main()
{
int i=5,j;
j = i++ + ++i + ++i + ++i;
printf("%d",j);
}

• j=5+7+8+9
  hence j= 29
• 9 years agoHelpfull: Yes(18) No(3)
• j = i++ + ++i + ++i + ++i;
  i++ will increment the value of i, but return the original value that i held before being incremented.
  ++i will increment the value of i, and then return the incremented value.
  j = 5+7+8+9
  j =29
• 9 years agoHelpfull: Yes(7) No(1)
• j = i++ + ++i + ++i + ++i;
  let post increment keep aside, then expression becomes
  j = i + ++i + ++i + ++i;
  then evaluate the expression consider the priorities
  i.e i-> 5 -> 6-> 7-> 8
  substituting the value in the final expression after implementing the all pre-increment operator i.e Right TO Left.
  finally i becomes 8, then substitute the value in the remaining expression
  j = 8 + 8 + 8 + 8
  i.e j=32
  the post increment should be consider finally to variable i,
  i becomes i++ final value of i is 9.
  the output of the program will be
  
  9 32
  
  better to consider memory blocks for variables for easy understanding.
• 9 years agoHelpfull: Yes(6) No(0)
• j=6+6+7+8
  =27
• 9 years agoHelpfull: Yes(2) No(2)
• It is compiler dependent, so no correct answer can be given,,
• 9 years agoHelpfull: Yes(2) No(0)
• j=i++ + ++i + ++i + ++i
  step1:
  ++i=6
  ++i=7
  ++i=8
  now i=8;
  step 2:
  y=i+i+i+i;
  y=28;
  step 3:
  i++=9
  i=9
• 9 years agoHelpfull: Yes(2) No(0)
• follows preorder - algebra -postorder
  output- 32
• 9 years agoHelpfull: Yes(1) No(1)
• 6+7+8+9=30
• 9 years agoHelpfull: Yes(1) No(1)
• 6+6+7+8=27
• 9 years agoHelpfull: Yes(1) No(1)
• j=5+7+8+9
  hence j= 29
• 9 years agoHelpfull: Yes(0) No(0)
• I THINK IT SHOULD BE COMPILLER ERROR BECAUSE I WILL STATRT WORKING FROM LEFT OR RIGHT IT DEPENDS ON COMPILLER
  
  
• 9 years agoHelpfull: Yes(0) No(1)
• j= 5+7+8+9=29
• 9 years agoHelpfull: Yes(0) No(2)
• the value in post increment is incremented after semi colon but in pre increment the value increases at the same place ........before semicolon of the second statement value of x is pre incremented 3 times so i=5+1+1+1 ie 8..now the statement means j=8+8+8+8 ie 32 and after semicolon the post incrementation will take place and value will i will become 9.......so output of program will be 32
• 9 years agoHelpfull: Yes(0) No(0)
• output will be 32
• 9 years agoHelpfull: Yes(0) No(1)
• j=5+7+8+9
  
  answer is j=39
• 9 years agoHelpfull: Yes(0) No(0)
• j=5+ 7+ 8 +9 =29
• 9 years agoHelpfull: Yes(0) No(0)
• j = 32
  i = 9
• 9 years agoHelpfull: Yes(0) No(1)
• ans is 32.
  this is register evaluation,it works with priority i.e.,pre operator has higher priority than post operator.
  In register evaluation data need to substituted after modifying all pre values.
  j=8+8+8+8=32 (3 pre operator )
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• 21...i++ means it will be incremented before the excecution of that statement and ++i means it will be incremented after the excecution of that statement so 6+5+5+5=21
• 9 years agoHelpfull: Yes(0) No(0)
• if copiler is 16 bit the ,,, ans = 16
  if compier is 32 bit ,,,,,,, ans = 32
• 9 years agoHelpfull: Yes(0) No(0)
• i=5
  
  ++i=7
  ++i=8
  ++i=9
  ans:5+7+8+9=29
• 9 years agoHelpfull: Yes(0) No(0)
• j=5+7+8+9=29
• 9 years agoHelpfull: Yes(0) No(0)
• Answer must be 27
  6+6+7+8=27
• 9 years agoHelpfull: Yes(0) No(0)
• j=29
  I dont know the reason
• 9 years agoHelpfull: Yes(0) No(0)
• Associatibity of increament operator is become from right to left so according to this rule value of this is 27
  
  
• 9 years agoHelpfull: Yes(0) No(0)
• j=5+7+8+9=29
• 9 years agoHelpfull: Yes(0) No(0)
• doesnot space between the incerment and + operator it gives an error bad operand
• 9 years agoHelpfull: Yes(0) No(0)
• u can not pridict the output unless compiler is provided
  same code will generate different output on different compilers like dev c,turbo c, and so on
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 27
  j = 5 + 7 + 8 + 9
• 9 years agoHelpfull: Yes(0) No(0)
• since arithmetic operation are evaluated from left to right, and unary are evaluated from right to left..
  j=(((i++ + ++i) + ++i )+ ++i)
  j=((( 6 + 6) + ++i)+ ++i) -->i=7
  j=((12 + 8)+ ++i) --> i=8
  j=20 + 9
  j=29
• 9 years agoHelpfull: Yes(0) No(0)
• 5+6+7+8
  28
• 9 years agoHelpfull: Yes(0) No(0)
• Undefined symbol J
• 9 years agoHelpfull: Yes(0) No(0)
• It will just like j=5+7+8+9(29)
• 7 years agoHelpfull: Yes(0) No(0)
• Actually, it depends on the compiler
  according to one compiler
  i is pre-incremented 3 times so i becomes 8 and that 8 will be placed in the place of i so
  j=8+8+8+8==>32
  according to another compiler
  j=5+7+8+9==>29
• 7 years agoHelpfull: Yes(0) No(0)