text: ## Student Interview Question for Applications Developers
  
  • 0
  
  Team: afai
  Country: India
  Interview Type: Written Test
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  z = 14. Here's why:
  
  When it comes to increment operators in a statement, ++x implies using the value of x after incrementing it by 1. On the order hand, x++ implies using the value of x as is and then incrementing it.
  
  So lets breakdown the statement into two steps such that the expression is written as such:
  
  ``z = (++y) + (++x)``
  
  ``z = z + (++x)``
  
  Equation 1:
  
  ``z = (++y) + (++x)``
  
  . Here, y is initially 8 and x is initially 1. Remember our rule with the operators, so we increment both values before using it. So this equation becomes (after the increment):
  
  ``z = 9 + 2 = 11``
  
  Equation 2:
  
  ``z = z + (++x)``
  
  .. Remember x is now 2 given we incremented it in equation 1 above and z = 11. Before we can use x in this equation, we have to increment if (from rules with the operators). So this equation becomes (after incrementing value of x):
  
  ``z = 11 + 3 = 14``
  
  Hope this makes sense :)
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  z = 14. Here's why:
  
  When it comes to increment operators in a statement, ++x implies using the value of x after incrementing it by 1. On the order hand, x++ implies using the value of x as is and then incrementing it.
  
  So lets breakdown the statement into two steps such that the expression is written as such:
  
  ``z = (++y) + (++x)``
  
  ``z = z + (++x)``
  
  Equation 1:
  
  ``z = (++y) + (++x)``
  
  . Here, y is initially 8 and x is initially 1. Remember our rule with the operators, so we increment both values before using it. So this equation becomes (after the increment):
  
  ``z = 9 + 2 = 11``
  
  Equation 2:
  
  ``z = z + (++x)``
  
  .. Remember x is now 2 given we incremented it in equation 1 above and z = 11. Before we can use x in this equation, we have to increment if (from rules with the operators). So this equation becomes (after incrementing value of x):
  
  ``z = 11 + 3 = 14``
  
  Hope this makes sense :)
  
  Comment hidden because of low score. Click to expand.
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  //9 + 2 + 3 = 14
  z = ++y + ++x + ++x;
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  //9 + 2 + 3 = 14
  {{z = ++y + ++x + ++x;}}
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  x=1, y=8, z
  z= (++(y=8)) + (++(x=1)) + (++(x=1))
  
  increase value of first x; it become x=2
  z=(9) + (x=2)) + (++(x=2))
  
  now increase value of next x; it become 3 so both places x represent value 3
  z= (9) + (x=3)) + (x=3))
  z=9+3+3
  z=15
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  ++ is an increment operator. increment operator before the variable has the higher priority than any other operator involved with the variable. It is called pre-increment operator. If increment operator is after the variable than it is called the post increment operator It has the lower priority.
  
  In the given code you have pre-increment operator's so when you apply plus operations first the value of the variable is increased then it is added to the another variable. like
  
  ``z  = (++y) + ++x + ++x;  //here first the value of y increase than it is added to the value of x like the normal plus operation. Similarly for the other x varialbes.``
  
  But if it was like this:
  
  ``z = ++y + x++ + ++x;``
  
  Here in case of first x it will add the value of x first to y and that value is added to the second x variable.But by the time control goes to the second x,the value of x is already increased by 1 in the post increment operator. So, the addition would be like this
  
  z = 9 + 1 + 3
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  (((++y) + (++x)) + (++x))
  Putting the brackets or drawing the expression tree will help to understand it better.
  The expression is evaluated from left to right.
  The operator ++ has higher precedence that +
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  It is a simple program with a little twist.
  Do not get confuse with the operator +, you need to split the equation
  
