HI I'm getting a segfault with this code :
#include <stdio.h>
#include <stdlib.h>
int main()
{
char **array;
int i = 0;
array = (char **)malloc(sizeof(char *) * 20);
for (i = 0; i < 20; i++)
{
array[i] = (char *)malloc(sizeof(char *) * 20);
array[i] = "____________________";
}
array[0][0] = '2';
return(0);
}
What am I doing wrong ?
I'm very confused when I need to modify 2d arrays, what are my best options here ?
Thanks a lot.
You get a segfault because with array[i] = "____________________"; you let point each array[i] to a string literal. Note that you do not copy the contents of the literal but let array[i] directly point to it. Consequently, with array[0][0] = '2' you are altering a string literal then, which is undefined behaviour (very likely to become apparent as a segfault in this case).
Use strcpy(array[i],"____________________); instead.
BTW: write array[i] = malloc(sizeof(char) * (20+1)); (note the sizeof(char) instead of sizeof(char*) here and the +1 for the string termination character).
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);
}
Why am I getting an "invalid initializer error" for the below code?
#include<stdio.h>
#include<malloc.h>
int main(){
int i = 2;
int j;
for(j = i ; j < 20; j++){
char *p[5] = malloc(20 * sizeof(char));
int len = 0;
p[0] = 'a';
}
return 0;
}
You have an Array of Pointers, and you're asigning only one Pointer to it:
char *p[5] = malloc(20 * sizeof(char));
Use this instead:
char *p = malloc(20 * sizeof(char));
If you actually want an Array of Pointers, but only set the first of them, this should be your code:
char *p[5] = {malloc(20 * sizeof(char))};
(Or, even better: char *p[5] = {0}; p[0] = malloc(20 * sizeof(char));)
If this is what you want, you have to asign the letter 'a' differently - use *(p[0]) = 'a'; instead.
Furthermore: you don't free your memory anymore. You should use free(p); to do this, otherwise it's possible that your program consumes more and more memory.
For small programs, this is no problem, because the OS frees the memory anyways after the Program has been closed.
But it's bad practice, and the reason for lots of bad software.
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?