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Our final task in gen_aux is to examine every element of r If the element is bracketed, we have to expand it into a sequence of words; otherwise, we append it to ret What may seem like magic on first reading is that this processing is exactly what we are doing in gen_aux and therefore, we can call gen_aux to do it! Such a call is called recursive, and it is one of those techniques that looks like it can't possibly work until you've tried it a few times To convince yourself that this function works, begin by noting that the function obviously works if word is not bracketed Next, assume that word is bracketed, but its rule's right-hand side has no bracketed words of its own It should still be easy to see that the program will work in this case, because when it makes each recursive call, the gen_aux that it calls will immediately see that the word is not bracketed Therefore, it will append the word to ret and return The next step is to assume that word refers to a slightly more complicated rule one that uses bracketed words in its right-hand side, but only words that refer to rules with no bracketed words of their own When you encounter a recursive call to gen_aux, do not try to figure out what it does Instead, remember that you have already convinced yourself that it works in this case, because you know that at worst, its argument is a category that does not lead to any further bracketed words Eventually, you will see that the function works in all cases, because each recursive call simplifies the argument We do not know any sure way to explain recursion Our experience is that people stare at recursive programs for a long time without understanding how they work Then, one day, they suddenly get it and they don't understand why they ever thought it was difficult Evidently, the key to understanding recursion is to begin by understanding recursion The rest is easy Having written gen_sentence, read_grammar, and the associated auxiliary functions, we'll want to use them:.



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int main() { // generate the sentence vector<string> sentence = gen_sentence(read_grammar(cin)); // write the first word, if any vector<string>::const_iterator it = sentencebegin(); if (!sentenceempty()) { cout << *it; ++it; } // write the rest of the words, each preceded by a space while (it != sentenceend()) { cout << " " << *it; ++it; } cout << endl; return 0; }

We read the grammar, generate a sentence from it, and then write the sentence a word at a time The only even minor complexity is that we put a space in front of the second and subsequent words of the sentence





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Not checking the validity of arguments is a common practice in C and C++ libraries Code that uses the library is responsible for making sure that all the arguments passed to library functions are valid If, however, you are used to the Java programming language, you may have to adjust to this particular aspect of the lack of safety in JNI programming

You can use the status bar to display status messages and read-only information about the object that the window represents In a window's status bar, ensure that each message fits the available space when the window is at its default size To avoid displaying obsolete information in a window's status bar, clear each status message when the next user action occurs in that window

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It is now time to write nrand We begin by noting that the standard library includes a function named rand (defined in <cstdlib>) That function takes no arguments, and returns a random integer in the range [0, RAND_MAX], where RAND_MAX is a large integer that is also defined in <cstdlib> Our job is to reduce the range [0, RAND_MAX], which includes both 0 and RAND_MAX, to [0, n), which includes 0 but not n, with the understanding that n <= RAND_MAX

You might think that it would suffice to compute rand() % n, which is the remainder when dividing the random integer by n In practice, this technique fails for two reasons The most important reason is pragmatic: rand() really returns only pseudo-random numbers Many C++ implementations' pseudo-random-number generators give remainders that aren't very random when the quotients are small integers For example, it is not uncommon for successive results of rand() to be alternately even and odd In that case, if n is 2, successive results of rand() % n will alternate between 0 and 1 There is another, more subtle reason to avoid using rand() % n: If the value of n is large, and RAND_MAX is not evenly divisible by n, some remainders will appear more often than others For example, suppose that RAND_MAX is 32767 (the smallest permissible value of RAND_MAX for any implementation) and n is 20000 In that case, there would be two distinct values of rand() that would cause rand() % n to be 10000 (namely, 10000 and 30000), but only one value of rand() that would cause rand() % n to be 15000 (namely, 15000) Therefore, the naive implementation of nrand would yield 10000 as a value of nrand(20000) twice as often as it would yield 15000 To avoid these pitfalls, we'll use a different strategy, by dividing the range of available random numbers into buckets of exactly equal size Then we can compute a random number and return the number of the corresponding bucket Because the buckets are of equal size, some random numbers may not fall into any bucket at all In that case, we keep asking for random numbers until we get one that fits The function is easier to write than to describe:

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