Control flow · about 40 min

Making decisions: if, elif and else

Why this matters

So far, every program has run every line, top to bottom, the same way for every input. Real programs make decisions: reject invalid input, charge shipping only under a certain order total, print an error if a file is missing. The if statement is how a program chooses between paths based on data. It is also where you meet Python's most distinctive feature: indentation defines structure.

The if statement

temperature = int(input())
if temperature > 30:
    print("It's hot.")
    print("Drink water.")
print("Have a nice day.")

With input 35:

It's hot.
Drink water.
Have a nice day.

With input 20:

Have a nice day.

The anatomy:

if is followed by a condition, any expression; Python evaluates it and checks whether it is truthy (recall truthiness from the expressions module: 0, "", None and empty collections are falsy, almost everything else is truthy).
A colon : ends the header line.
The block (or body) is the indented lines below. They run only if the condition is truthy.
The first line indented back to the if's level ends the block. print("Have a nice day.") is not part of the if; it always runs.

Indentation is syntax

In many languages, braces { } mark blocks and indentation is just a courtesy to readers. In Python, the indentation is the block marker. The tokenizer (remember the compile pipeline) turns an increase in indentation into an INDENT token and a decrease into a DEDENT token, and the grammar uses those exactly as other languages use braces.

if temperature > 30:        <- header, ends with :
    print("It's hot.")      <- INDENT: block starts
    print("Drink water.")   <- same level: same block
print("Have a nice day.")   <- DEDENT: block ended

The convention, which you should follow, is 4 spaces per level. Any consistent amount works, but every line in a block must be indented by exactly the same amount. Never mix tabs and spaces; configure your editor to insert spaces when you press Tab.

else: the other path

else gives a block that runs when the condition is falsy. Exactly one of the two blocks runs:

age = int(input())
if age >= 18:
    print("adult")
else:
    print("minor")

else has no condition of its own, and it lines up with its if.

elif: many paths

For more than two cases, chain conditions with elif (short for "else if"):

n = int(input())
if n > 0:
    print("positive")
elif n < 0:
    print("negative")
else:
    print("zero")

Python tests the conditions in order and runs the block of the first one that is truthy. Once a branch runs, the rest of the chain is skipped entirely; their conditions are not even evaluated. If no condition is true, the else block runs (if there is one; it is optional).

      n > 0 ? --yes--> print("positive") ---------------+
        | no                                            |
      n < 0 ? --yes--> print("negative") ---------------+
        | no                                            |
      else    -------> print("zero") -------------------+
                                                        v
                                               continue after chain

Because only the first match runs, order matters when conditions overlap. Grading from high to low lets each test assume the higher ones failed:

score = int(input())
if score >= 90:
    grade = "A"
elif score >= 80:      # we already know score < 90 here
    grade = "B"
elif score >= 70:
    grade = "C"
else:
    grade = "F"
print(grade)

Reversing the order (testing >= 70 first) would give every passing student a C.

Combining conditions

Conditions can use and, or, not and chained comparisons, all from the expressions module:

x = int(input())
if 1 <= x <= 10:
    print("in range")
if x < 0 or x > 100:
    print("out of bounds")
if not (x % 2 == 0):
    print("odd")

Prefer positive, readable conditions: if x % 2 == 1 reads better than if not (x % 2 == 0).

Nesting

A block can contain any statements, including another if. Each level adds 4 more spaces:

username = input()
password = input()
if username == "ada":
    if password == "engine":
        print("welcome")
    else:
        print("wrong password")
else:
    print("unknown user")

Nesting is sometimes the clearest structure, but deep nesting is hard to read. Often it can be flattened with and or with elif:

if username != "ada":
    print("unknown user")
elif password != "engine":
    print("wrong password")
else:
    print("welcome")

Both versions behave the same. The second reads as a list of cases, which is easier to scan.

pass: an empty block

A block cannot be empty; Python requires at least one statement. When you need a placeholder, use pass, a statement that does nothing:

if x > 0:
    pass   # TODO: handle positives
else:
    print("not positive")

The conditional expression

When you just need to choose between two values, there is a compact expression form:

age = 20
status = "adult" if age >= 18 else "minor"
print(status)
print(f"{age} is {'even' if age % 2 == 0 else 'odd'}")

adult
20 is even

Read A if condition else B as "A, if the condition holds, otherwise B". It is an expression (it produces a value), so it can go inside an f-string, a function call or an assignment. Use it for simple choices; for anything with side effects or more than two outcomes, a full if statement is clearer.

Common mistakes

Forgetting the colon.

if x > 3
    print("big")

  File "main.py", line 1
    if x > 3
            ^
SyntaxError: expected ':'

Inconsistent indentation.

if x > 3:
    print("big")
      print("really big")

  File "main.py", line 3
    print("really big")
IndentationError: unexpected indent

And a missing indent after a header gives IndentationError: expected an indented block after 'if' statement on line 1.

Using = instead of ==. if x = 3: is a SyntaxError (Python helpfully suggests Maybe you meant '==' or ':=' instead of '='?). = assigns; == compares.

Misreading or. if answer == "y" or "yes": is always true, because it means (answer == "y") or ("yes"), and the non-empty string "yes" is truthy. Write if answer == "y" or answer == "yes":, or if answer in ("y", "yes"):.

