Ok, this is a C programming homework question. But I'm truly stuck.
I ask the user to input words, and then I insert the input into an array, but I can't have any control over the number of words the user types.
I guess what I'm asking is how do you declare a an array in C without declaring its length and without asking the user what the length should be.
I know this has something to do with malloc, but if you could give me some examples of how to do this, I would really appreciate it.
You can malloc a block of memory large enough to hold a certain number of array items.
Then, before you exceed that number, you can use realloc to make the memory block bigger.
Here's a bit of C code that shows this in action, reallocating an integer array whenever it's too small to hold the next integer.
#include <stdio.h>
#include <stdlib.h>
int main (void) {
int *xyzzy = NULL; // Initially NULL so first realloc is a malloc.
int currsz = 0; // Current capacity.
int i;
// Add ten integers.
for (i = 0; i < 10; i++) {
// If this one will exceed capacity.
if (i >= currsz) {
// Increase capacity by four and re-allocate.
currsz += 4;
xyzzy = realloc (xyzzy, sizeof(int) * currsz);
// Should really check for failure here.
}
// Store number.
xyzzy[i] = 100 + i;
}
// Output capacity and values.
printf ("CurrSz = %d, values =", currsz);
for (i = 0; i < 10; i++) {
printf (" %d", xyzzy[i]);
}
printf ("\n");
return 0;
}
You can realloc it every time like:
int size = 0;
char **array = malloc(0);
while(/* something */)
{
char *string = // get input
size++;
array = realloc(array, size * sizeof(char*));
array[size - 1] = string;
}
Or in chunks if you care about speed.
Yes, you want malloc. Checkout this tut.
http://www.cprogramming.com/tutorial/dynamic_memory_allocation.html
This site is good in general for learning.
Here is an example of using realloc, it is basically exactly what you are asking to do.
http://www.cplusplus.com/reference/clibrary/cstdlib/realloc/
0) obviously you will need multiple buffers, so you will need a list like structure: perhaps a record with char array 100 chars and a pointer to next structure
1) You need to capture the words char by char and store them in your buffer
2) once the buffer is full you allocate another record, chain it with the previous one and keep going until you are out of mem or the process is over.
That should be better performance than realloc function. I believe malloc is trying to give contious block of memory. Therefore the list like structure will be faster and work better.
Related
I wanted to create a function that deletes from an array of segments the ones that are longer than a given number, by freeing the memory I don't need anymore. The problem is that the function I've created frees also all the memory allocated after the given point. How can I limit it, so that it frees just one pointer without compromising the others?
Here is the code I've written so far:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
typedef struct
{
double x1;
double y1;
double x2;
double y2;
} Segment;
double length(Segment* s)
{
return sqrt(pow(s->x1 - s->x2, 2) + pow(s->y1 - s->y2, 2));
}
// HERE IS THE PROBLEM!!
void delete_longer(Segment* as[], int n, double max_len)
{
for(int i = 0; i < n; i++)
{
if(length(as[i]) > max_len)
{
as[i] = NULL; // Those two lines should be swapped, but the problem remains
free(as[i]);
}
}
}
int main()
{
const int SIZE = 5;
Segment** arr = (Segment**)calloc(SIZE, sizeof(Segment*));
for(int i = 0; i < SIZE; i++)
{
arr[i] = (Segment*)malloc(sizeof(Segment));
}
srand(time(0));
for(int i = 0; i < SIZE; i++)
{
arr[i]->x1 = rand() % 100;
arr[i]->x2 = rand() % 100;
arr[i]->y1 = rand() % 100;
arr[i]->y2 = rand() % 100;
printf("Lungezza: %d\n", (int)length(arr[i]));
}
delete_longer(arr, SIZE, 80);
for(int i = 0; i < SIZE && arr[i]; i++)
{
printf("Lunghezza 2: %d\n", (int)length(arr[i]));
}
return 0;
}
First of all the free function should come after the instruction that sets the pointer to NULL, but that's not the main cause of the problem.
