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speller.c is one of the files that make up Problem Set 5, but it is already written by CS50 Staff and you may not alter it! Instead, you must implement your solution by altering three other component files: dictionary.c, dictionary.h, and Makefile.
3
votes
Accepted
PSET5 - SPELLER - CODE COMPILES BUT FAILS CHECK50
Time for a little tough love here. From time to time, new programmers in this class are under the incorrect impression that getting a program to compile means success. It doesn't. Just because a prog …
1
vote
Accepted
pset5 speller - check functions seems to only return false?
Hmmm... This looks suspicious:
// if there is no words at the position hash_code in the table
if (table[hash_code] == NULL)
{
new_word = table[hash_code];
}
Shouldn't this b …
1
vote
Accepted
Week 5: Speller. Hash function segmentation fault
I suspect that it is failing anytime it tries to hash a word with only one letter.
Look carefully at the if statement:
if (sizeof(word) < 2)
What, exactly, is word?
unsigned int hash(const char *word …
1
vote
Accepted
pset5 - seg fault when using large dictionaries or texts
Take a close look at the hash function. What happens if a word contains a non-alpha, such as an apostrophe? What values are returned?
Try inserting this into your hash function to diagnose:
printf("S …
1
vote
Accepted
PSET 5 speller fails when solving a memory issue
I'm not really clear on exactly what you're doing. If you're compiling it one way with the code commented out and another way with it not commented out, while the next, identical code block is presen …
0
votes
Accepted
HELP! Speller segfault
Interesting problem. ;-)
I'll point you in the right direction with a couple of questions.
Which line is triggering the seg fault?
What word is being processed when it seg faults?
What is the value …
1
vote
Pset5 Speller - Weird Load Function Bug (Edited w/ Code)
I suspect that this problem exists with each and every linked list that the table[] elements point to. (i.e., if the hash value is the first letter of each word, the first word in the dictionary file …
1
vote
Accepted
Speller Segmentation Fault Even After Freeing Malloc'd Nodes
I see a couple related issues. Let's look at the while loop in load().
char *word = malloc(46 * sizeof(char));
while(fscanf(dict, "%s", word) != EOF)
{
fscanf(dict, "%s", word);
node *tmp = m …
1
vote
Accepted
Pset 5: Speller - Double free or corruption (fasttop) error
Let's look carefully at this one. First, one has to figure out where the error occurs. Valgrind shows it to be line 148 in unload.
node *tmp = table[i]->next;
Since this isn't a free() call …
1
vote
Accepted
Everything is correct in my Check50 but in program I have other ERRORS - PSET05 Speller
Valgrind is catching something that check50 doesn't look for. There is an issue with your hash function. It can't deal with one-letter words like "a".
Consider that when the word is 1 letter long, wor …
2
votes
Accepted
PSET5 Speller - Problem with Check Function
The most obvious problem is the use of the pointer headPointer. The code updates this pointer to point at the new node just added. The problem with that is that table[] will never get updated, so the …
0
votes
Accepted
Pset5 speller segmentation fault when loading
Hmmm, help50 wasn't very helpful. The problem sorta lies with line 94, but not really. All that valgrind and help50 is telling you is that the problem memory was allocated on line 94. That gives you …
0
votes
Accepted
PSET5 Speller case conversions
I see a number of issues here. Let's go through them in order.
First, look at the for loop:
for (int i=0; i < N; i++)
{
int w = tolower(word[i]) - 'a';
}
What are you trying to accomplish here? This …
0
votes
Accepted
PSET 5 Speller Issues with Check
[EDIT - previous answer totally unrelated to the problem.]
Now that I have the full code to test, the problem became quickly obvious. The full code produces a segmentation fault when executed. This ca …
0
votes
Accepted
Pset 5 Speller fscanf returning 0
The problem is that this code is executing the fscanf and processing whatever is read before it checks for EOF.
There are several subtleties that are interacting here, leading to your result. First, t …