I'm trying to collect input integers one by one. My array starts with size 1 and I want to expand it by 1, with every input collected (is this a smart thing to do?)
Anyway, this is the code I managed to come up with, but it's not working as intended.
After this process, sizeof(array) always returns 8, which I assume means that the array is only being resized once. (sizeof(int) is 4 bits)
Trying to output the array results in multiple instances of the first input variable.
OUTPUT CODE
for(int s=0;s<sizeof(array)/sizeof(int);s++){
printf("%i\n",array[i]);
}
ORIGINAL CODE:
int i;
int size = 1;
int *array = malloc(size * sizeof(int));
int position = 0;
do{
i = getchar() - 48;
f (i != -16 && i != -38 && i != 0) {
array[position] = i;
position++;
size++;
*array = realloc(array, size * sizeof(int));
}
} while (i != 0);
UPDATED STILL NOT WORKING CODE
int i;
int size = 1;
int *array = malloc(size * sizeof(int));
int position = 0;
do{
i = getchar() - 48;
f (i != -16 && i != -38 && i != 0) {
array[position] = i;
position++;
size++;
array = realloc(array, size * sizeof(int));
}
} while (i != 0);
array = realloc(...)
not *array. Per the realloc docs, realloc returns the pointer, which you can store directly in your array pointer.
Edit One thing that will make your life easier: use char constants instead of raw numbers. E.g.,
i = getchar();
if(i != ' ' && i != '\n' && i != '0') {
/* 48-16 48-38 48-0 right? */
array[position] = i - '0'; /* '0' = 48 */
One thing that jumps out at me: inside your loop, this line:
*array = realloc(array, size * sizeof(int));
should instead be:
array = realloc(array, size * sizeof(int));
In the original version, you were sticking the result of realloc in the first element of the array by dereferencing the pointer first. Without the asterisk, you're reassigning the array itself.
(With some copy-paste from my comment:) sizeof(array) returns 8 because it equals sizeof(int*) (array is type int*) which is 8 (you're probably compiling as 64-bit). sizeof doesn't work how you think for pointers to arrays.
Similarly, your output code is wrong, for the same reason. You only print the first two elements because sizeof(array)/sizeof(int) will always be 8/4=2. It should be
for(int s=0;s<size;s++){
printf("%i\n",array[s]);
}
(note also changed index variable i to s)
where size is the variable from your other code chunk(s). You cannot find the length of the array from sizeof if it's dynamically allocated with pointers; that's impossible. Your code must "remember" the size of your array.
Related
How am I supposed to use dynamic memory allocations for arrays?
For example here is the following array in which i read individual words from a .txt file and save them word by word in the array:
Code:
char words[1000][15];
Here 1000 defines the number of words the array can save and each word may comprise of not more than 15 characters.
Now I want that that program should dynamically allocate the memory for the number of words it counts. For example, a .txt file may contain words greater that 1000. Now I want that the program should count the number of words and allocate the memory accordingly.
Since we cannot use a variable in place of [1000], I am completely blank at how to implement my logic. Please help me in this regard.
You use pointers.
Specifically, you use a pointer to an address, and using a standard c library function calls, you ask the operating system to expand the heap to allow you to store what you need to.
Now, it might refuse, which you will need to handle.
The next question becomes - how do you ask for a 2D array? Well, you ask for an array of pointers, and then expand each pointer.
As an example, consider this:
int i = 0;
char** words;
words = malloc((num_words)*sizeof(char*));
if ( words == NULL )
{
/* we have a problem */
printf("Error: out of memory.\n");
return;
}
for ( i=0; i<num_words; i++ )
{
words[i] = malloc((word_size+1)*sizeof(char));
if ( words[i] == NULL )
{
/* problem */
break;
}
}
if ( i != num_words )
{
/* it didn't allocate */
}
This gets you a two-dimensional array, where each element words[i] can have a different size, determinable at run time, just as the number of words is.
