lvalue required as left operand of assignment error when using C

Consider the following array:

int arr[5] = {1,2,3,4,5};

We know that name of the array points to address of first element of array i.e. address of 1 here.

So, the following code:

printf("%u",arr);

will give address of 1 in this case. This address of 1 is constant and cannot be changed. So, in a C program, it is not possible to change the value of arr.

The following code will give an error: lvalue required as left operand

#include

int main(){
     int arr[5] = {1,2,3,4,5};  
     arr++;
     return 0;
}

Even if you give a lvalue by modifying the code as below, it will still throw a compilation error.
error: incompatible types when assigning to type 'int[5]' from type 'int *'  

int main(){
     int arr[5] = {1,2,3,4,5};  
     arr = arr +1;
     return 0;
}

Thus, important thing to understand here is, we are missing out on a key concept.
Here, arr is a constant just like a const variable which cannot be changed and any attempt to change that value will result in a compilation error.

It will be more comprehensible if the error "lvalue required as left operand" would have read "arr is a constant and cannot be changed".

Having said that, the following operation is valid.

int main(){
     int arr[5] = {1,2,3,4,5};  
     printf("%d\n", *(arr + 1));
     return 0;
}

This is because, we are fetching the value at index 1 of array arr and not increasing the value of arr.

This is similar to int y = x +5;
We are not increasing the value of x here. We are just adding 5 to value of 5 and assigning it to y.