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Implementation of the Dictionary class:
int Dictionary::find_word(char *s)
{
char word[81];
for (int i = 0; i < nwords; ++i)
if (stricmp(words[i].get_word(word),s) == 0)
return i;
return -1;
}
void Dictionary::add_def(char *word, char **def)
if (nwords < Maxwords)
words[nwords].put_word(word);
while (*def != 0)
words[nwords].add_meaning(*def++);
++nwords;
int Dictionary::get_def(char *word, char **def)
char meaning[81];
int nw = 0;
int word_idx = find_word(word);
if (word_idx >= 0)
while (words[word_idx].get_meaning(nw,meaning) != 0)
def[nw] = new char[strlen(meaning)+1];
strcpy(def[nw++],meaning);
def[nw] = 0;
return nw;
There is a pebble merchant. He sells the pebbles, that are used for shining the floor. His main duty is to take the length of the room’s sides. But he sometimes mistakes doing that
#questiStarting with a blank solution, write a program to prompt the user for an employee number, hourly rate and hours worked. Compute and display the employee number, gross weekl
#questionhow can i add new customer and alot him a new unique id ..
coding
#include #include #include void* memorycopy (void *des, const void *src, size_t count) { size_t n = (count + 7) / 8; char* destination = (char *) des; char* source =
An anagram is a type of word play, the result of rearranging the letters of a word or phrase to produce a new word or phrase. For example the anagram of tea is tea, tae, eat, eta,
I want Packet capture analysis Project Description: Need assistance in debugging some packet capture Skills required is C Programming
program to check whether a given point lies inside a rectangle or not
No. it's not required The C++ language guarantees that delete p will do nothing if p is equivalent to NULL. As you might get the test backwards, and as most testing methodologie
The for Loop For loop is the controlled form of loop. The general format of this : for( initialize ; test ; update) { statements;
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