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3 Задача с зачёта - #27
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Palezehvat wants to merge 3 commits into
Open
Palezehvat wants to merge 3 commits into
Palezehvat wants to merge 3 commits into
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yurii-litvinov
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yurii-litvinov
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Автомат правильный. сделано не совсем то, что просили и не очень аккуратно
| #include <stdbool.h> | ||
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| typedef enum Stage { | ||
| first, |
| fourth, | ||
| fifth, | ||
| sixth | ||
| } Stage; |
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Названия состояний не мнемоничны, не отражают их назначения и по сути ничем не лучше чисел. Как в анекдоте, const int one = 1;
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| if (symbol >= 'A' && symbol <= 'Z' || symbol == '+' || symbol == '-' | ||
| || symbol >= '0' && symbol <= '9' || symbol == '.' || symbol == '%') { |
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Сложные логические условия, тем более если планируется их переиспользовать, лучше выносить в функцию
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| char buffer[1000] = { '\0' }; | ||
| char result[10000] = { '\0' }; |
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Это вообще не нужно, задачей было по строке сказать "да/нет", возвращать что-либо не просили
| } else { | ||
| memset(buffer, 0, strlen(buffer)); | ||
| indexBuffer = 0; | ||
| stage = first; |
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Не-а, тут надо было вернуть false, потому что строка уже не соответствует регэкспу
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| #include <stdio.h> | ||
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| //Variable for error checking |
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Suggested change
| //Variable for error checking | |
| // Variable for error checking |
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