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I have a smart microwave oven. Say I need to heat my tea for a minute and a half. I type in 90. When I press the Start button, the 90 transforms into 1:30, a minute thirty. It’s the same quantity of time, but expressed differently. Such magic is merely clever programming.
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I screwed up some code in a most delightful manner, all without knowing what I did. In fact, it was a student who ran my less-than-brilliant code and discovered the obvious flaw. But I was clueless.
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The strcasecmp() function isn’t part of the standard C library, though many compilers feature it. As having the function is handy, I wrote about crafting your own version back in 2019. Alas, that implementation is flawed.
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Any mortal programmer would have stopped with last week’s Lesson, where a tally of unique and duplicate words is output. This is the desired result, right? Yes, but it’s an un-orderly list.
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Encoding means nothing if you can’t decode, which is the task for this month’s Exercise: Transform the encoded hex bytes back into characters, unwinding the formula used to create them. The challenge isn’t really as much with coding the math as it is with translating hex bytes back into integer values — characters.
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The next step in the unique words program is to tally the count of each word. From last week’s Lesson, the word list is sorted, which makes the task of counting duplicates easy.
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The task for last month’s Exercise was to encode a string. As you may have feared, the task for this month’s Exercise is to decode that string. Let me review:
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In our last episode, the unique words code is able to parse and list individual words in the buffer. To find unique and duplicate words, the next step is to sort the list.
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From last week’s Lesson, the text in a buffer is parsed, creating pointers to each word in the string. Alas, the addresses of these words (the pointers) aren’t saved, which is stupid. To handle the job, and to keep the Unique Words project moving forward, a dynamic array of pointers must be allocated.
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Continuing with my Unique Words project from last week’s Lesson: Once the buffer contains text, the next step is to parse the words: to split the long string of text stored in memory into separate word chunks. For this task, I turn to my old pall, the strtok() function.
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