what is the wrong with array address with post increment operator? [duplicate]

Question

#include<stdio.h>
 int main(){
    char arr[] = "Pickpocketing my peace of mind";
    int i;
    printf("%c",*arr);
    arr++;
    printf("%c",*arr);

   return 0;
 }

I am getting error: lvalue required as increment operand. My question is , if we increment a variable like int i=10; i++ will not give any error. but y this?

It is illegal because someone decided that this is illegal.

Lasse V. Karlsen
– Lasse V. Karlsen

2018-08-08 11:17:40 +00:00
Commented Aug 8, 2018 at 11:17 — Lasse V. Karlsen
– Lasse V. Karlsen, Commented Aug 8, 2018 at 11:17

haccks · Accepted Answer · 2018-08-08 10:45:36Z

2

You are trying to modify an array which is not permissible. You can't modify an array but its value. As you can see that error message itself speaks that arrays are not an lvalue so you can't modify it.
Do not confuse an array with a pointer.

char arr[] = "Pickpocketing my peace of mind"; 
char *ptr = arr;
ptr++;            // OK. ptr is an lvalue. It can be modified.
arr++;            // ERROR. arr in not an lvalue. Can't be modified.

answered Aug 8, 2018 at 10:45

haccks

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2 Comments

yashu Over a year ago

Hi Thank you, But my confusion is, arr will give the base address, which is the address of the first element, if I do arr++, it should point to next element address in the array right?

haccks Over a year ago

You can't modify an array. I made it clear in the answer.

Rizwan · Accepted Answer · 2018-08-08 11:07:49Z

0

Moving an array pointer is not permitted. If you want to move to next elements using ++ on array pointers just copy the array pointer in to some other pointer variable. Like

char arr[] = "Pickpocketing my peace of mind";
char *movablePtr = arr;

Now you can move movablePtr by movablePtr++ / movablePtr--.

Also I to add when you say arr[5] actually it means *(arr + 5 * sizeof(char)) when array is of char type.

edited Aug 8, 2018 at 11:07

answered Aug 8, 2018 at 10:54

Rizwan

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6 Comments

yashu Over a year ago

Thank you so much, Rizwan...

Ilja Everilä Over a year ago

It'd seem like you're indicating that arrays are pointers, but they are not. They are just converted to a pointer to their first element in most contexts.

Rizwan Over a year ago

May be I was not very clear in expressing, but indeed you are right Arrays are not pointers. I explained it in this way just to give some insight.

Ilja Everilä Over a year ago

Also arr[5] is not equivalent to what you describe, but *(arr + 5).

Rizwan Over a year ago

I think you mean *(arr + 5 * sizeof(char)) in this case.

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