Error : Index was outside the bounds of the array. [duplicate]

Question

This question already has an answer here:

• What is an IndexOutOfRangeException / ArgumentOutOfRangeException and how do I fix it? 4 answers

I'm aware of what the issue is stating but I am confused to how my program is outputting a value that's outside of the array..

I have an array of ints which is 0 - 8 which means it can hold 9 ints, correct? I have an int which is checked to make sure the users input value is 1-9. I remove one from the integer (like so)

if (posStatus[intUsersInput-1] == 0) //if pos is empty
{
    posStatus[intUsersInput-1] += 1; 
}//set it to 1

then I input 9 myself and I get the error. It should access the last int in the array, so I don't see why I'm getting an error. Relevant code:

public int[] posStatus;       

public UsersInput()    
{    
    this.posStatus = new int[8];    
}

int intUsersInput = 0; //this gets try parsed + validated that it's 1-9    

if (posStatus[intUsersInput-1] == 0) //if i input 9 it should go to 8?    
{    
    posStatus[intUsersInput-1] += 1; //set it to 1    
} 

Error:

"Index was outside the bounds of the array." "Index was outside the bounds of the array."

Answer 1

You have declared an array that can store 8 elements not 9.

this.posStatus = new int[8]; 

It means postStatus will contain 8 elements from index 0 to 7.

Answer 2

public int[] posStatus;       

public UsersInput()    
{    
    //It means postStatus will contain 9 elements from index 0 to 8. 
    this.posStatus = new int[9];   
}

int intUsersInput = 0;   

if (posStatus[intUsersInput-1] == 0) //if i input 9, it should go to 8?    
{    
    posStatus[intUsersInput-1] += 1; //set it to 1    
} 

Answer 3

//if i input 9 it should go to 8?

You still have to work with the elements of the array. You will count 8 elements when looping through the array, but they are still going to be array(0) - array(7).