The C library holds functions that help you round-off floating point values. I can name a few, but then it would make this month’s rounding exercise too easy. So I won’t!
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\nRounding is a science. It involves three approaches:<\/p>\n
The issue of how to deal with 0.5 exactly is a source of debate, a topic I don’t want to get into.<\/p>\n
For this month’s Exercise, you are to craft your own rounding functions. Two floating point values are given in the code skeleton, shown below. Your job is to round those values straight up, straight down, or split the rounding based on the fractional part of the number. You make that happen by filling in code for the roundup()<\/em>, rounddown()<\/em>, and roundhalf()<\/em> functions.<\/p>\n Here’s a major hint, which I’m providing gratis because I’m forbidding you from using any existing C library functions to handle the rounding process: You can easily lop off the fractional part of a floating point variable by typecasting it to an int<\/em>.<\/p>\n Also, once you figure out the roundup()<\/em> function, doing rounddown()<\/em> is a snap. And then doing the roundhalf()<\/em> function is equally as easy.<\/p>\n Click here<\/a> to view my solution, although I recommend that you first try to solve the puzzle on your own.<\/p>\n","protected":false},"excerpt":{"rendered":" Round up. Round down. Round halfsies. Continue reading \r\n#include <stdio.h>\r\n\r\nfloat roundup(float n)\r\n{\r\n}\r\n\r\nfloat rounddown(float n)\r\n{\r\n}\r\n\r\nfloat roundhalf(float n)\r\n{\r\n}\r\n\r\nint main()\r\n{\r\n float v1,v2;\r\n\r\n v1 = 1234.56789;\r\n v2 = 98765.4321;\r\n\r\n printf(\"Value v1 = %f\\n\",v1);\r\n printf(\"Value v2 = %f\\n\",v2);\r\n printf(\"Rounded up:\\t%f\\t%f\\n\",\r\n roundup(v1),\r\n roundup(v2));\r\n printf(\"Rounded down:\\t%f\\t%f\\n\",\r\n rounddown(v1),\r\n rounddown(v2));\r\n printf(\"Rounded half:\\t%f\\t%f\\n\",\r\n roundhalf(v1),\r\n roundhalf(v2));\r\n\r\n return(0);\r\n}<\/pre>\n