If you’ve used any legacy C libraries before, you’ve probably used these things called bit fields, even if you didn’t know what they were.<\/p>\n
Bit fields are a way to efficiently store multiple boolean values in one or more bytes. If you can imagine an 8 bit integer as binary, each bit would have to be either a 1 or a 0, corresponding to an “On” or “Off” value. With just one 8 bit variable, you can store 8 on\/off values for your program. The syntax on how to set, unset, and retrieve these values can be a little confusing however. The method for using these helpful little bit fields involves some binary math with the binary operators AND, OR, and NOT. In C based languages these are written as &, |, and, ~, respectively. I’ll give an example in C that will hopefully make this math clearer. First we need to define some values as our flags. Each flag has to have a value that is a power of 2. For example:<\/p>\n
enum FLAGS\r\n{\r\n\tFLAG_A = 0x01,\/\/ 1\r\n\tFLAG_B = 0X02,\/\/ 2\r\n\tFLAG_C = 0X04,\/\/ 4\r\n\tFLAG_D = 0X08,\/\/ 8\r\n\tFLAG_E = 0X10,\/\/ 16\r\n\tFLAG_F = 0X20 \/\/ 32\r\n\r\n\t\/\/ so on and so forth\r\n};\r\n<\/pre>\nThese could just as easily be written as the actual integer values, however I’ll stick to HEX notation for this example. Next we need a variable of at least 6 bits to hold our flags’ on or off state. Let’s use an integer type, and set all the flags to “Off.”<\/p>\n
static int THE_OPTION = 0x00;\r\n<\/pre>\nNow we’re ready to set, unset, and test some flags! To set the option to “On” for a certain flag, we must turn the bit “On” in our variable. To do this we use the OR operation. I explain why in the example below.<\/p>\n
FLAG_C = 4, in 8-bit binary this is 00000100\r\nTHE_OPTION = 0, in 8-bit binary\t 00000000\r\n\r\nif we OR them:\r\n\r\nTHE_OPTION |= FLAG_C\r\n\r\nthe binary math looks like:\r\n\r\n\t00000000\r\nOR\t00000100\r\n-------------\r\n\t00000100\r\n\r\n<\/pre>\nThis is a simple example but the point holds. To apply a flag, we apply the OR operator. To remove an option, we use the AND operator.<\/p>\n
THE_OPTION = 0B00000100; \/\/binary notation for 4\r\n\r\nTHE_OPTION &= ~FLAG_C; \/\/ THE_OPTION AND-EQUALS NOT FLAG_C\r\n\r\nthe binary math looks like:\r\n\r\n\t00000100\r\nAND\t11111011\r\n-------------\r\n\t00000000\r\n\r\n<\/pre>\nTo check the status of a flag in THE_OPTION, one can simply check:<\/p>\n
if(THE_OPTION & FLAG_F)\r\n{\r\n\t\/\/do things\r\n}\r\n<\/pre>\nThese can also be chained, for instance<\/p>\n
#include \"stdio.h\"\r\nenum FLAGS\r\n{\r\n\tFLAG_A = 0x01,\/\/ 1\r\n\tFLAG_B = 0X02,\/\/ 2\r\n\tFLAG_C = 0X04,\/\/ 4\r\n\tFLAG_D = 0X08,\/\/ 8\r\n\tFLAG_E = 0X10,\/\/ 16\r\n\tFLAG_F = 0X20 \/\/ 32\r\n\r\n\t\/\/ so on and so forth\r\n};\r\n\r\nstatic int THE_OPTION = 0x00;\r\nvoid test_options()\r\n{\r\n\tchar line[512], *p;\r\n\tp = &line;\r\n\tp += sprintf(p,\"THE_OPTION contains \");\r\n\tif(FLAG_A & THE_OPTION)\r\n\t{\r\n\t\tp += sprintf(p, \"FLAG_A and \");\r\n\t}\r\n\tif(FLAG_B & THE_OPTION)\r\n\t{\r\n\t\tp += sprintf(p, \"FLAG_B and \");\r\n\t}\r\n\tif(FLAG_C & THE_OPTION)\r\n\t{\r\n\t\tp += sprintf(p, \"FLAG_C and \");\r\n\t}\r\n\tif(FLAG_D & THE_OPTION)\r\n\t{\r\n\t\tp += sprintf(p, \"FLAG_D and \");\r\n\t}\r\n\tif(FLAG_E & THE_OPTION)\r\n\t{\r\n\t\tp += sprintf(p, \"FLAG_E and \");\r\n\t}\r\n\tif(FLAG_F & THE_OPTION)\r\n\t{\r\n\t\tp += sprintf(p, \"FLAG_F and \");\r\n\t}\r\n\tprintf(\"%s, that is all.\\n\", line);\r\n}\r\nint main()\r\n{\r\n\r\n\tTHE_OPTION = (FLAG_A | FLAG_C | FLAG_F);\r\n\ttest_options();\r\n\r\n\tTHE_OPTION &= ~FLAG_A;\r\n\ttest_options();\r\n\r\n\tTHE_OPTION &= ~(FLAG_C | FLAG_F);\r\n\ttest_options();\r\n\r\n\tTHE_OPTION = 0XFF;\r\n\ttest_options();\r\n\treturn 0;\r\n}\r\n\r\n\r\n\r\n<\/pre>\nThat will print<\/p>\n
THE_OPTION contains FLAG_A and FLAG_C and FLAG_F and , that is all.\r\nTHE_OPTION contains FLAG_C and FLAG_F and , that is all.\r\nTHE_OPTION contains , that is all.\r\nTHE_OPTION contains FLAG_A and FLAG_B and FLAG_C and FLAG_D and FLAG_E and FLAG_F and , that is all.\r\n<\/pre>\n