For a program, I would like to make an array copy of the arguments sent in by command line using malloc().
So for example if I do ./a.out one two three
I want an array with {a.out, one, two, three} in it.
However, I have some issues getting my program to work. Here's what I have:
static char** duplicateArgv(int argc, char **argv)
{
char *array;
int j = 0;
// First allocate overall array with each element of char*
array = malloc(sizeof(char*) * argc);
int i;
// For each element allocate the amount of space for the number of chars in each argument
for(i = 1; i < (argc + 1); i++){
array[i] = malloc(strlen(*(argv + i)) * sizeof(char));
int j;
// Cycle through all the chars and copy them in one by one
for(j = 0; j < strlen(*(argv + i)); j++){
array[i][j] = *(argv + i)[j];
}
}
return array;
}
As you might imagine, this doesn't work. I apologize ahead of time if this somehow totally doesn't make sense, as I just started learning about pointers. Also, I'm not quite sure how to write code to free up every element in the *array after I do what I need to the copy.
Could anyone give me some tips on what I should look into to make it do what I want?
Thanks for any help!
You're not allocating or copying the terminating NULL characters:
This line needs to be changed to this for the NULL.
array[i] = malloc((strlen(*(argv + i)) + 1) * sizeof(char));
And the loop should be changed to this:
for(j = 0; j <= strlen(*(argv + i)); j++){
Also, the code can be better optimized if you saved the result of the strlen() call since you call it in so many places.
Try the loop as this:
// For each element allocate the amount of space for the number of chars in each argument
for(i = 0; i < argc; i++){
int length = strlen(argv[i]);
array[i] = malloc((length + 1) * sizeof(char));
int j;
// Cycle through all the chars and copy them in one by one
for(j = 0; j <= length; j++){
array[i][j] = argv[i][j];
}
}
first you need to allocate a vector of char*, not just a char*
char **array;
array = malloc(sizeof(char*)*(argc+1)); // plus one extra which will mark the end of the array
now you have an array[0..argc] of char* pointers
then for each argument you need to allocate space for the string
int index;
for (index = 0; index < argc; ++index)
{
arrray[index] = malloc( strlen(*argv)+1 ); // add one for the \0
strcpy(array[index], *argv);
++argv;
}
array[index] = NULL; /* end of array so later you can do while (array[i++]!=NULL) {...} */
With
char *array;
you define an object of type char*. That is: an object which value can point to a char (and the next char, ..., ...)
You need
char **array;
With this new type, the value of array points to a char*, ie another pointer. You can allocate memory and save the address of that allocated memory in a char*, you't do that with a char.
Related
is there a simple one liner I can use in C to allocate arrays in (pointer of arrays)
This line creates 10 pointers of arrays
char *out[10];
I can't do this
char *out[100]=(char[10][100])malloc(sizeof(char)*10*100);
error: cast specifies array type
same error with
char *out[10]=(char*[10])malloc(sizeof(char)*10*100);
do I need to do it in loop like this
int main()
{
char *out[10];
int x=0;
while(x<10)
{
*(out+x)=malloc(sizeof(char)*100);// is this line correct?
x++;
}
*out[0]='x';
printf("%c\n",out[0][0]);
free(out);
return 0;
}
but this cause warning that
req.c:75:3: warning: attempt to free a non-heap object ‘out’ [-Wfree-nonheap-object]
75 | free(out);
so do I need to allocate and free each array in (array of pointers) in loop
Can't I do allocation and free arrays in array of pointer in one line instead of loop?
or is there anything thing in my loop wrong too
To allocate an array of pointers to strings, you need to do:
char** out = malloc(sizeof(char*[10]));
The whole point of using this form is that each pointer in that array of pointers can be allocated with individual size, as is common with strings. So it doesn't make sense to allocate such with a "one-liner", or you are using the wrong type for the task.
