Question

Consider the following procedures for string manipulation. Procedure Call Explanation concat(str1, str2) Returns a single string consisting of str1 followed by str2. For example, concat("key", "board") returns "keyboard". substring(str, start, length) Returns a substring of consecutive characters from str, starting with the character at position start and containing length characters. The first character of str is located at position 1. For example, substring("delivery", 3, 4) returns "live". len(str) Returns the number of characters in str. For example, len("pizza") returns 5. Assume that the string oldString contains at least 4 characters. A programmer is writing a code segment that is intended to remove the first two characters and the last two characters from oldString and assign the result to newString. For example, if oldString contains "student", then newString should contain "ude". Which of the following code segments can be used to assign the intended string to newString ? Select two answers. newString substring(oldString, 3, len(oldString) - 4) Option A,newString ← substring(oldString, 3, len(oldString) - 4) newString substring(oldString, 3, len(oldString) - 2) Option B,newString ← substring(oldString, 3, len(oldString) - 2) tempString substring(oldString, 3, len(oldString) - 2) newString substring(tempString, 1, len(tempString) - 2) Option C,tempString ← substring(oldString, 3, len(oldString) - 2) newString ← substring(tempString, 1, len(tempString) - 2) tempString1 substring(oldString, 1, 2) tempString2 substring(oldString, len(oldString) - 2, 2) newString concat(tempString1, tempString2)

Answer 1

The correct choices are A and C.

- A: newString ← substring(oldString, 3, len(oldString) - 4)
- Starts at character 3 and takes (n - 4) characters → keeps characters 3 through n-2 (removes first two and last two).

- C: tempString ← substring(oldString, 3, len(oldString) - 2)
newString ← substring(tempString, 1, len(tempString) - 2)
- tempString is the substring from position 3 to the end (length n-2). Then taking its first (n-4) characters removes the last two of tempString, leaving characters 3 through n-2 of the original.

B is wrong because it takes too many characters (it would include the original last two). D is wrong because it constructs a string of the first two + last two instead of removing them.