3.5 if Statement

• Pseudocode: Suppose that a passing grade on an examination is 60. The pseudocode

If student’s grade is greater than or equal to 60
      Display 'Passed'

• If the condition is true, 'Passed' is displayed. Then, the next pseudocode statement in order is “performed.”
• If the condition is false, nothing is displayed, and the next pseudocode statement is “performed.”
• Indentation emphasizes that 'Passed' is displayed only if the condition is true.

Corresponding if Statement

grade = 85

if grade >= 60:
    print('Passed')

Passed

Suite Indentation

• Indenting a suite is required.

if grade >= 60:
print('Passed')  # statement is not indented properly

  File "<ipython-input-3-eb5359d7857b>", line 2
    print('Passed')  # statement is not indented properly
        ^
IndentationError: expected an indented block

• Statements in a suite must have the same indentation.

if grade >= 60:
    print('Passed')
  print('Good job!)

  File "<tokenize>", line 3
    print('Good job!)
    ^
IndentationError: unindent does not match any outer indentation level

if Statement Flowchart

• The decision (diamond) symbol contains a condition that can be either True or False.
• Two flowlines emerging from it:
  • One indicates the direction to follow when the condition in the symbol is True.
  • The other indicates the direction to follow when the condition is False.

Every Expression Can Be Treated as True or False

if 1:
    print('Nonzero values are true, so this will print')

Nonzero values are true, so this will print

if 0:
    print('Zero is false, so this will not print')

An Additional Note on Confusing == and =

• Using == instead of = in an assignment statement can lead to subtle problems.
• Writing grade == 85 when we intend to define a variable with grade = 85 would cause a NameError.
• Logic error: If grade had been defined before the preceding statement, then grade == 85 would evaluate to True or False, depending on grade’s value, and not perform the intended assignment.

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