C: Realloc() after reading second line in file results in garbage values - c

I'm attempting to read sequences from a FASTA file into a table of structs that I've created, which each contain a character array member called "seq". My code seems to work well for the first loop, but when I realloc() memory for the second sequence, the pointer seems to point to garbage values and then the strcat() method gives me a segfault.
Here's the whole FASTA file I'm trying to read from:
>1
AAAAAAAAAAGWTSGTAAAAAAAAAAA
>2
LLLLLLLLLLGWTSGTLLLLLLLLLLL
>3
CCCCCCCCCCGWTSGTCCCCCCCCCCC
Here's the code (sorry that some of the variable names are in french):
typedef struct _tgSeq { char *titre ; char *seq ; int lg ; } tgSeq ;
#define MAX_SEQ_LN 1000
tgSeq* readFasta(char *nomFile) {
char ligne[MAX_SEQ_LN];
tgSeq *lesSeq = NULL;
int nbSeq=-1;
FILE *pF = fopen(nomFile, "r");
while(fgets(ligne, MAX_SEQ_LN, pF) != NULL) {
if(ligne[0] == '>') {
/*create a new sequence*/
nbSeq++;
//reallocate memory to keep the new sequence in the *lesSeq table
lesSeq = realloc(lesSeq, (nbSeq)*sizeof(tgSeq));
//allocate memory for the title of the new sequence
lesSeq[nbSeq].titre = malloc((strlen(ligne)+1)*sizeof(char));
//lesSeq[nbSeq+1].titre becomes a pointer that points to the same memory as ligne
strcpy(lesSeq[nbSeq].titre, ligne);
//Now we create the new members of the sequence that we can fill with the correct information later
lesSeq[nbSeq].lg = 0;
lesSeq[nbSeq].seq = NULL;
} else {
/*fill the members of the sequence*/
//reallocate memory for the new sequence
lesSeq[nbSeq].seq = realloc(lesSeq[nbSeq].seq, (sizeof(char)*(lesSeq[nbSeq].lg+1+strlen(ligne))));
strcat(lesSeq[nbSeq].seq, ligne);
lesSeq[nbSeq].lg += strlen(ligne);
}
}
// Close the file
fclose(pF);
return lesSeq;
}
For the first line (AAAAAAAAAAGWTSGTAAAAAAAAAAA), lesSeq[nbSeq].seq = realloc(lesSeq[nbSeq].seq, (sizeof(char)*(lesSeq[nbSeq].lg+1+strlen(ligne)))); gives me an empty character array that I can concatenate onto, but for the second line (LLLLLLLLLLGWTSGTLLLLLLLLLLL) the same code gives me garbage characters like "(???". I'm assuming the problem is that the reallocation is pointing towards some sort of garbage memory, but I don't understand why it would be different for the first line versus the second line.
Any help you could provide would be greatly appreciated! Thank you!

The problem here is the first realloc gets the value of nbSeq as 0 which does not allocate any memory.
Replace
int nbSeq=-1;
with
int nbSeq=0;
Access the index with lesSeq[nbSeq - 1]

Some programmer dude already pointed out that you do not allocate enough memory.
You also seem to expect some behaviour from realloc that will not happen.
You call realloc with NULL pointers. This will make it behave same as malloc.
For the first line (AAAAAAAAAAGWTSGTAAAAAAAAAAA), ...= realloc(); gives me an empty character array that I can concatenate onto, but for the second line (LLLLLLLLLLGWTSGTLLLLLLLLLLL) the same code gives me garbage characters like "(???".
You should not expect any specifiy content of your allocated memory. Especially the memory location is not set to 0. If you want to rely on that, you can use calloc.
Or you simply assign a 0 to the first memory location.
You do not really concatenaty anything. Instead you allocate new memory where you could simply use strcpy instead of strcat.

Related

When i try to go through the file does it overwrite the array I am trying to save to?

char * names[10];
int i = 0;
char theRow[20];
while (fgets(theRow, 80 , fp) != NULL) {
names[i] = theRow;
i++;
}
The output of the array I am trying to save to is all the same depending on the last element.
for ex. file has (hi, I , 2, 3, karl)
the array then becomes (karl,karl,karl,karl,karl)
Did i miss something or what can it be?
With names[i] = nameRow;, you are storing the same pointer (to the same block of memory) again and again, but with every line you read, you overwrite the contents if this single chunk of memory.
You need to copy the contents you read in into your buffer nameRow, before you overwrite it then in the next iteration of the loop.
names[i] = strdup(nameRow);
Function strdup does two things, (1) it reserves memory large enough to hold a copy of nameRow (using malloc), and (2) it copies the content then.
This also makes clear that you'll have to free each entry copied later.
And, though widely adopted, strdup is not part of standard C; so in order to be platform independent, you'll have to write...
names[i] = malloc(strlen(nameRow)+1);
strcpy(names[i],nameRow);