Overlapping conditions in the wrong order. As in the grade example, a broad condition placed first swallows the cases meant for later branches.

Recap

if condition: runs its indented block only when the condition is truthy; elif adds more conditions; else catches everything left.
In an if/elif/else chain, exactly one block runs at most: the first whose condition is true.
Indentation defines blocks; use 4 spaces and be consistent. Python turns indentation changes into INDENT and DEDENT tokens.
pass is an empty statement for placeholder blocks.
A if condition else B is an expression that picks one of two values.

Going further: run import dis; dis.dis("if x > 0:\n y = 1\nelse:\n y = 2") and find the POP_JUMP_IF_FALSE instruction: at bytecode level, an if is a comparison followed by a conditional jump over the block.

Exercises

Exercise 1 · write code

Positive, negative or zero

Read a whole number and print exactly one word: positive, negative or zero.

File: main.py

Starting point

n = int(input())
if n > 0:
    print("positive")

Example

input

expected output

positive

Example

input

-7

expected output

negative

Hint 1

You need three branches: if, elif and else.

Hint 2

Only the first branch whose condition is true runs.

Reference solution

n = int(input())
if n > 0:
    print("positive")
elif n < 0:
    print("negative")
else:
    print("zero")

Exercise 2 · quiz

Which branch runs?

What does this print when score is 95?

score = 95
if score >= 50:
    print("pass")
elif score >= 90:
    print("excellent")
else:
    print("fail")

pass
excellent
pass and excellent
fail

Answer

pass

Python checks the conditions top to bottom and runs only the first branch whose condition is true; the rest are skipped without even being evaluated. 95 >= 50 is true, so pass is printed and the elif is never reached. When conditions overlap, put the most specific one first: test >= 90 before >= 50.

Exercise 3 · write code

Letter grades

Read a whole-number score and print a letter grade:

score	output
90 to 100	`A`
80 to 89	`B`
70 to 79	`C`
60 to 69	`D`
0 to 59	`F`

If the score is below 0 or above 100, print invalid score instead.

File: main.py

Starting point

score = int(input())
if score >= 60:
    print("D")
elif score >= 90:
    print("A")
else:
    print("F")

Example

input

expected output

Example

input

expected output

Hint 1

Check for invalid input first, so the grade branches only see valid scores.

Hint 2

Order the grade checks from highest to lowest; then each elif only needs a lower bound.

Reference solution

score = int(input())
if score < 0 or score > 100:
    print("invalid score")
elif score >= 90:
    print("A")
elif score >= 80:
    print("B")
elif score >= 70:
    print("C")
elif score >= 60:
    print("D")
else:
    print("F")

Exercise 4 · write code

Leap years

Read a year (a positive whole number) and print YEAR is a leap year or YEAR is not a leap year, with the year in place of YEAR. For input 2024:

2024 is a leap year

The Gregorian rule: a year is a leap year if it is divisible by 4, except years divisible by 100, which are not, except years divisible by 400, which are. So 2024 and 2000 are leap years; 1900 and 2023 are not.

Use a conditional expression (X if condition else Y) or if/else, whichever you find clearer.

File: main.py

Starting point

year = int(input())
if year % 4 == 0:
    print(f"{year} is a leap year")
else:
    print(f"{year} is not a leap year")

Example

input

expected output

2024 is a leap year

Example

input

expected output

2023 is not a leap year

Hint 1

year % 4 == 0 tests divisibility by 4.

Hint 2

One way to write the rule: divisible by 4 and (not divisible by 100 or divisible by 400).

Hint 3

Test 1900 (not leap) and 2000 (leap): they are where simple solutions go wrong.

Reference solution

year = int(input())
leap = year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
word = "a" if leap else "not a"
print(f"{year} is {word} leap year")

Exercise 5 · write code

Classify a triangle

Read three whole numbers on one line, separated by spaces: the side lengths of a triangle. Print one of:

not a triangle if any side is zero or negative, or if the sides cannot form a triangle (some side is greater than or equal to the sum of the other two)
equilateral if all three sides are equal
isosceles if exactly two sides are equal
scalene if all sides differ

For input 3 4 5 the output is scalene. You can read the three numbers with:

parts = input().split()
a = int(parts[0])
b = int(parts[1])
c = int(parts[2])

File: main.py

Starting point

parts = input().split()
a = int(parts[0])
b = int(parts[1])
c = int(parts[2])
if a == b == c:
    print("equilateral")
else:
    print("scalene")

Example

input

3 4 5

expected output

scalene

Example

input

2 2 2

expected output

equilateral

Hint 1

Rule out invalid triangles first, then classify.

Hint 2

Since equilateral is checked before isosceles, a == b or b == c or a == c is enough for isosceles.

Hint 3

Degenerate input such as 1 2 3 (1 + 2 equals 3) is not a triangle.

Reference solution

parts = input().split()
a = int(parts[0])
b = int(parts[1])
c = int(parts[2])
if a <= 0 or b <= 0 or c <= 0 or a >= b + c or b >= a + c or c >= a + b:
    print("not a triangle")
elif a == b == c:
    print("equilateral")
elif a == b or b == c or a == c:
    print("isosceles")
else:
    print("scalene")