In case you don't need individual sizes but are rather looking for a char [10][100] 2D array with static size, then the correct way to allocate such is:
char (*out)[100] = malloc(sizeof(char[10][100]));
You can allocate the full array in one single step and have pointers inside that array:
char *out[10];
data = malloc(100); //sizeof(char) is 1 by definition
for (int x=0; x<10; x++) {
out[i] = data + x * 10;
}
*out[0] = 'x';
printf("%c\n",out[0][0]);
free(data); // you must free what has been allocated
int i;
char** out = (char**)malloc(sizeof(char*)*10);
for(i = 0; i<10;i++)
out[i] = (char*)malloc(sizeof(char)*100);
out[1][1] = 'a';
OR with same dimensions
#include <stdio.h>
#include <stdlib.h>
void main()
{
int r = 10, c = 100; //Taking number of Rows and Columns
char *ptr, count = 0, i;
ptr = (char*)malloc((r * c) * sizeof(char)); //Dynamically Allocating Memory
for (i = 0; i < r * c; i++)
{
ptr[i] = i + 1; //Giving value to the pointer and simultaneously printing it.
printf("%c ", ptr[i]);
if ((i + 1) % c == 0)
{
printf("\n");
}
}
free(ptr);
}
I am still new with C and I am trying to empty a 2d char array. Here is the declaration:
char arg_array = (char**)calloc(strlen(buf), sizeof (char**));
for(i = 0; i<(strlen(buf)); i++)
{
arg_array[i] = (char*) calloc (strlen(buf), sizeof(char*));
}
Here is where I try to empty it:
void make_empty(char **arg_array)
{
int i;
for(i = 0; i <= BUFSIZ; i++)
{
arg_array[i][0] = '\0';
}
return;
}
Any help is appreciated
So, am I doing it right because this seems to give me segfaults when I try to add data to the array again and then print it?
Empty is just to have it empty - how can I explain more? lol
Try this:
void make_empty(char **arg_array, int rows, int cols)
{
int i,j;
for(i = 0; i <rows; i++)
{
for(j=0; j<cols;j++)
{
arg_array[i][j] = '\0';
}
}
return;
}
Where rows is number of rows and cols number of cols of your array.
P.S. This function clears the whole array as you should always do. As I commented before, putting '\0' as a first char in string does not clear the whole row, it only makes the rest of it ,,invisible'' for functions like printf. Check this link for more information: http://cplusplus.com/reference/clibrary/cstdio/printf/
There is no need to empty it. Often in C, memory allocation is done with malloc which simply returns to you a block of memory which is deemed owned by the caller. When calloc is called, as well as returning you a block of memory, the memory is guaranteed to be initialized to 0. This means for all intents and purposes it is already 'empty'.
Also I'm not quite sure if your code does what you are intending. Let me explain what it does at the moment:
char arg_array = (char**)calloc(strlen(buf), sizeof (char**));
This line is simply wrong. In C, there is no need to cast pointers returned from calloc because they are of type void *, which is implicitly casted to any other pointer type. In this case, you are storing it in a char type which makes no sense. If you do this:
char ** arg_array = calloc(strlen(buf), sizeof (char**));
Then it allocates an array of pointers of strlen(buf) length. So if buf is "hello" then you have now allocated an array which can store 5 pointers.
for(i = 0; i<(strlen(buf)); i++)
{
arg_array[i] = calloc (strlen(buf), sizeof(char*));
}
Again, I have removed the redundant cast. What this does is populates the array allocated earlier. Each index of the array now points to a char string of strlen(buf) * sizeof(char *) length. This is probably not what you want.
Your question is more clear to me now. It appears you want to remove the strings after populating them. You can do it two ways:
Either free each of the pointers and allocate more space later as you did before
Or set the first character of each of the strings to a null character
To free the pointers:
for(i = 0; i<(strlen(buf)); i++)
{
free(arg_array[i]);
}
To set the first character of each string to a null character:
for(i = 0; i<(strlen(buf)); i++)
{
arg_array[i][0] = '\0';
}
That is the same code as what you have originally and should be fine.
As proof, the following code will run without errors:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main(void)
{
char * buf = "hello";
char ** arg_array = calloc(strlen(buf), sizeof (char**));
unsigned int i;
for(i = 0; i < strlen(buf); i++) {
arg_array[i] = calloc(strlen(buf),
sizeof(char *));
}
for(i = 0; i < strlen(buf); i++) {
arg_array[i][0] = '\0';
}
for(i = 0; i < strlen(buf); i++) {
free(arg_array[i]);
}
free(arg_array);
return EXIT_SUCCESS;
}
If your code is segfaulting, the problem is coming from somewhere else. Did you overwrite the arg_array variable? Are you sure BUFSIZE is equal to strlen(buf)?