You will need to free() all of the resultant memory by looping over the array when you're done with it:
for ( i = 0; i < num_words; i++ )
{
free(words[i]);
}
free(words);
If you don't, you'll create a memory leak.
You could also use calloc. The difference is in calling convention and effect - calloc initialises all the memory to 0 whereas malloc does not.
If you need to resize at runtime, use realloc.
Malloc
Calloc
Realloc
Free
Also, important, watch out for the word_size+1 that I have used. Strings in C are zero-terminated and this takes an extra character which you need to account for. To ensure I remember this, I usually set the size of the variable word_size to whatever the size of the word should be (the length of the string as I expect) and explicitly leave the +1 in the malloc for the zero. Then I know that the allocated buffer can take a string of word_size characters. Not doing this is also fine - I just do it because I like to explicitly account for the zero in an obvious way.
There is also a downside to this approach - I've explicitly seen this as a shipped bug recently. Notice I wrote (word_size+1)*sizeof(type) - imagine however that I had written word_size*sizeof(type)+1. For sizeof(type)=1 these are the same thing but Windows uses wchar_t very frequently - and in this case you'll reserve one byte for your last zero rather than two - and they are zero-terminated elements of type type, not single zero bytes. This means you'll overrun on read and write.
Addendum: do it whichever way you like, just watch out for those zero terminators if you're going to pass the buffer to something that relies on them.
While Ninefingers provided an answer using an array of pointers , you can also use an array of arrays as long as the inner array's size is a constant expression. The code for this is simpler.
char (*words)[15]; // 'words' is pointer to char[15]
words = malloc (num_words * sizeof(char[15]);
// to access character i of word w
words[w][i];
free(words);
If you're working in C:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define WORD_LEN 15
int resizeArray(char (**wordList)[WORD_LEN], size_t *currentSize, size_t extent)
{
int result = 1;
char (*tmp)[WORD_LEN] = realloc(*wordList,
(*currentSize + extent) * sizeof **wordList);
if (tmp)
{
*currentSize += extent;
*wordList = tmp;
}
else
result = 0;
return result;
}
int main(void)
{
char *data[] = {"This", "is", "a", "test",
"of", "the", "Emergency",
"Broadcast", "System", NULL};
size_t i = 0, j;
char (*words)[WORD_LEN] = NULL;
size_t currentSize = 0;
for (i = 0; data[i] != NULL; i++)
{
if (currentSize <= i)
{
if (!resizeArray(&words, ¤tSize, 5))
{
fprintf(stderr, "Could not resize words\n");
break;
}
}
strcpy(words[i], data[i]);
}
printf("current array size: %lu\n", (unsigned long) currentSize);
printf("copied %lu words\n", (unsigned long) i);
for (j = 0; j < i; j++)
{
printf("wordlist[%lu] = \"%s\"\n", (unsigned long) j, words[j]);
}
free(words);
return 0;
}
If you intend to go for C++, STL is very useful for something dynamic allocation and is very easy. You can use std::vector ..
In modern C (C99) you have an additional choice, variable length arrays, VLA, such as that:
char myWord[N];
In principle you could also do such a thing in two dimensions, but if your sizes get too big, you may risk a stack overflow. In your case the easiest thing would be to use a pointer to such an array and to use malloc / realloc to resize them:
typedef char Word[wordlen];
size_t m = 100000;
Word* words = malloc(m * sizeof(Word));
/* initialize words[0]... words[m-1] here */
for (size_t i = 0; i < m; ++i) words[i][0] = '\0';
/* array is too small? */
m *= 2;
void *p = realloc(words, m*sizeof(Word));
if (p) words = p;
else {
/* error handling */
}
.
free(words);
This code should work (modulo typos) if wordlen is a constant or a variable, as long as you keep everything inside one function. If you want to place it in a function you should declare your function something like
void myWordFunc(size_t wordlen, size_t m, char words[m][wordlen]);
that is the length parameters must come first to be known for the declaration of words.