What causes the behaviour I described was the fact that the second for loop in the main stops after finding the first NULL pointer. Instead I should have written:
for(int i = 0; i < SIZE ; i++)
{
if(arr[i])
printf("Lunghezza 2: %d\n", (int)length(arr[i]));
}
You have two main problems:
In the delete function you write:
as[i] = NULL;
free(as[i]);
This is the wrong order. You must first free the memory and then set the element to null. But note that this is not the cause of your perceived problem, it only causes a memory leak (i.e. the memory of as[i] becomes inaccessible). You should write:
free(as[i]);
as[i] = NULL;
Your second problem is in your for loop, which now stops at the first null element. So not all the memory after it is deleted, you just don't print it. The loop should be for example:
for(int i = 0; i < SIZE; i++)
{
printf("Lunghezza 2: %d\n", arr[i]?(int)length(arr[i]):0);
}
Note: I agree with the discussion that free(NULL) may be implementation dependent in older implementations of the library function. In my personal opinion, never pass free a null pointer. I consider it bad practice.
There's no way to change the size of an array at runtime. The compiler assigns the memory statically, and even automatic arrays are fixed size (except if you use the last C standard, in which you can specify a different size at declaration time, but even in that case, the array size stands until the array gets out of scope). The reason is that, once allocated, the memory of an array gets surrounded of other declarations that, being fixed, make it difficult ot use the memory otherwise.
The other alternative is to allocate the array dynamically. You allocate a fixed number of cells, and store with the array, not only it's size, but also its capacity (the maximum amount of cell it is allow to grow) Think that erasing an element of an array requires moving all the elements behind to the front one place, and this is in general an expensive thing to do. If your array is filled with references to other objects, a common technique is to use NULL pointers on array cells that are unused, or to shift all the elements one place to the beginning.
Despite the technique you use, arrays are a very efficient way to access multiple objects of the same type, but they are difficult to shorten or enlengthen.
Finally, a common technique to handle arrays in a way you can consider them as variable length is to allocate a fixed amount of cells (initially) and if you need more memory to allocate double the space of the original (there are other approaches, like using a fibonacci sequence to grow the array) and use the size of the array and the actual capacity of it. Only in case your array is full, you call a function that will allocate a new array of larger size, adjust the capacity, copy the elements to the new copy, and deallocate the old array. This will work until you fill it again.
You don't post any code, so I shall do the same. If you have some issue with some precise code, don't hesitate to post it in your question, I'll try to provide you with a working solution.
The following code compiled fine yesterday for a while, started giving the abort trap: 6 error at one point, then worked fine again for a while, and again started giving the same error. All the answers I've looked up deal with strings of some fixed specified length. I'm not very experienced in programming so any help as to why this is happening is appreciated. (The code is for computing the Zeckendorf representation.)
If I simply use printf to print the digits one by one instead of using strings the code works fine.
#include <string.h>
// helper function to compute the largest fibonacci number <= n
// this works fine
void maxfib(int n, int *index, int *fib) {
int fib1 = 0;
int fib2 = 1;
int new = fib1 + fib2;
*index = 2;
while (new <= n) {
fib1 = fib2;
fib2 = new;
new = fib1 + fib2;
(*index)++;
if (new == n) {
*fib = new;
}
}
*fib = fib2;
(*index)--;
}
char *zeckendorf(int n) {
int index;
int newindex;
int fib;
char *ans = ""; // I'm guessing the error is coming from here
while (n > 0) {
maxfib(n, &index, &fib);
n -= fib;
maxfib(n, &newindex, &fib);
strcat(ans, "1");
for (int j = index - 1; j > newindex; j--) {
strcat(ans, "0");
}
}
return ans;
}
Your guess is quite correct:
char *ans = ""; // I'm guessing the error is coming from here
That makes ans point to a read-only array of one character, whose only element is the string terminator. Trying to append to this will write out of bounds and give you undefined behavior.
