Interesting numbers have a mathematical property that makes them handy in a certain way. For programming, these numbers present useful binary patterns that may be obvious in hexadecimal but meaningless in decimal. For non-programmers and non-mathematicians, interesting numbers offer visual patterns that most likely have no useful application beyond their look.
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\nFor this month’s Exercise, the challenge is to write code that loops from 100 to 1000 and outputs the values 111, 222, 333, 444, 555, 666, 777, 888, and 999.<\/p>\n
For my first attempt, I chose to use character comparisons. Each value is passed to the interesting()<\/em> function, where it’s converted to a string. Each character in the string is compared with each other and if they all match, the function returns 1 (TRUE):<\/p>\n By the way, the sprintf()<\/em> function in this solution is considered unsafe because the buffer can hold only three characters. For my solution, no value larger than 3 characters is passed, so it works. In the wild, however, values must be checked for overflow.<\/p>\n My second solution uses math, so it’s straightforward:<\/p>\n When Click here<\/a> to view the full source code for my first solution on Github.<\/p>\n The second solution is shown above, but you can click here<\/a> to view it on Github as well.<\/p>\n I hope you devised an interesting solution to the puzzle, which I confess is a rather simple challenge. In the larger programming picture, however, pattern-matching is a worthy field of study with various interesting applications.<\/p>\n","protected":false},"excerpt":{"rendered":" Interesting numbers have a mathematical property that makes them handy in a certain way. For programming, these numbers present useful binary patterns that may be obvious in hexadecimal but meaningless in decimal. For non-programmers and non-mathematicians, interesting numbers offer visual … Continue reading \r\nint interesting(int v)\r\n{\r\n char buffer[4];\r\n\r\n sprintf(buffer,\"%d\",v);\r\n\r\n if( buffer[0]==buffer[1] && buffer[1]==buffer[2] )\r\n return(1);\r\n else\r\n return(0);\r\n}<\/pre>\n\r\n#include <stdio.h>\r\n\r\nint main()\r\n{\r\n int x;\r\n\r\n for(x=100;x<1000;x++)\r\n {\r\n if( (x % 111) == 0 )\r\n printf(\"%d is interesting\\n\",x);\r\n }\r\n\r\n return(0);\r\n}<\/pre>\nx % 111<\/code> is equal to zero, the number contains three of the same digits and that value is output. Simple.<\/p>\n