If the 15 in your example is variable, use one of the available answers (from Ninefingers or John Boker or Muggen).
If the 1000 is variable, use realloc:
words = malloc(1000 * sizeof(char*));
// ... read 1000 words
if (++num_words > 1000)
{
char** more_words = realloc(words, 2000 * sizeof(char*));
if (more_words) {printf("Too bad");}
else {words = more_words;}
}
In my code above, the constant 2000 is a simplification; you should add another variable capacity to support more than 2000 words:
if (++num_words > capacity)
{
// ... realloc
++capacity; // will reallocate 1000+ words each time; will be very slow
// capacity += 1000; // less reallocations, some memory wasted
// capacity *= 2; // less reallocations but more memory wasted
}
Here is a little information on dynamically allocating 2d arrays:
http://www.eskimo.com/~scs/cclass/int/sx9b.html
char ** words = malloc( 1000 * sizeof(char *));
int i;
for( i = 0 ; i < 1000 ; i++)
*(words+i) = malloc(sizeof(char) * 15);
//....
for( i = 0 ; i < 1000 ; i++)
free(*(words+i));
free(words);
I have to arrays of int for example arr1={0,1,1,0,0}, arr2={1,0,1,1,1} and I need to return 1 char* created by malloc that will be shown like this : "01100,10111".
when I do for loop it doesn't work, how can I do it ?
char* ans = (char*)malloc((size * 2+1) * sizeof(int));
for (int i = 0; i < size; i++)
ans[i] = first[i];
ans[size] = ",";
for (int i = size+1; i < 2*size+1; i++)
ans[i] = second[i];
Among the multitude of problems:
Your allocation size is wrong. It should include space for the separating comma and the terminating nullchar. sizeof(int) is wrong regardless, it should be sizeof(char) and as-such can be omitted (sizeof(char) is always 1).
Your storage is wrong. You want to store characters, and your values should be adjusted relative to '0'.
Your indexing of the second loop is wrong.
In reality, you don't need the second loop in the first place:
char* ans = malloc(size * 2 + 2);
for (int i = 0; i < size; i++)
{
ans[i] = '0' + first[i];
ans[size+1+i] = '0' + second[i];
}
ans[size] = ',';
ans[2*size+1] = 0;
That's it.
1.
char* ans = (char*)malloc((size * 2+1) * sizeof(int));
What is size here? It is not defined and declared in the provided code.
You do not need to cast the return value of malloc() to char. In fact, you do not need to cast the return value of malloc() anymore. It is a habit from the early C days.
Why do you need a char pointer here at all exactly? If you want to print 01100,10111 there is no need to use a char pointer for the output of the integer values.
2.
for (int i = 0; i < size; i++)
ans[i] = first[i];
Again what is size here?
What is first here? If it isn´t a pointer this statement is invalid.
3.
ans[size] = ",";
This operation is invalid. You are trying to assign a string to a pointer.
By the way, I don´t know what you trying to do with this statement. You can incorporate the comma separate in the output of 01100,10111, without your intend to include it int the memory of the int arrays itself.
4.
for (int i = size+1; i < 2*size+1; i++)
ans[i] = second[i];
Same as above: What is value and the type of size?
What is second? If it isn´t it a pointer this statement is invalid.
5.
To answer to the question title:
(How to) Copy two arrays of int to one char* in C
This isn´t possible. You can´t copy two arrays with its data to a pointer to char.
There are at least four issues with your code.
You malloc the wrong size, you want to use sizeof(char).
You need to zero terminate it, so you need to add extra room for the terminating zero
char* ans = (char*)malloc((size * 2+2) * sizeof(char));
second[size * 2+1] = 0;
Also the indexing of the second loop is wrong. You are accessing second array out of bounds. Make the loop more like the first.
We also need to convert the integer value to a char in the loops.
for (int i = 0; i < size; i++)
ans[size+i+1] = second[i] + '0';
I'm attempting to store arrays of integers that I read from a file (with a separate function) in a 2D array but I keep having issues with Segmentation fault. I know it's an issue with my pointers but I can't figure out exactly what I'm doing wrong.
