

score 1 · Accepted Answer · answered Feb 27 '12 at 18:32

The operand of the pre- and postfix versions of ++ and -- must be a modifiable lvalue. Unfortunately, array expressions such as arr are non-modifiable lvalues, hence your error message.

If you want to walk the array using a pointer expression, you'll have to declare a second pointer and set it to point to the first element of the array.

int *p = arr; // or &arr[0] - in this context they evaluate to the same thing

C programming; lvalue required

5 Answers5