One solution is to dynamically allocate memory for the string, and if you don't know the size beforehand then you need to reallocate to increase the size. If you do this, don't forget to add space for the string terminator, and to free the memory once you're done with it.
Basically, you have two approaches when you want to receive a string from function in C
Caller allocates buffer (either statically or dynamically) and passes it to the callee as a pointer and size. Callee writes data to buffer. If it fits, it returns success as a status. If it does not fit, returns error. You may decide that in such case either buffer is untouched or it contains all data fitting in the size. You can choose whatever suits you better, just document it properly for future users (including you in future).
Callee allocates buffer dynamically, fills the buffer and returns pointer to the buffer. Caller must free the memory to avoid memory leak.
In your case the zeckendorf() function can determine how much memory is needed for the string. The index of first Fibonacci number less than parameter determines the length of result. Add 1 for terminating zero and you know how much memory you need to allocate.
So, if you choose first approach, you need to pass additional two parameters to zeckendorf() function: char *buffer and int size and write to the buffer instead of ans. And you need to have some marker to know if it's first iteration of the while() loop. If it is, after maxfib(n, &index, &fib); check the condition index+1<=size. If condition is true, you can proceed with your function. If not, you can return error immediately.
For second approach initialize the ans as:
char *ans = NULL;
after maxfib(n, &index, &fib); add:
if(ans==NULL) {
ans=malloc(index+1);
}
and continue as you did. Return ans from function. Remember to call free() in caller, when result is no longer needed to avoid memory leak.
In both cases remember to write the terminating \0 to buffer.
There is also a third approach. You can declare ans as:
static char ans[20];
inside zeckendorf(). Function shall behave as in first approach, but the buffer and its size is already hardcoded. I recommend to #define BUFSIZE 20 and either declare variable as static char ans[BUFSIZE]; and use BUFSIZE when checking available size. Please be aware that it works only in single threaded environment. And every call to zeckendorf() will overwrite the previous result. Consider following code.
char *a,*b;
a=zeckendorf(10);
b=zeckendorf(15);
printf("%s\n",a);
printf("%s\n",b);
The zeckendorf() function always return the same pointer. So a and b would pointer to the same buffer, where the string for 15 would be stored. So, you either need to store the result somewhere, or do processing in proper order:
a=zeckendorf(10);
printf("%s\n",a);
b=zeckendorf(15);
printf("%s\n",b);
As a rule of thumb majority (if not all) Linux standard C library function uses either first or third approach.
I've taking many attempts at solving this problem but failed every time.
I have an array
char *array[1024] = {};
Now I would like to add an item to the array and would also access the items by numbers
For example:
array[0] would be the first item
array[1] would be the second
array[2] would be the third item
But also I would like to know how many items are in the array so I could use something like
for(int i = 0; i <= totalitemsinarray; i++) {
print(array[i]);
}
You cannot change the size of an array in C. You can however allocate a sufficiently large array and then fill it up with entries. First, declare an array with a sufficient size, say, 1024.
char *array[1024];
Then declare a variable fill that counts the number of used slots in array. Initialize it to 0 as 0 slots are used in the beginning. Then, each time you insert an item, increment fill:
array[fill++] = ...;
...
array[fill++] = ...;
Make sure that you never attempt to insert more than 1024 items into the array, C doesn't check that for you.
For a more flexible approach, use malloc() to allocate memory for the array and then periodically enlarge it with realloc() when it's full. If you increase the array size in exponential steps (say, multiply with Φ = 0.5 + 0.5 √2 ≈ 1.61), this runs in O(1) amortised time per entry inserted.