C project using pointers to send arrays

In this program I have to create a system of pointers that will, after reading in only 80 characters of a larger input, supply a pointer-to-pointer-to-character operation. The result is then sent to a function determining the number of words in the total input and the average amount of letters they contain. My problem is that I cannot dont't know how to create the pointer system without generating a exc_bad_access warning. Additionally I cannot find a combination of malloc and free that is suiting my needs. Any help with this would be greatly appreciated.
inputPtr = (char*)malloc(81 * sizeof(char));
while (fgets(wordy, 81, stdin) != NULL) {
numChar = strlen(wordy);
inputPtr = wordy;
for (i = 0; i < groupRange; ++i) {
sentPtr[i] = &inputPtr;
}
if (numChar == 80) {
groupRange++;
}
free(inputPtr);
printwords(*sentPtr, numChar);
}
Lets take a close look at these three lines from your code:
inputPtr = (char*)malloc(81 * sizeof(char));
...
inputPtr = wordy;
...
free(inputPtr);
The first allocates memory, and assign the pointer to that memory to the variable inputPtr.
The second line reassigns inputPtr so it no longer points to the memory you have allocated. You will lose that memory and have a memory leak.
Finally the last line, where you attempt to free what inputPtr is pointing to, and exactly what it is pointing to I don't know but it probably isn't memory you have allocated with malloc. That leads to undefined behavior.
Exactly how to solve your problem I'm not sure about, but a good start would be to not allocate memory dynamically, and then of course not call free.

C - Getting a Debug Assertion Failed Error by reading a csv-file

i am just doing a program in C, which has to read different files in csv-format. Every file consists of two lines.
The first line describes what topics are saved, while the second lines contains the data of the topics. Every file has
6 columns. The data are information like dates, source and category. I actually wrote a program that get the path and
gives the content in one dynamical char array back, but there is always a Debug Assertion Error that crashes it all the time. My Code is:
char* readfile(char csvpath[]) {
FILE *csv;
int c;
int countcontent = 100; //counter for the length of the content array
int counter = 0; //counter of the amount of the inserted chars
char *temp; //temp = buffer
char *content = (char*)calloc(100,sizeof(char)); //content of the file
csv = fopen(csvpath,"r");
while(c = fgetc(csv) != EOF) { //while file isnt at the end
if(countcontent <= counter) {
realloc(content,100*sizeof(char));
countcontent += 100;
}
temp = (char*)calloc(20,sizeof(char));
fgets(temp,20,csv);
content = concat(content,temp); //concat is my own function and add the 2. string behind the 1.
counter+= 20;
}
fclose(csv);
return content;}
Actually i ignore that there are two different lines, cause i want to delete the first one at the end anyway, because no data is saved there. But can you help me to find the solution for this error?
Your problem is this line
realloc(content,100*sizeof(char));
The realloc function returns a pointer to the reallocated memory. Think about what would happen if the realloc call can't just resize the already allocated chunk, and has to actually allocate a completely new chunk of memory, because it can't automatically update the pointer you pass to it.
Another problem with that statement is that it doesn't matter how much more memory you need, you will always allocate 100 bytes. The size argument you provide to realloc is the new size.
Oh, and remember that realloc can fail, and return a NULL pointer. If you don't want to loose your original pointer, you need to have a temporary variable to store the returned pointer, and check it for NULL.

Segmentation fault on malloc

After running this function many (not sure exactly how many) times, it seg faults on a simple memory allocation. Why would this suddenly happen? I did notice something strange in GDB. In the function that calls it, normally there's 6-digit long hex value for wrd (wrd = 0x605140 for example), however on the call where it crashes, the hex value is only two digits long. (wrd=0x21). I also checked the wrd->length, and it's 3.
The line that it crashes on is...
char *word_temp = malloc(wrd->length * sizeof(char));
EDIT:
Here's the code that creates the wrd...
while(fgets(input, 100, src) != 0)
{
int i = 0;
while(input[i] != '\0')
{
i++;
}
struct word *wrd = malloc(sizeof(struct word));
wrd->letters = input;
wrd->length = i;
If I'm getting an overflow, how do I fix that?
Looks like wrd->length does not include the terminating '\0'.
Fix 1, allocate word_temp like this:
char *word_temp = malloc( wrd->length + 1 );
Fix 2, include the '\0' by modifying you length count loop:
int i = 0;
while(input[i++] != '\0') {}
This will increase i one more time than code in the question, which is easy to see if you consider case of input being empty.
Note that you need to do either fix 1 or fix 2, not both. Choose which ever works with rest of your code.
You probably have a second issue with this line:
wrd->letters = input;
It does not copy input, it copies the pointer. If you change contents of input, contents of wrd->letters changes too, because they point to same memory location. Also if input is a local char array, then once it goes out of scope, wrd->letters becomes a dangling pointer, which will be overwritten by other data, and modifying it after that will result in memory corruption.
Possible fix (depending on rest of your code) is to use strdup:
wrd->letters = strdup(input);
Remember that it is now allocated from heap, so when done, you must remember to do
free(wrd->letters);
About wrd being 0x21, that indicates either memory corruption, or that you actually have two separate wrd variables, and one one is left uninitialized.
For example, maybe wrd is a function parameter struct word *wrd, in which case you only modify the local value in function, it does not get passed back to the caller. To modify the pointer of caller, you need to have pointer to pointer: struct word **wrd and then do (*wrd) = malloc... and (*wrd)->letters... etc.

