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 '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++);
  }

• Meaning  First Old value gets printed first and then 'i' is incremented.
• So output of this code is = [ 0 1 2 3 4 ]

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 = [ 3 4 5 6 7]

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);
  }

• Old value of 'i is assigned to 'i' itself.
• And again value is incremented after assigning.
• So while printing i Output will be = [1 2 3 4 5].
• But this Type is not supported by many compiles .

Note : 

1. Same things happens while using Post-Decrement.
2. Post Increment is Unary Operator.
3. Post-Increment Operator has Highest Priority

---