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added 2 commits
February 6, 2025 13:35
give information about code
add feature calculator add calculator test
moongua404
reviewed
Feb 9, 2025
moongua404
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moongua404
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되게 깔끔하고 간결한 코드를 짜신 것 같아요! 첫 주차인데도 정규형을 적극적으로 활용하셔서 더 그런 것 같아요. 1주차 미션도 수고 많으셨습니다ㅎㅎ
| assertSimpleTest(() -> | ||
| assertThatThrownBy(() -> runException("//#\\n0#30")) | ||
| .isInstanceOf(IllegalArgumentException.class) | ||
| ); |
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반복되는 테스트는 Parameterized Test 등으로 처리할 수 있어요!
| String patternString = "^(?://(.{1})\\\\n(.*)|(?!//).*)$"; | ||
| Pattern pattern = Pattern.compile(patternString); | ||
| Matcher matcher = pattern.matcher(string); | ||
| boolean isContain = matcher.matches(); |
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정규표현식을 씀에도 함수 분리가 안돼서 불필요하게 필드가 낭비되는 것 같아요
| throw new IllegalArgumentException("Not in rule. Can't caculate this expression"); | ||
| } | ||
| } | ||
| } |
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주석에 설명적인 부분은 README.md에 정리하고 흐름적인 부분은 함수 분리로 함수명을 통해 전달하면 좋을 것 같아요
| String[] splitExpression = expression.split(delimiter); | ||
| int sum = 0; | ||
| //소수를 표현하기 위한 .이 있으면 소수가 아니라 구분자로 인식될 것이기에 소수는 없다고 생각해도 된다. | ||
| for (int i =0; i < splitExpression.length; i++){ |
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for 문 별다른 이유가 없다면 인덱스 접근보다 원소로 접근하는게 효율적일 것 같아요!
piamin04
reviewed
Feb 11, 2025
| int sum = 0; | ||
| //소수를 표현하기 위한 .이 있으면 소수가 아니라 구분자로 인식될 것이기에 소수는 없다고 생각해도 된다. | ||
| for (int i =0; i < splitExpression.length; i++){ | ||
| if (!splitExpression[i].matches("^[1-9][0-9]*$")) { |
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현재 splitExpression[i].matches("^[1-9][0-9]*$")로 숫자를 검증하고 있는데,
매번 정규표현식을 컴파일하는 것보다 숫자 검증을 위한 정규표현식을 별도의 메서드로 분리하는건 어떤가요..?
szoon2426
reviewed
Feb 12, 2025
| String patternString = "^(?://(.{1})\\\\n(.*)|(?!//).*)$"; | ||
| Pattern pattern = Pattern.compile(patternString); | ||
| Matcher matcher = pattern.matcher(string); | ||
| boolean isContain = matcher.matches(); |
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pattern이랑 matcher라는걸 방금 알았어요 ㄷㄷ 저는 코드로 하나하나 다짰는데 너무 쉽고 간편한 방법이 있었네요
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calculator 기능 구현 및 test 추가