There is no way to do what you're asking directly with C. One option could be if you knew that only certain values were valid. For example, you have an array of char *s so often people use NULL as a flag/invalid value. In that case you could initialize your array to have all NULLs and use that to know the size of the array:
char *array[1024];
memset(array, 0, sizeof(array));
/* .... */
for (int i = 0; i < sizeof(array)/sizeof(char*); i++) {
if (array[i]) {
printf("%s\n", array[i]);
}
}
char *array[1024] = {};
First, that is an array with 1024 char pointers/strings. Those elements can be 0s or plain garbage. If you don't plan to set them all you may want to nullify the array.
For the matter of storing the values and the count you might want to have a look at structs. For example:
typedef struct elem {
int count;
char *value;
} elem;
Then elem.count would be the number and elem.value would be the value accordingly.
And then initialize them in a for loop.
The only really valid way to approach this, is to dynamically grow the array. Allocate the array on the heap, and manage two counts: 1. the count of currently used elements, and 2. the count of elements for which you currently have memory allocated. Something like this:
//the setup
size_t arrayLength = 0, allocatedSize = 8;
int* array = malloc(sizeof(*array) * allocatedSize);
//grow the array -> first check that we have space to add an element
if(arrayLength == allocatedSize) {
array = realloc(array, allocatedSize *= 2);
assert(array);
}
assert(arrayLength < allocatedSize);
//grow the array -> add an element
array[arrayLength++] = ...;
You see, the realloc() call is not too much hassle, but it will protect you from bugs when the requirements change. My experience is that any fixed limit in the code, as insanely large as it may seem to be, will eventually be exceeded, and miserable failure will result. The only safeguard is to use as much memory as needed everywhere.
In the C language, is there a way to automatically grow an array.
For example:
int arr [100] [10];
If the array is full is it possible to have it "automatically" become larger? Or is that only possible if you're using C++. How would you write this in C?
There is no such feature in C: you have to declare the array using pointers, detect the "array is full" condition manually, call malloc, make a copy into an extended array, and free the original one. Even the variable-length arrays would not work, because they let you set their size only once per the array lifetime.
In C++, you can use std::vector<std::vector<int> > instead of a plain array. You still need to detect the "array is full" condition, but the std::vector<T> container takes care of all re-allocations and extensions on resizing for you.
"automatic" growth of any array in C is not possible. If you declare an array statically:
int arr[10];
you have however many memory locations as you indicated. If you want to be able to change it during runtime you need to declare it dynamically using malloc() and make it larger using realloc()
A quick example for you:
int main(void){
int input, count = 0, length = 2;
int * arr = malloc(sizeof(int) * length); // array of size 2
while((input = getchar()) != 'q') //get input from the user
{
getchar(); //get rid of newlines
arr[count] = input;
if(count + 1 == length){ // if our array is running out of space
arr = realloc(arr, length * length); // make it twice as big as it was
length *= length;
}
count++;
}
for(length = 0; length < count; length++) // print the contents
printf("%d\n", arr[length]);
free(arr);
return 0;
}
I know it could be done using malloc, but I do not know how to use it yet.
For example, I wanted the user to input several numbers using an infinite loop with a sentinel to put a stop into it (i.e. -1), but since I do not know yet how many he/she will input, I have to declare an array with no initial size, but I'm also aware that it won't work like this int arr[]; at compile time since it has to have a definite number of elements.
Declaring it with an exaggerated size like int arr[1000]; would work but it feels dumb (and waste memory since it would allocate that 1000 integer bytes into the memory) and I would like to know a more elegant way to do this.
This can be done by using a pointer, and allocating memory on the heap using malloc.
Note that there is no way to later ask how big that memory block is. You have to keep track of the array size yourself.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char** argv)
{
/* declare a pointer do an integer */
int *data;
/* we also have to keep track of how big our array is - I use 50 as an example*/
const int datacount = 50;
data = malloc(sizeof(int) * datacount); /* allocate memory for 50 int's */
if (!data) { /* If data == 0 after the call to malloc, allocation failed for some reason */
perror("Error allocating memory");
abort();
}
/* at this point, we know that data points to a valid block of memory.
