Difference between ++i and i++ [duplicate]

Answer 1

Note: you are invoking undefined behavior (modifying a variable twice between sequence points)

Answer 2

++i happens before calculating the whole expression and i++ happens after. In the first example one of increments happen before calculating the value, so "i" becomes 21 and you get 21 + 21. In the last example both happen before, so "i" becomes 22 and you get 22 + 22.

Answer 3

This seemed really interesting, so I took a peek at the disassembly (MSVC++2008)

     k = (++i)+(++i);
0122413C  mov         eax,dword ptr [i] 
0122413F  add         eax,1 
01224142  mov         dword ptr [i],eax 
01224145  mov         ecx,dword ptr [i] 
01224148  add         ecx,1 
0122414B  mov         dword ptr [i],ecx 
0122414E  mov         edx,dword ptr [i] 
01224151  add         edx,dword ptr [i] 
01224154  mov         dword ptr [k],edx 

As you can see, it increments i twice and then adds i to itself. The same thing happens if there are multiple instances of (++i).

Anyway, since the Standard doesn't guarantee anything, modifying i more than once will lead to undefined behaviour.

Answer 4

++i will give result i=i+1. If i=10 then in k = (++i)+(++i); expression (++i) will give incremented value that means first increment will happen but in case of i++ the increment will be affected on i after the expression.

So i=10

k = (i++) + (++i);

 10    11   12

 10    +  12=22  

k = (++i) + (++i);

 11   11  12

  11   +  12=23

Answer 5

i++ is post-increment, ++i is pre-increment. i++ will increment i after the statement is completed.

To illustrate in your examples:

Example 1:

k = 0
i = 10

i += 1
k = i + i  // 11 + 11
i += 1

Example 2:

k = 0
i = 10

i += 1
i += 1
k = i + i  // 12 + 12

Answer 6

I'm guessing that both pre-increment operators are running before the addition statement gets calculated. hence 24.

All depends on how the compiler sees what you're doing, but I'm guessing that's what you're seeing.