Write the output of the following: a) num1 = 4 num2 = num1 + 1 num1 = 2 print (num1, num2) b) num1, num2 = 2,

Exercises · Q5

Q. Write the output of the following:

a)
num1 = 4
num2 = num1 + 1
num1 = 2
print (num1, num2)

b)
num1, num2 = 2, 6
num1, num2 = num2, num1 + 2
print (num1, num2)

c)
num1, num2 = 2, 3
num3, num2 = num1, num3 + 1
print (num1, num2, num3)

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Concept understanding — Variable Assignment Order

Variable Assignment Order — First Principles

Imagine you are following a recipe. The recipe says: "Take the number of eggs, call it $x$ . Then take $x$ , add 2, and call the result $y$ ." You would never dream of using $y$ before you have defined $x$ — that would be nonsense. Variable assignment order is simply the rule that a variable must be given a value before it can be used in any expression that defines another variable.

The Intuition

Think of a variable as a labelled box. You cannot take something out of a box that is empty. When you write:

x = 5

y = x + 3

you first put the number 5 into box $x$ , then you look into box $x$ , take out the 5, add 3, and put the result (8) into box $y$ . The order is: define first, use second.

Now consider what happens if you reverse the order:

y = x + 3

x = 5

At the moment you try to compute $y$ , box $x$ is empty. The expression $x + 3$ has no meaning — it is like asking "what is nothing plus 3?" In most programming languages and in mathematical reasoning, this is an error. The variable $x$ has not yet been assigned.

The Precise Statement

Important

Variable Assignment Order: In any sequence of assignments (or definitions), a variable must appear on the left-hand side of an assignment (i.e., be defined) before it appears on the right-hand side of any later assignment (i.e., before it is used). Violating this order produces an undefined variable error.

Formally, for a sequence of statements $S_1, S_2, \dots, S_n$ , if a variable $v$ is used in $S_k$ (on the right-hand side), then there must exist some $S_j$ with $j < k$ where $v$ appears on the left-hand side.

Why This Matters in Exams

In Indian board exams (CBSE, ICSE, state boards), especially in Computer Science and Mathematics, you will encounter questions that test this concept in two ways:

Tracing code: Given a sequence of assignments, you must compute the final values. The order is everything.
Spotting errors: A question may present a jumbled or incorrect sequence and ask you to identify why it fails.

A Worked Example

Consider this sequence:

a = 10
b = a + c
c = 5

At the second line, $c$ is used but has never been assigned. This violates assignment order. The correct order would be:

a = 10
c = 5
b = a + c

Now $b$ becomes $10 + 5 = 15$ .

A Common Pitfall …

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a) num2 is computed from the old num1 (4+1=5) before num1 changes → prints 2 5.

b) The right side (num2, num1 + 2) is evaluated first with the old values (6, 4) → prints 6 4.

c) The right side needs num3 + 1 but num3 does not exist yet → NameError (nothing printed).

Part	Output
a	`2 5`
b	`6 4`

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a) prints 2 5 (num2 kept the old num1+1); b) prints 6 4 (the right-hand tuple is built from the OLD values before any assignment); c) crashes with NameError — num3 appears on the right side before it exists.

The idea — assignment order. Two rules decide all three parts: (1) an assignment uses the values variables have at that moment — later changes don't back-propagate; (2) in a parallel assignment the entire right-hand side is evaluated first, then unpacked left to right.

num1 = 4
num2 = num1 + 1     # num2 = 5 (uses num1's value NOW)
num1 = 2            # num2 is unaffected
print (num1, num2)

Statement	num1	num2
num1 = 4	4	—
num2 = num1 + 1	4	5
num1 = 2	2	5

Output: 2 5

num1, num2 = 2, 6
num1, num2 = num2, num1 + 2   # RHS = (6, 2+2) = (6, 4), THEN assigned
print (num1, num2)

Statement	num1	num2
num1, num2 = 2, 6	2	6
RHS builds (6, 4), unpacks	6	4

Output: 6 4

num1, num2 = 2, 3
num3, num2 = num1, num3 + 1   # RHS needs num3 -> it does not exist yet!
print (num1, num2, num3)
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