Trying to understand pointers as a beginner in C-
I've got this struct:
typedef struct {
int size; // dimension of array
int **arr; // pointer to heap allocated array
} someStruct;
So I use malloc to generate this struct, and an array, and initialize all the values to zero-
someStruct *m = (someStruct*)malloc(sizeof(someStruct));
m->size = n;
m->arr = (int**)malloc(n * sizeof(int));
// initialize array
for (int i = 0; i < n; i++) {
*(m->arr + i) = (int*)malloc(n * sizeof(int));
// set value to 0
for (int j = 0; j < n; j++) {
*(*(m->arr + i) + j) = 0;
}
}
After this I basically continue to access the array in later stages using the same kind of pointer logic-
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int num = *(*(m->arr + i) + j);
printf("num: %d\n", num);
}
}
Here's the problem- when I try to use this method of access, I'm clearly not getting the right answer- my print output look like this:
num: -2043774080
num: 22031
num: 0
num: 0
...
num: 0
num: 0
Here's the really weird part- this seeming bug of the 'weird' random numbers only comes when I'm creating and accessing an array of size 5-
I've come to believe that the whole
*(*(m->arr + i) + j)
method of access must be wrong- any help on this would be really useful. Thanks in advance, I apologize if this was already answered, my searching was unable to find it.
You should give complete code, but I think I was able to figure out your intent. You have one glaring problem, and many style issues. Here is what I think your code should look like:
typedef struct {
int size; // dimension of array
int **arr; // pointer to heap allocated array
} MagicSquare;
:
:
// no need to dynamically allocate this, it is small
MagicSquare m;
m.size = n;
m.arr = malloc(n * sizeof(int*)); // note it is sizeof(int*), not (int)
// initialize array
for (int i = 0; i < n; i++) {
m.arr[i] = malloc(n * sizeof(int));
// set value to 0
for (int j = 0; j < n; j++) {
m.arr[i][j] = 0;
}
}
:
:
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf("num: %d\n", m.arr[i][j]);
}
}
Note that if you want to initialize the allocated memory to zero, you should just use calloc, which does this initialization for you:
// initialize array
for (int i = 0; i < n; i++) {
m.arr[i] = calloc(n,sizeof(int));
}
Related
I am relatively new to c, and I still have not been able to find a good way of passing and returning a multi-dimensional array from a function. I found the following code, however it doesn't seem like a good way to do things because it passes the array and then to use it, it creates a duplicate with the malloc function. Is there a way to do it without the copying and malloc function, or a better way to pass and return an 2d array from a function in c in general? Thanks.
#include <stdio.h>
#include <stdlib.h>
int **matrix_sum(int matrix1[][3], int matrix2[][3]){
int i, j;
int **matrix3;
matrix3 = malloc(sizeof(int*) * 3);
for(i = 0; i < 3; i++) {
matrix3[i] = malloc(sizeof(int*) * 3);
}
for(i = 0; i < 3; i++){
for(j = 0; j < 3; j++){
matrix3[i][j] = matrix1[i][j] + matrix2[i][j];
}
}
return matrix3;
}
int main(){
int x[3][3], y[3][3];
int **a;
int i,j;
printf("Enter the matrix1: \n");
for(i = 0; i < 3; i++){
for(j = 0; j < 3; j++){
scanf("%d",&x[i][j]);
}
}
printf("Enter the matrix2: \n");
for(i = 0; i < 3; i++){
for(j = 0; j < 3; j++){
scanf("%d",&y[i][j]);
}
}
a = matrix_sum(x,y); //asigning
printf("The sum of the matrix is: \n");
for(i = 0; i < 3; i++){
for(j = 0; j < 3; j++){
printf("%d",a[i][j]);
printf("\t");
}
printf("\n");
}
//free the memory
for(i = 0; i < 3; i++) {
free(a[i]);
}
free(a);
return 0;
}
The array are not copied for agurments. Just pointers to the first elements of them (int[3]) are passed.
To avoid malloc(), you should add another argument to specify the array where the result should be stored.
#include <stdio.h>
void matrix_sum(int matrix3[][3], int matrix1[][3], int matrix2[][3]){
int i, j;
for(i = 0; i < 3; i++){
for(j = 0; j < 3; j++){
matrix3[i][j] = matrix1[i][j] + matrix2[i][j];
}
}
}
int main(){
int x[3][3], y[3][3], a[3][3];
int i,j;
printf("Enter the matrix1: \n");
for(i = 0; i < 3; i++){
for(j = 0; j < 3; j++){
scanf("%d",&x[i][j]);
}
}
printf("Enter the matrix2: \n");
for(i = 0; i < 3; i++){
for(j = 0; j < 3; j++){
scanf("%d",&y[i][j]);
}
}
matrix_sum(a,x,y);
printf("The sum of the matrix is: \n");
for(i = 0; i < 3; i++){
for(j = 0; j < 3; j++){
printf("%d",a[i][j]);
printf("\t");
}
printf("\n");
}
return 0;
}
An array can be declared and used through a reference (pointer)
for instance
char array[] = {'h','e','l', 'l', 'o', '\0'};
char *pointer = array;
the way pointers work can be understood by calling sizeof() on a given type
printf("char:%d\nchar_pointer: %d\n", sizeof(char), sizeof(char*));
which results in the following output.
