May the 4th be with you! It’s Jedi day, or Star Wars day, or whatever. Here’s a binary puzzle for you, one which demonstrates how the bitwise OR operator can be used to set bits in an integer value.
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\r\n#include <stdio.h>\r\n\r\nchar *binbin(int n);\r\n\r\nint main()\r\n{\r\n int input,result,bit,x;\r\n\r\n bit = 1;\r\n printf(\"Type a number between 0 and 255: \");\r\n scanf(\"%d\",&input);\r\n printf(\"%s\\n========\\n\",binbin(input));\r\n for(x=0;x<8;x++)\r\n {\r\n result = bit | input;\r\n printf(\"%s\\n\",binbin(result));\r\n bit*=2;\r\n }\r\n return(0);\r\n}\r\n\r\nchar *binbin(int n)\r\n{\r\n static char bin[9];\r\n int x;\r\n\r\n for(x=0;x<8;x++)\r\n {\r\n bin[x] = n & 0x80 ? '1' : '0';\r\n n <<= 1;\r\n }\r\n bin[x] = '\\0';\r\n return(bin);\r\n}\r\n<\/pre>\nThe code asks for input, a teensy int value (8 bits). That value is then manipulated using the logical OR for each bit position, 0 through 7.<\/p>\n
Here's how the output looks when zero is input:<\/p>\n