Here is my function (takes an integer and compares it with an integer read from a file before storing it in my 2D array).
int **getStopTimes(int stop_id) {
int **result = malloc(sizeof(*result));
char const* const fileName = "stop_times_test.txt";
FILE* txt = fopen(fileName, "r");
char line[256];
int count = 0;
while (fgets(line, sizeof(line), txt) != NULL) {
int *formattedLine = getStopTimeData(line); //getStopTimeData returns a pointer to an array of ints, memory is allocated in the function
if (formattedLine[1] == stop_id) {
result[count] = formattedLine;
count++;
}
}
fclose(txt);
return result;
}
And my main:
int main(int argc, char *argv[]) {
int **niceRow = getStopTimes(21249);
for (int i=0; i<2; i++) { //Only looping 3 iterations for test purposes
printf("%d,%d,%d,%d\n",niceRow[i][0], niceRow[i][1], niceRow[i][2], niceRow[i][3]);
}
free(niceRow);
return 0;
}
getStopTimeData function thats being called (Pulls certain information from an array of chars and stores/returns them in an int array):
int *getStopTimeData(char line[]) {
int commas = 0;
int len = strlen(line);
int *stopTime = malloc(4 * sizeof(*stopTime)); //Block of memory for each integer
char trip_id[256]; //Temp array to build trip_id string
char stop_id[256]; //Temp array to build stop_id string
int arrival_time; //Temp array to build arrival_time string
int departure_time; //Temp array to build departure_time string
int counter;
for(int i = 0; i <len; i++) {
if(line[i] == ',') {
commas++;
counter = 0;
continue;
}
switch(commas) { //Build strings here and store them
case 0 :
trip_id[counter++] = line[i];
if(line[i+1] == ',') trip_id[counter] = '\0';
break;
case 1: //Convert to hours past midnight from 24hr time notation so it can be stored as int
if(line[i] == ':' && line[i+3] == ':') {
arrival_time = (line[i-2]-'0')*600 + (line[i-1]-'0')*60 + (line[i+1]-'0')*10 + (line[i+2]-'0');
}
break;
case 2 :
if(line[i] == ':' && line[i+3] == ':') {
departure_time = (line[i-2]-'0')*600 + (line[i-1]-'0')*60 + (line[i+1]-'0')*10 + (line[i+2]-'0');
}
break;
case 3 :
stop_id[counter++] = line[i];
if(line[i+1] == ',') stop_id[counter] = '\0';
break;
}
}
//Assign and convert to ints
stopTime[0] = atoi(trip_id);
stopTime[1] = atoi(stop_id);
stopTime[2] = arrival_time;
stopTime[3] = departure_time;
return stopTime;
}
This line:
int **result = malloc(sizeof(*result));
allocates just memory for one single pointer. (*result is of type int *, so it's a pointer to data -- the sizeof operator will tell you the size of a pointer to data ... e.g. 4 on a 32bit architecture)
What you want to do is not entirely clear to me without seeing the code for getStopTimeData() ... but you definitely need more memory. If this function indeed returns a pointer to some ints, and it handles allocation correctly, you probably want something along the lines of this:
int result_elements = 32;
int **result = malloc(sizeof(int *) * result_elements);
int count = 0;
[...]
if (formattedLine[1] == stop_id) {
if (count == result_elements)
{
result_elements *= 2;
result = realloc(result, result_elements);
}
result[count] = formattedLine;
count++;
}
Add proper error checking, malloc and realloc could return (void *)0 (aka null) on out of memory condition.
Also, the 32 for the initial allocation size is just a wild guess ... adapt it to your needs (so it doesn't waste a lot of memory, but will be enough for most use cases)
The upper answer is good,
just to give you an advice try to avoid using 2D array but use a simple array where you can store all your data, this ensures you to have coalescent memory.
