Домашнее задание 12.1. Двоичное представление. Разгуляева А.И.#12
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Домашнее задание 12.1. Двоичное представление. Разгуляева А.И.#12
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p-senichenkov
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Dec 2, 2025
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Нет тестов, CMake и CI
src/hw_12-1_binary/hw_12-1_binary.c
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| #include <stdlib.h> | ||
| #include <string.h> | ||
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| // функция для определения необходимого количества бит (двоичное представление зависит от количества бит) |
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Во-первых это оверкил. Во-вторых, это неприменимо на практике.
Нужно выбрать какую-то фиксированную длину представления и везде её использовать.
Bumps [actions/checkout](https://github.com/actions/checkout) from 5 to 6. - [Release notes](https://github.com/actions/checkout/releases) - [Changelog](https://github.com/actions/checkout/blob/main/CHANGELOG.md) - [Commits](actions/checkout@v5...v6) --- updated-dependencies: - dependency-name: actions/checkout dependency-version: '6' dependency-type: direct:production update-type: version-update:semver-major ... Signed-off-by: dependabot[bot] <support@github.com>
p-senichenkov
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Dec 26, 2025
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| } | ||
| // если число отрицательное (старший бит = 1, дополнительный код) | ||
| else { |
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По стайлгайду else форматируется не так
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| // определяем необходимое количество бит | ||
| int bits = 8; |
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Очень странный комментарий. И bits лучше сделать символьной константой.
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| // заполняем строку нулями | ||
| for (int i = 0; i < bits; i++) { | ||
| binary[i] = '0'; |
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Удобнее было бы работать с bool, или хотя бы 0 и 1, а не символами
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