rw7026
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AnswersAn overseas customer is reporting a crash of a released driver which has undergone formal in-house testing. This has impacted their delivery schedule and is looking to cost them millions of dollars in lost revenue. What steps would you take to alleviate this situation and close it off?
- rw7026 in Canada for Automobile| Report Duplicate | Flag | PURGE
QNX Software Engineer in Test Testing
Before you delving into the test cases, ask the interviewer questions like where is the toaster being used? for restaurant, home use? e.g. if for restaurant, you need to more focus on testing its reliability, like toast many breads with different modes being set. Also, check the toaster's reaction when it is being misused. Like turn off the power while toasting a bread. Changing the mode while toasting. Push off the bread while toasting.
- rw7026 May 11, 2013I think:
1. compute: sqrt(1000000000) = 31622.78 and sqrt(9000000000) = 94868.33;
2. from 31623 to 94868:
first compute the squire of every integer between this range;
check if the value is non repeated 10 digits, if yes, then put it in an array.
3. The result is the array
The total times of computation equals to 94868 - 31623 = 63245.
haha! Exactly! My answer is: I would first try to reproduce the crash. If it can be reproduced, then I will first tell customer to change his/her driver to the last version which does not have crash issue. In the meantime, I will notice the customer that I will try to fix this issue and get back to him/her later. Then, I will look into the code, do a regression test, try to fix the bug. If I could, I will update a new build and do the rest of the testings. If I couldn't, I will describe the problem and report it to develop team to get it solved.
Do you think it is a good answer? Thank you!
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for rand7, each number between 1 and 7 has roughly 15% probability to be selected.
- rw7026 May 12, 2013for rand5, each number between 1 and 5 has 20% probability to be selected.
What about we use rand5() to control the probabilities for 1-7 to be selected?
Like for rand5, the probability of a number is not being selected is 80%.
code:
int a = rand5(), b = rand5();
if (a!=1 && b != 1) return 1;
else if (a!=1 && b != 2) return 2;
else if (a!=1 && b != 3) return 3;
else if (a!=1 && b != 4) return 4;
else if (a!=1 && b != 5) return 5;
else if (a!=2 && b != 1) return 6;
else if (a!=3 && b != 1) return 7;