After that, you can access your 1D array with an easy trick to see it like a 2D array
Consider that your 2D array has a line_size
To access it like a matrix or a 2d array you need to find out the corresponding index of your 1d array for given x,y values
index = x + y * line size;
In the opposite way:
you know the index, you want to find x and y corresponding to this index.
y = index / line_size;
x = index mod(line_size);
Of course, this "trick" can be used if you already know your line size
New to C and Valgrind and manual memory management and I'm having trouble locating an error that I'm getting when I run Valgrind. I have this function which gets strings from the user:
char **get_fragments_from_user(){
// No more than 20k strings containing at most 1k characters
char **strings = malloc(20000 * sizeof(char *));
char tempstring[MAX_INPUT]; //MAX_INPUT = 1001
int count = 0;
while(true){
printf("\n> ");
fgets(tempstring, MAX_INPUT, stdin);
if((strlen(tempstring) > 0) && (tempstring[strlen(tempstring) - 1] == '\n')){
tempstring[strlen(tempstring) - 1] = '\0';
}
if(tempstring[0] == 'q') break;
strings[count] = malloc(sizeof(char) * (strlen(tempstring)+1));
strcpy(strings[count], tempstring);
count++;
}
int i = 0;
char **fstrings = malloc((count)*sizeof(char *)); // count+1 needed? Something I tried removing while debugging
for(i = 0; i < count; i++){
fstrings[i] = malloc(sizeof(char) * (strlen(strings[i])+1));
strcpy(fstrings[i], strings[i]);
free(strings[i]);
}
free(strings);
return fstrings;
}
The idea here is simply to get strings and put them in an array. I initially allocate an array that is large enough to fit the maximum number of strings that could ever be entered (20,000), but I then resize the array so that I don't allocate more memory than the each string needs. I am a little embarrassed with the above code, since its less clean than anything I would have written in another language, but that was my first pass through.
I then get "Invalid read of size 8" from Valgrind when I try to calculate the number of strings in the array using this function:
int lengthOf(char **arr){
int i = 0;
while(arr[i] != NULL){
i++;
}
return i;
}
I'm pretty sure this is due to a dereferenced pointer or something, but I can't find it for the life of me and I've been looking at this code for an hour or so.
So, I believe the problem was that I wasn't allocating enough memory to store the whole array.
Instead of doing:
malloc(count * sizeof(char *));
I should have been allocating count+1, so either:
malloc((count + 1) * sizeof(char *))
or
calloc((count + 1), sizeof(char *));
I'm trying to realloc the size of my array but it doesn't change. I can separate it to two issues:
Coordinate *closeCoordinatesArray = malloc(sizeof(Coordinate) * 0);
Coordinate nextCoordinate = coordinatesMainArray[nextCoordinateIndex];
for (int p = 0; p < 4; p++) {
switch (p) {
case 0:
if (((nextCoordinate.x - 1) >= 0) && ((nextCoordinate.y - 1) >= 0)) {
int sizeOfArray = sizeof(*closeCoordinatesArray);
int sizeOfFirstObject = sizeof(closeCoordinatesArray[0]);
int closeCoordinatesArrayLength = (sizeOfArray / sizeOfFirstObject);
closeCoordinatesArray = realloc(closeCoordinatesArray,sizeof(Coordinate) * (closeCoordinatesArrayLength + 1));
sizeOfArray = sizeof(*closeCoordinatesArray);
sizeOfFirstObject = sizeof(closeCoordinatesArray[0]);
closeCoordinatesArrayLength = (sizeOfArray / sizeOfFirstObject);
After i malloc it in the first row, it shows that its length is 1. I would assume it should be 0 (i want it to be 0).
After a realloc it in case 0 I check the length and it's still 1.
What am i doing wrong?
It's important to remember that sizeof is evaluated at compile time; therefore it could not know about a variable length array. All it gives you is the size of the implicit pointer associated with the array. (Remember that in C pointers and arrays are equivalent).