Pre increment and post increment

Question

I'm having trouble understanding how Post Increment (++), Pre Increment (--) and addition/subtraction work together in an example.

x++ means add 1 to the variable.

x-- means subtract 1 from the variable.

But I am confused with this example:

int x = 2, y = 3, z = 1;`

y++ + z-- + x++;

I assume this means 3(+1) + 1(-1) + 2(+1) Which means the result should be 7.

But when I compile it, I get 6. I don't understand.

int main() {
  int x=2, y=3, z=1;
  int result;

  result = y++ + z-- + x++;    //this returns 6

  cout << result << endl;
  return 0;
}

Solution 1

 result = y++ + z-- + x++;
           3     1      2  = 6

if you perform this again

 result1 = y++ + z-- + x++;
          4     0      3  = 7

reason

operator++ returns the original value, before incrementing the variable.

and

++operator returns the incremented value

-- is same as above just its decrement

Solution 2

Because the postfix operator++ returns the original value, before incrementing the variable. The prefix operator++ increments the varialbe and returns a reference to it. The behaviour can be easily illustrated with an example:

#include <iostream>

int main()
{
  int n = 1;
  std::cout << n++ << "\n"; // prints 1
  std::cout << n << "\n";   // prints 2

  int m = 10;
  std::cout << "\n";
  std::cout << ++m << "\n"; // prints 11 
  std::cout << m << "\n";   // prints 11
}

Solution 3

when you write x++ it uses the current value of x and then increases it by one.

you want to write ++x instead if you want to increase first.

Solution 4

Pre-increment operator(++p) first increase the value and assign it and post increment operator(p++) first assign the value and then perform increment operation.Here all variable are post increment i.e it initially assign its value (on buffer) then increase (for y and x by 1) and decrease z by 1. i.e initially assign 3 + 1 + 2 in buffer(addition is performed on buffer value) and then perform increment/decrements as x= 3,y=4 and z=0

for more information you can read answer on What is the correct answer for cout << c++ << c;? and Why are these constructs (using ++) undefined behavior? questions

Solution 5

The position of ++ matter.

If ++ precedes the variable, e.g. ++counter, the value returned is the value in counter after it has been incremented. If ++ follows the variable, e.g. counter++, the value returned is the value in counter before it has been incremented.

(source)

View more solutions

Comments

[Siddharth-Verma]

It will first use the value and then* increment

[jrok]

The code in OP is well defined and can be explained. The links don't answer the question.

[Toby Speight]

What language is that? It certainly doesn't look like any C++ I've seen...