Remember, however, that this memory is not initialized in any way -- it contains garbage.
Let's start by clearing it. */
memset(data, 0, sizeof(int)*datacount);
/* now our array contains all zeroes. */
data[0] = 1;
data[2] = 15;
data[49] = 66; /* the last element in our array, since we start counting from 0 */
/* Loop through the array, printing out the values (mostly zeroes, but even so) */
for(int i = 0; i < datacount; ++i) {
printf("Element %d: %d\n", i, data[i]);
}
}
That's it. What follows is a more involved explanation of why this works :)
I don't know how well you know C pointers, but array access in C (like array[2]) is actually a shorthand for accessing memory via a pointer. To access the memory pointed to by data, you write *data. This is known as dereferencing the pointer. Since data is of type int *, then *data is of type int. Now to an important piece of information: (data + 2) means "add the byte size of 2 ints to the adress pointed to by data".
An array in C is just a sequence of values in adjacent memory. array[1] is just next to array[0]. So when we allocate a big block of memory and want to use it as an array, we need an easy way of getting the direct adress to every element inside. Luckily, C lets us use the array notation on pointers as well. data[0] means the same thing as *(data+0), namely "access the memory pointed to by data". data[2] means *(data+2), and accesses the third int in the memory block.
The way it's often done is as follows:
allocate an array of some initial (fairly small) size;
read into this array, keeping track of how many elements you've read;
once the array is full, reallocate it, doubling the size and preserving (i.e. copying) the contents;
repeat until done.
I find that this pattern comes up pretty frequently.
What's interesting about this method is that it allows one to insert N elements into an empty array one-by-one in amortized O(N) time without knowing N in advance.
Modern C, aka C99, has variable length arrays, VLA. Unfortunately, not all compilers support this but if yours does this would be an alternative.
Try to implement dynamic data structure such as a linked list
Here's a sample program that reads stdin into a memory buffer that grows as needed. It's simple enough that it should give some insight in how you might handle this kind of thing. One thing that's would probably be done differently in a real program is how must the array grows in each allocation - I kept it small here to help keep things simpler if you wanted to step through in a debugger. A real program would probably use a much larger allocation increment (often, the allocation size is doubled, but if you're going to do that you should probably 'cap' the increment at some reasonable size - it might not make sense to double the allocation when you get into the hundreds of megabytes).
Also, I used indexed access to the buffer here as an example, but in a real program I probably wouldn't do that.
#include <stdlib.h>
#include <stdio.h>
void fatal_error(void);
int main( int argc, char** argv)
{
int buf_size = 0;
int buf_used = 0;
char* buf = NULL;
char* tmp = NULL;
char c;
int i = 0;
while ((c = getchar()) != EOF) {
if (buf_used == buf_size) {
//need more space in the array
buf_size += 20;
tmp = realloc(buf, buf_size); // get a new larger array
if (!tmp) fatal_error();
buf = tmp;
}
buf[buf_used] = c; // pointer can be indexed like an array
++buf_used;
}
puts("\n\n*** Dump of stdin ***\n");
for (i = 0; i < buf_used; ++i) {
putchar(buf[i]);
}
free(buf);
return 0;
}
void fatal_error(void)
{
fputs("fatal error - out of memory\n", stderr);
exit(1);
}
This example combined with examples in other answers should give you an idea of how this kind of thing is handled at a low level.
One way I can imagine is to use a linked list to implement such a scenario, if you need all the numbers entered before the user enters something which indicates the loop termination. (posting as the first option, because have never done this for user input, it just seemed to be interesting. Wasteful but artistic)
Another way is to do buffered input. Allocate a buffer, fill it, re-allocate, if the loop continues (not elegant, but the most rational for the given use-case).
I don't consider the described to be elegant though. Probably, I would change the use-case (the most rational).