ITP_JAN_25 | HASAN DEMIROZ | Module-Structuring-and-Testing-Data | Sprint 3 #424
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| if (numerator < denominator) return true; | ||
| if (Math.abs(numerator) < Math.abs(denominator)) | ||
| return true; | ||
| else if (Math.abs(numerator) > Math.abs(denominator)) return false; |
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if you use >==, that would include situation 2,3 into one, right?
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I have updated the code and used less line of codes. Thanks.
| // ====> complete with your assertion | ||
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| // Stretch: | ||
| // What other scenarios could you test for? | ||
| const numeratorZero = isProperFraction(0, 3); |
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should this return true, 0/3 is fraction or not? But it's nice you come up with this corner case.
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The function updated and it returns false for the 0/3 is fraction or not case.
| if (rank === "A") return 11; | ||
| if (["J", "Q", "K", "10"].includes(rank)) return 10; |
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what happend if the input is something like "a♠" , 'j♠", should it still be accounted as the right input or not?
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The code has updated. Before the statement it makes it uppercase the it.
| @@ -1,6 +1,9 @@ | |||
| function isProperFraction(numerator, denominator) { | |||
| if (numerator < denominator) return true; | |||
| // add your completed function from key-implement here | |||
| else if (numerator > denominator) return false; | |||
| else if(numerator === denominator) return false; | |||
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is that the function you already included in the previous file? where you have the math.abs(). also considering the situation of numerator and denominator could be zero.
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The code has updated and combined the else if cases.
| }); | ||
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| test("should return 'th' for greater than 3", () => { | ||
| for (let number = 4; number < 20; number++) { |
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11th has been handled, but how about 21st, 31st?
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The code has updated according to edge cases and works fine now.
| const char = "c"; | ||
| const count = countChar(str,char); | ||
| expect(count).toEqual(0); | ||
| }); |
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there is a corner case which could potentially failed . when findchar is empty string "".
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