How to allocate memory for an array of strings of unknown length in C

I have an array, say, text, that contains strings read in by another function. The length of the strings is unknown and the amount of them is unknown as well. How should I try to allocate memory to an array of strings (and not to the strings themselves, which already exist as separate arrays)?
What I have set up right now seems to read the strings just fine, and seems to do the post-processing I want done correctly (I tried this with a static array). However, when I try to printf the elements of text, I get a segmentation fault. To be more precise, I get a segmentation fault when I try to print out specific elements of text, such as text[3] or text[5]. I assume this means that I'm allocating memory to text incorrectly and all the strings read are not saved to text correctly?
So far I've tried different approaches, such as allocating a set amount of some size_t=k , k*sizeof(char) at first, and then reallocating more memory (with realloc k*sizeof(char)) if cnt == (k-2), where cnt is the index of **text.
I tried to search for this, but the only similar problem I found was with a set amount of strings of unknown length.
I'd like to figure out as much as I can on my own, and didn't post the actual code because of that. However, if none of this makes any sense, I'll post it.
EDIT: Here's the code
int main(void){
char **text;
size_t k=100;
size_t cnt=1;
int ch;
size_t lng;
text=malloc(k*sizeof(char));
printf("Input:\n");
while(1) {
ch = getchar();
if (ch == EOF) {
text[cnt++]='\0';
break;
}
if (cnt == k - 2) {
k *= 2;
text = realloc(text, (k * sizeof(char))); /* I guess at least this is incorrect?*/
}
text[cnt]=readInput(ch); /* read(ch) just reads the line*/
lng=strlen(text[cnt]);
printf("%d,%d\n",lng,cnt);
cnt++;
}
text=realloc(text,cnt*sizeof(char));
print(text); /*prints all the lines*/
return 0;
}
The short answer is you can't directly allocate the memory unless you know how much to allocate.
However, there are various ways of determining how much you need to allocate.
There are two aspects to this. One is knowing how many strings you need to handle. There must be some defined way of knowing; either you're given a count, or there some specific pointer value (usually NULL) that tells you when you've reached the end.
To allocate the array of pointers to pointers, it is probably simplest to count the number of necessary pointers, and then allocate the space. Assuming a null terminated list:
size_t i;
for (i = 0; list[i] != NULL; i++)
;
char **space = malloc(i * sizeof(*space));
...error check allocation...
For each string, you can use strdup(); you assume that the strings are well-formed and hence null terminated. Or you can write your own analogue of strdup().
for (i = 0; list[i] != NULL; i++)
{
space[i] = strdup(list[i]);
...error check allocation...
}
An alternative approach scans the list of pointers once, but uses malloc() and realloc() multiple times. This is probably slower overall.
If you can't reliably tell when the list of strings ends or when the strings themselves end, you are hosed. Completely and utterly hosed.
C don't have strings. It just has pointers to (conventionally null-terminated) sequence of characters, and call them strings.
So just allocate first an array of pointers:
size_t nbelem= 10; /// number of elements
char **arr = calloc(nbelem, sizeof(char*));
You really want calloc because you really want that array to be cleared, so each pointer there is NULL. Of course, you test that calloc succeeded:
if (!arr) perror("calloc failed"), exit(EXIT_FAILURE);
At last, you fill some of the elements of the array:
arr[0] = "hello";
arr[1] = strdup("world");
(Don't forget to free the result of strdup and the result of calloc).
You could grow your array with realloc (but I don't advise doing that, because when realloc fails you could have lost your data). You could simply grow it by allocating a bigger copy, copy it inside, and redefine the pointer, e.g.
{ size_t newnbelem = 3*nbelem/2+10;
char**oldarr = arr;
char**newarr = calloc(newnbelem, sizeof(char*));
if (!newarr) perror("bigger calloc"), exit(EXIT_FAILURE);
memcpy (newarr, oldarr, sizeof(char*)*nbelem);
free (oldarr);
arr = newarr;
}
Don't forget to compile with gcc -Wall -g on Linux (improve your code till no warnings are given), and learn how to use the gdb debugger and the valgrind memory leak detector.
In c you can not allocate an array of string directly. You should stick with pointer to char array to use it as array of string. So use
char* strarr[length];
And to mentain the array of characters
You may take the approach somewhat like this:
Allocate a block of memory through a call to malloc()
Keep track of the size of input
When ever you need a increament in buffer size call realloc(ptr,size)

Resources