Slice the string.

Length and case.

In the lab, choose the String handling program. It uses all four operations to count the vowels in a word; step through it and watch each letter being taken out with SUBSTRING.

DECLARE Name : STRING
Name ← "Ada Lovelace"
OUTPUT LENGTH(Name)
OUTPUT UCASE(Name)
OUTPUT LCASE(Name)

This outputs 12, ADA LOVELACE and ada lovelace. The space counts as a character, so LENGTH is 3 + 1 + 8 = 12.

• LENGTH: returns an INTEGER, useful as the end value of a loop and for length checks.
• UCASE and LCASE: return a new string; Name itself is unchanged unless you assign the result back.
• Comparing: "smith" and "Smith" are different strings. Convert both to the same case before comparing.

The same idea in Python (for comparison only: the exam uses pseudocode)

name = "Ada Lovelace"
print(len(name))      # 12
print(name.upper())   # ADA LOVELACE
print(name.lower())   # ada lovelace

SUBSTRING.

SUBSTRING needs three things: the string, the position to start at, and how many characters to take.

DECLARE Word : STRING
Word ← "COMPUTER"
OUTPUT SUBSTRING(Word, 4, 3)
OUTPUT SUBSTRING(Word, LENGTH(Word), 1)

The same idea in Python (for comparison only: the exam uses pseudocode)

word = "COMPUTER"
print(word[3:6])      # PUT: Python counts from 0 and stops before 6
print(word[-1])       # R

Strings in a loop.

To look at every character, loop from 1 to LENGTH and take a substring of length 1 each time.

DECLARE Word : STRING
DECLARE Count : INTEGER
DECLARE Index : INTEGER
DECLARE Letter : CHAR
INPUT Word
Count ← 0
FOR Index ← 1 TO LENGTH(Word)
    Letter ← LCASE(SUBSTRING(Word, Index, 1))
    IF Letter = 'e'
      THEN
        Count ← Count + 1
    ENDIF
NEXT Index
OUTPUT Count

The same idea in Python (for comparison only: the exam uses pseudocode)

word = input()
count = 0
for letter in word.lower():
    if letter == "e":
        count = count + 1
print(count)