The process of converting the sample program from one that relies upon constants to one that exclusively uses variables is nearly complete. Only two steps are left:
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By obtaining the widest string in the array (from last week’s Lesson<\/a>), and assuming that output is sent to a screen that’s 80-columns wide, you can calculate the maximum number of columns possible and then determine the rows required for output. Both values, columns and rows, are then used for the sorted-column process outlined in an earlier Lesson<\/a> in this series.<\/p>\n The first step is to determine how many columns of strings can fit comfortably across the screen. The assumed width is 80 characters. So, logically, you divide 80 by the widest string in the array. The problem with this approach is that the width value could divide evenly into 80. That would leave no space between the columns. Therefore, you must also add padding to the end of each string displayed, then calculate the number of columns:<\/p>\n Above, the widest()<\/em> function, presented in last week’s Lesson, obtains the longest string length from array The if-else<\/em> structure adds padding to the length of the widest string. If the string is an odd number, determined when To determine how many columns of output can fit across the screen, I use this calculation:<\/p>\n The To obtain the number of rows, I use this equation:<\/p>\n The number of For example, suppose the array has 36 items and the Now you have all four former-constants translated into variables. It’s possible at this point to present the final code, but I have one more trick to show you, which I’ll save for next week’s Lesson<\/a>, the final installment of this series.<\/p>\n","protected":false},"excerpt":{"rendered":" With almost all the pieces in place, it’s time to calculate the number of columns and rows for program output. Continue reading \r\n\/* set string width for printf() format *\/<\/span>\r\n width = widest(fruit,items);\r\n if( width % 2 )\r\n width += 3; \/* assert even width *\/<\/span>\r\n else\r\n width += 2; \/* at least two spaces padding *\/<\/span><\/pre>\nfruit[]<\/code>. The items<\/code> variable represents the number of elements in fruit[]<\/code>. The value returned, the length of the longest string, is stored in variable width<\/code>.<\/p>\nwidth%2<\/code> is true, width<\/code> is increased by 3. Otherwise width<\/code> is increased by 2. That way at least 2 spaces of padding are added to each column in the output.<\/p>\ncolumns = OUTPUT_WIDTH\/width;<\/code><\/p>\nOUTPUT_WIDTH<\/code> constant is set to 80. So if the widest string is 11 characters, the code would pad that out to 14 characters: 80\/14 is just under six, so the integer result is 5. The output for an array with the widest string at 11 characters would be displayed in 5 columns.<\/p>\nrows = items\/columns+1;<\/code><\/p>\nitems<\/code> in the array divided by columns<\/code> obtains the number of rows<\/code> necessary in a table — but, 1 is added to the result. This increment accounts for the rounding error in OUTPUT_WIDTH\/width<\/code> for the columns<\/code> calculation.<\/p>\nOUTPUT_WIDTH\/width<\/code> calculation results in 5 columns: 36\/5 results in 7 rows (due to integer rounding). Yet 7 rows of 5 columns shows only 35 items. Therefore, you must increment the value of rows<\/code> to ensure that all array elements are displayed in the table.<\/p>\n