  ``````z = ++y + ++x + ++x;
  as
  z= (++y) + (++x) + (++x);
  
  Now as per your program the declared values are
  x=1 and y=8
  Now applying the value of y =8 since y is preincremented  (++y) in the first part of the equation, the value of y becomes 9
  the value of x is 1 in the second part of the equation the value of x will be preincremented (x++) so value of x will become 2
  when it comes to the third part of the equation the incremented value of x ie 2 will be again preincremented to 3.
  So the total will be
  z=(++y)+(++x)+(++x);
  z=9+2+3
  z=14.``````
  
  Hope you are cleared
  
  Name:
  
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explanation: The provided text is a discussion or explanation of a programming question, specifically about the evaluation of a statement involving pre-increment operators in a programming language (likely C, C++, or Java). The text does not contain a direct question but rather an explanation of how the expression `z = ++y + ++x + ++x;` is evaluated, given initial values of `x = 1` and `y = 8`.
  
  The text explains the rules of pre-increment operators and breaks down the expression step by step, leading to the conclusion that `z = 14`.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ## Student Interview Question for Applications Developers
  
  • 0
  
  Team: afai
  Country: India
  Interview Type: Written Test
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  z = 14. Here's why:
  
  When it comes to increment operators in a statement, ++x implies using the value of x after incrementing it by 1. On the order hand, x++ implies using the value of x as is and then incrementing it.
  
  So lets breakdown the statement into two steps such that the expression is written as such:
  
  ``z = (++y) + (++x)``
  
  ``z = z + (++x)``
  
  Equation 1:
  
  ``z = (++y) + (++x)``
  
  . Here, y is initially 8 and x is initially 1. Remember our rule with the operators, so we increment both values before using it. So this equation becomes (after the increment):
  
  ``z = 9 + 2 = 11``
  
  Equation 2:
  
  ``z = z + (++x)``
  
  .. Remember x is now 2 given we incremented it in equation 1 above and z = 11. Before we can use x in this equation, we have to increment if (from rules with the operators). So this equation becomes (after incrementing value of x):
  
  ``z = 11 + 3 = 14``
  
  Hope this makes sense :)
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  z = 14. Here's why:
  
  When it comes to increment operators in a statement, ++x implies using the value of x after incrementing it by 1. On the order hand, x++ implies using the value of x as is and then incrementing it.
  
  So lets breakdown the statement into two steps such that the expression is written as such:
  
  ``z = (++y) + (++x)``
  
  ``z = z + (++x)``
  
  Equation 1:
  
  ``z = (++y) + (++x)``
  
  . Here, y is initially 8 and x is initially 1. Remember our rule with the operators, so we increment both values before using it. So this equation becomes (after the increment):
  
  ``z = 9 + 2 = 11``
  
  Equation 2:
  
  ``z = z + (++x)``
  
  .. Remember x is now 2 given we incremented it in equation 1 above and z = 11. Before we can use x in this equation, we have to increment if (from rules with the operators). So this equation becomes (after incrementing value of x):
  
  ``z = 11 + 3 = 14``
  
  Hope this makes sense :)
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  //9 + 2 + 3 = 14
  z = ++y + ++x + ++x;
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  //9 + 2 + 3 = 14
  {{z = ++y + ++x + ++x;}}
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  x=1, y=8, z
  z= (++(y=8)) + (++(x=1)) + (++(x=1))
  
  increase value of first x; it become x=2
  z=(9) + (x=2)) + (++(x=2))
  
  now increase value of next x; it become 3 so both places x represent value 3
  z= (9) + (x=3)) + (x=3))
  z=9+3+3
  z=15
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  ++ is an increment operator. increment operator before the variable has the higher priority than any other operator involved with the variable. It is called pre-increment operator. If increment operator is after the variable than it is called the post increment operator It has the lower priority.
  
  In the given code you have pre-increment operator's so when you apply plus operations first the value of the variable is increased then it is added to the another variable. like
  
  ``z  = (++y) + ++x + ++x;  //here first the value of y increase than it is added to the value of x like the normal plus operation. Similarly for the other x varialbes.``
  
  But if it was like this:
  
  ``z = ++y + x++ + ++x;``
  
  Here in case of first x it will add the value of x first to y and that value is added to the second x variable.But by the time control goes to the second x,the value of x is already increased by 1 in the post increment operator. So, the addition would be like this
  
  z = 9 + 1 + 3
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  (((++y) + (++x)) + (++x))
  Putting the brackets or drawing the expression tree will help to understand it better.
  The expression is evaluated from left to right.
  The operator ++ has higher precedence that +
  
  Comment hidden because of low score. Click to expand.
  0
  of 0 vote
  
  It is a simple program with a little twist.
  Do not get confuse with the operator +, you need to split the equation
  
  ``````z = ++y + ++x + ++x;
  as
  z= (++y) + (++x) + (++x);
  
  Now as per your program the declared values are
  x=1 and y=8
  Now applying the value of y =8 since y is preincremented  (++y) in the first part of the equation, the value of y becomes 9
  the value of x is 1 in the second part of the equation the value of x will be preincremented (x++) so value of x will become 2
  when it comes to the third part of the equation the incremented value of x ie 2 will be again preincremented to 3.
  So the total will be
  z=(++y)+(++x)+(++x);
  z=9+2+3
  z=14.``````
  
  Hope you are cleared
  
  Name:
  
  Writing Code? Surround your code with {{{ and }}} to preserve whitespace.
  
  ### Books
  
  is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.
  
  ### Videos
  
  CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a discussion or explanation of a programming question, specifically about the evaluation of a statement involving pre-increment operators in a programming language (likely C, C++, or Java). The text does not contain a direct question but rather an explanation of how the expression `z = ++y + ++x + ++x;` is evaluated, given initial values of `x = 1` and `y = 8`.
  
  The text explains the rules of pre-increment operators and breaks down the expression step by step, leading to the conclusion that `z = 14`.
  
  Label: no-question