Prefix and Postfix operator

Question

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j is initialized to 1 therefore 1st operand (j++) takes the value of 1. Concurrently 2nd operand also takes the value of 1 but it as a prefix(++j) operator value 1 is quickly convert into 2. Therefore k (1+2) must be 3, but in the output k=4 how can this be explained.

using System;
namespace Code
{
class Program
{
static void Main(string[] args)
{
int i, j = 1, k;

for (i = 0; i < 5; i++)
{
k = j++ + ++j;
Console.Write(k + " ");
}
Console.ReadKey();
}
}
}
//4 8 12 16 20

Vulpes · Answer

That's correct Maha. j's memory slot receives values of: 1 - initially 2 - postfix ++ 3 - ++ prefix

Maha · Answer

Thank you for your explanation.

Maha · Answer

I understand from the explanation about 1st example that ++j is getting the value from j++. That means ++j is not getting the value from its initialisation where j is given the value of 1. That is why k=4. I wish to know whether it is the case.

Vulpes · Answer

Regarding the first example you're forgetting that j is incremented twice first by the postfix operator (after the 1st operand is evaluated) and then by the prefix operator (before the 2nd operand is evaluated). The second operand therefore has a value of 1 + 1 + 1 = 3 and adding 1 to this gives k = 4. Regarding the second example the first operand (j++) is (as before) 1 and the second operand (++m) is 2 and 1 + 2 = 3. j is also incremented by the postfix operator and therefore has a value of 2 when the second iteration of the for loop begins. So for this iteration the first operand is 2 and the second operand (++m) is 3 giving k = 2 + 3 = 5. And so on for the other iterations.

Maha · Answer

Hi thank you for your explanation. How can this be explained? using System; namespace ConsoleApplication6 { class Program { static void Main(string[] args) { int i j = 1 k m = 1 ; for (i = 0; i { k = j++ + ++m ; Console.Write(k +  ); } Console.ReadKey(); } } } //3 5 7 9 11

Iftikar Hussain · Answer

Hi In postfix increment the equation is evaluated first and then the increment operation takes place In prefix increment first of all the increment operation takes place and then the equation is evaluated. b=++a implies first increment a and now solve the equation hence b contains value 2; b=a++ implies first evaluate the operation b=a; i.e. b=1 now. and after performing this assignment a is incremented. In the memory a=2 whereas b is assigned the previous value of a as the expression was evaluated before increment Regards Iftikar