Postfix in array[i++] seems effect my expecting result

**sheng8** · August 15th, 2014, 06:15 PM

Hi All,
I have a newbie question which makes me struggle couple hours and search online for nothing. will you please kindly help me on this? i have no one to ask since i am study by my own. thank you for your time and support.

		char array[] = new char[10];
 
		for (int i = 0; i < 10;){
			array[i++] = (char)('A' + i); 
			System.out.println("char for array is: " + array[0]);
		}

I don't understand that in the beginning of the for loop, "i" was init to 0 and (char) ('A' + i) remain 'A' which was assigned to array[0].
However, the result for array[0] is B. Not A.

can anyone welling to share the opinion? appreciate for your help.

**Junky** · August 15th, 2014, 08:21 PM

You are incrementing i inside the loop. You should try to avoid changing the value of your loop control variable inside the loop as it can have unexpected side effects as you have found out. Increment in the loop header like it should be and it will work.

**sheng8** · August 15th, 2014, 09:58 PM

Thank you for the suggestion and appreciate for the help. i did try to avoid the increment i in the loop and it works fine.
if something like non-define element and try to print it out, it may shows unexpected value which i understand. but in my case, i really want to know the reason of storing B in array[0] instead of A.

and i also remove the loop as follow, same thing happen:

		char array[] = new char[10];
	        int i = 0;
		array[i++] = (char)('A' + i); 
		System.out.println("char for array is: " + array[0]);

It will be nice to have the opinion, thanks again and wish the best

**Cornix** · August 16th, 2014, 01:42 AM

If you write i++ it means you increment the variable i by one.
This code:

int i = 0;
intArray[i++] = i;

will store the value 1 at position 0 in the array. This is because i is incremented by 1 before its assigned.

**sheng8** · August 16th, 2014, 02:34 AM

i learn the evaluate left-hand operand first rule, thank you

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