The reason why I am stuck on this is because of how C handles arrays, which is through pointers which I am new at. This is my case(not exactly my problem but a simplistic case):
char *keywords[k_num_of_keywords];
for(int i = 0; i < k_num_of_keywords; i++) {
char temp[] = "float";
keywords[i] = temp;
}
In this case, it would put float in all the indexes, but more importantly, the same address of wherever the string "float" is located. My case is slightly different, because I want to automate initializing a new char array for each index, but still use the same variable, rather than the same address in each index. How can I accomplish this in a for loop?
You should allocate memory for every element for "keywords", for example:
char *keywords[k_num_of_keywords];
for(int i = 0; i < k_num_of_keywords; i++) {
char temp[] = "float";
int len = strlen(temp);
keywords[i] = (char *)malloc(sizeof(char) * (len + 1));
memcpy(keywords[i], temp, len + 1);
}
I have char * lines[1000] string that can hold 1000 characters. How to create 100 arrays of that string. I get error with this code down.
char * lines[1000];
lines = (lines*)malloc(100 * sizeof(lines));
main.c:19:20: error: expected expression before ')' token
The simplest way is:
char lines[100][1000];
Alternatively:
char* lines[100];
int i;
for (i = 0; i < 100; i++) {
lines[i] = malloc(1000);
}
...
for (i = 0; i < 100; i++) {
free(lines[i]);
}
The latter is a bit more flexible in that -- with minor modifications -- it permits you to allocate a different amount of memory for every string.
It looks like you want an array strings, each string holding at most 1000 characters. There are some issues with your code.
You've declared an array of char *s but what you really want is a pointer to an array of chars. For that, your declaration should be
char (*lines)[1000];
On the other hand, you shouldn't forget about the NULL bytes at the end of strings, and should probably instead declare
char (*lines)[1001];
To set the pointer, you'll want to use
lines = (char (*)[1001]) malloc(100 * sizeof(char[1001]));
or
lines = (char (*)[1001]) malloc(100 * sizeof(*lines));
the latter working because, with lines a pointer to an array of chars, *lines is a char[1001]. Remember to make sure you didn't get a NULL pointer back.
At the end, you should free the memory you've malloced with
free(lines);
You can write a for-loop as:
char * lines[1000];
int i = 0;
for (i = 0; i < 1000; i++)
{
lines[i] = (char*)malloc(100 * sizeof(lines));
}
Don't forget to free-up the memory pointed by all the pointers
for (i = 0; i < 1000; i++)
{
free(lines[i])
}
Why don't you create a 2 dimensional array?
so I'm trying to write a function that concats a char**args to a char*args
What I have so far is":
char *concat(char **array)
{
int size = 0;
int i=0;
int j=0;
int z=0;
while (array[i]!=NULL)
{
printf(" %s \n", array[i]);
size = size + sizeof(array[i])-sizeof(char); //get the total size, minus the
//size of the null pointer
printf("%d \n",size);
i++;
}
size = size+1; //add 1 to include 1 null termination at the end
char *newCommand = (char*) malloc(size);
i=0;
while(i<sizeof(newCommand))
{
j=0;
z=0;
while (array[j][z]!='\0')
{
newCommand[i] = array[j][z];
i++;
z++;
}
j++;
}
newCommand[sizeof(newCommand)-1]='\0';
return newCommand;
}
this doesn't seem to work. Anyone know what's wrong?
I'd do it like this (untested):
int size = 0;
int count = 0;
while (array[count]) {
size += strlen(array[i]);
count++;
}
char *newCommand = malloc(size + 1);
char *p = newCommand;
newCommand[0] = 0; // Null-terminate for the case where count == 0
for (int i = 0; i < count; i++) {
strcpy(p, array[i]);
p += strlen(array[i]);
}
First, your size calculation was wrong. You wanted the size of the strings, but sizeof(array[i]) gives you the size of a single element in your array which is a pointer and thus 4 (32-bit) or 8 (64-bit). You need to use strlen instead.
Next, your manual copying was also off. It's easier to do it with a moving pointer and strcpy (which is to be avoided normally but we've calculated the sizes with strlen already so it's OK here). The use of strcpy here also takes care of null termination.
Main issue is that you keep using sizeof() with a pointer argument, whereas I think you are trying to get the size of the corresponding array.
sizeof() can only give you information that's available at compile time, such as the sizes of raw types like char and int, and the sizes of arrays with a fixed length such as a char[10]. The sizes of the strings pointed to by a char* is only computable at run time, because it depends on the exact values passed to your function.
For sizeof(newCommand) you probably need size, and for sizeof(array[i]), you probably need strlen(array[i]).