ДЗ № 12.1 Двоичное представление, Шестаков Николай#16
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NicholayShestakov wants to merge 9 commits intomainfrom
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ДЗ № 12.1 Двоичное представление, Шестаков Николай#16NicholayShestakov wants to merge 9 commits intomainfrom
NicholayShestakov wants to merge 9 commits intomainfrom
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WoWaster
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Dec 16, 2025
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| if ((1 << i) & decimal) { | ||
| binary[31 - i] = '1'; | ||
| } else { | ||
| binary[31 - i] = '0'; |
Collaborator
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Ой, ну, так тоже можно, но не проще ли хранить прямо числа 0 и 1 в char (или вообще сделать `bool)?
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| char* decimalToBinary(int32_t decimal) | ||
| { | ||
| char* binary = malloc(32 * sizeof(char)); |
Collaborator
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Хорошо бы это обернуть в структуру, чтобы пользователь ненароком в другие функции не передал массив из 2 или 122 элементов
| return binary; | ||
| } | ||
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| char* binarySum(char* binaryFirst, char* binarySecond) |
Collaborator
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Здесь получается не очень честно, потому что у Вас может получиться двойка. А если перейти на язык
электроники, то без этого можно обойтись :) https://en.wikipedia.org/wiki/Adder_(electronics)#Full_adder
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| if (binary[0] == '1') { | ||
| ++decimal; | ||
| decimal *= -1; |
Collaborator
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Может просто вычесть 2^31?
| #include <stdlib.h> | ||
| #include <string.h> | ||
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| void test() |
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Хорошо бы иметь побольше тестов, но ладно
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