Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

Algorithms | Misc | Question 10

  • Last Updated : 28 Jun, 2021

The procedure given below is required to find and replace certain characters inside an input character string supplied in array A. The characters to be replaced are supplied in array oldc, while their respective replacement characters are supplied in array newc. Array A has a fixed length of five characters, while arrays oldc and newc contain three characters each. However, the procedure is flawed

void find_and_replace(char *A, char *oldc, char *newc) {
    for (int i = 0; i < 5; i++)
       for (int j = 0; j < 3; j++)
           if (A[i] == oldc[j]) A[i] = newc[j];
}

The procedure is tested with the following four test cases
(1) oldc = “abc”, newc = “dab”
(2) oldc = “cde”, newc = “bcd”
(3) oldc = “bca”, newc = “cda”
(4) oldc = “abc”, newc = “bac”
The tester now tests the program on all input strings of length five consisting of characters ‘a’, ‘b’, ‘c’, ‘d’ and ‘e’ with duplicates allowed. If the tester carries out this testing with the four test cases given above, how many test cases will be able to capture the flaw?

(A) Only one
(B) Only two
(C) Only three
(D) All four


Answer: (B)

Explanation: The test cases 3 and 4 are the only cases that capture the flaw. The code doesn’t work properly when an old character is replaced by a new character and the new character is again replaced by another new character. This doesn’t happen in test cases (1) and (2), it happens only in cases (3) and (4).

Quiz of this Question

My Personal Notes arrow_drop_up
Recommended Articles
Page :