How to solve expression containing post increment Operator ?

Post increment in expression solving

Initial Values :

int i=0,j=0;
int k = 0;
int temp=3;

Depending Upon These Initial Values Read Carefully following Code Snippets ,

1.Post Increment Verity No 1

for(k=0;k<5;k++)
  {
  j = i++;
  printf("%d",j);
  }

i++ is Post Increment.
Meaning : First Old value of variable ‘i’ gets assigned to ‘j’ and then ‘i’ gets incremented.
So output of this code is = [ 0 1 2 3 4 ]

2.Post Increment Verity No 2

for(k=0;k<5;k++)
  {
  printf("%d",i++);
  }

3.Post Increment Verity No 3

for(k=0;k<5;k++)
  {
  j = i++ + temp;
  printf("%d",j);
  }

Meaning First Old value of ‘i’ is added with temp and result of addition is assigned to ‘j’
temp = 3
So output of this code is = [ 3 4 5 6 7]

4.Post Increment Verity No 4

for(k=0;k<5;k++)
  {
  j = i++ < temp;
  printf("%d\n",j);
  }

Meaning First Old value of ‘i’ is compared with ‘temp’ . Then result is assigned to j, and new value is used in second iteration .
temp = 3
So output of this code is = [ 1 1 1 0 0]

5.Post Increment Verity No 5

for(k=0;k<5;k++)
  {
  j = i++;
  printf("%d",i);
  }

Old value of ‘i gets reflected only within same expression
In the above snippet Old value is assigned to ‘j’ . but in the next statement always ‘new value’ is used.
So while printing i Output will be = [1 2 3 4 5].

6.Post Increment Verity No 6

for(k=0;k<5;k++)
  {
  i = i++;
  printf("%d",i);
  }

Note :