char:1
char_pointer: 4
these results mean that even though a char has 1byte, its pointer needs 4 in order to be stored in memory thus, they are not the same type.
now in order to pass an array as an argument to a function you have many options
void function1(array[4])
{
//this function can receive an array of four elements and only four elements;
//these types of functions are useful if the algorithm inside the function only works
//with a given size. e.g. matrix multiplication
}
//or
void function2(char array[], int size)
{
//this function can receive an array of elements of unknown size, but you can
//circumvent this by also giving the second argument, the size.
int i;
for(i = 0; i <= size; i++)
{
printf("%c", array[i]);
}
}
In order to use or call any of these functions you could pass the array or a pointer to the array
function2(array, 5);
function2(pointer, 5);
//these results are the same
The same applies to a multidimensional array
void function3(char** multi_dim_array, array_size_first_dim, array_size_second_dim);
//and you can call it by using the same syntax as before;
void main(int argc, char[] argv*)
{
char** multi_dim = malloc(sizeof(char*) * 3);
int i;
for(i = 0; i<=3 ; i++)
{
multi_dim[i] = malloc(sizeof(char) * 4);
}
function3(multi_dim, 3,4);
}
if you want to return a multidimensional array you can just return a pointer
char **malloc_2d_array(int dim1, int dim2)
{
char ** array = malloc(sizeof(char*)*dim1);
int i;
for(i = 0; i<=dim2; i++)
{
array[i] = malloc(sizeof(char) * dim2);
}
return array;
}
as a final note, the reason the code you found, copies the array, is because of functional programming(a way of programming if you will) where a function call cant modify its input, thus it will always create a new value;
First of all this is not gonna be a technical explanation. I am just gonna try and explain what works not why.
For passing a multidimensional array you can use either an array with a defined size as you did in your example code:
void matrix_sum(int matrix3[][3])
Or if you don't want to use a defined size and want to take care of memory usage you can use a pointer to a pointer. For this case you also need to pass the size (unless you are passing NULL-terminated strings). Like this:
void matrix_sum(int **matrix, int size)
BUT for this case you can't call the function with a "normal" array. You need to use a pointer to a pointer or a pointer to an array.
int **matrix;
// make sure to allocate enough memory for this before initializing.
or:
int *matrix[];
For returning an array you can just return a pointer to a pointer like you did in your code example.
But you don't need to return an array, because if you change a value in an array, (in a different function) the value will stay changed in every other function.
A short example for this:
#include <stdio.h>
void put_zeros(int matrix[][3])
{
int i;
int j;
i = 0;
while (i < 3)
{
j = 0;
while (j < 3)
{
matrix[i][j] = 0;
j++;
}
i++;
}
}
void print_matrix(int matrix[][3])
{
int i;
int j;
i = 0;
while (i < 3)
{
j = 0;
while (j < 3)
{
printf("%d ", matrix[i][j]);
j++;
}
printf("\n");
i++;
}
}
int main(void)
{
int matrix_first[3][3] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
print_matrix(matrix_first);
put_zeros(matrix_first);
print_matrix(matrix_first);
}
This will print "1 2 3 4 5 6 7 8 9" because that's the first value we assigned.
After calling put_zeros it will contain and print "0 0 0 0 0 0 0 0 0" without the put_zeros returning the array.
1 2 3
4 5 6
7 8 9
0 0 0
0 0 0
0 0 0
I have to create a dynamic and 3 dimensional integer-Array in C.
But I have to create the pointers separately and use malloc. I know how to create 2d Array with malloc but I get confused using the following method, and i couldn't really find other question similar to this.
2x2x2 Integer-Array
First step:
int *firstD;
firstD = (int) malloc(2*sizeof(int));
Second step:
int *secondD;
secondD = (int) malloc(2 * firstD * sizeof(int));
Third step:
int *thirdD;
thirdD = (int) malloc(2 * secondD * sizeof(int));
I think maybe I have to add pointers in the starting (int*) and increase it every stepp by one more pointer?
Allocate an array to store pointers to arrays of row pointers.
Allocate arrays to store row pointers.
Allocate arrays to store each rows.
#include <stdlib.h>
int main(void) {
int size1 = 2, size2 = 2, size3 = 2;
int*** array;
array = malloc(sizeof(int**) * size1); // 1
for (int i = 0; i < size1; i++) {
array[i] = malloc(sizeof(int*) * size2); // 2
for (int j = 0; j < size2; j++) {
array[i][j] = malloc(sizeof(int) * size3); // 3
}
}
// free arrays
for (int i = 0; i < size1; i++) {
for (int j = 0; j < size2; j++) {
free(array[i][j]);
}
free(array[i]);
}
free(array);
return 0;
}
I wrote type names explicitly in the above example, but I suggest using dereferencing to obtain size of elements to allocate is better to prevent causing typos.
#include <stdlib.h>
int main(void) {
int size1 = 2, size2 = 2, size3 = 2;
int*** array;
// allocate arrays
array = malloc(sizeof(*array) * size1); // 1
for (int i = 0; i < size1; i++) {
array[i] = malloc(sizeof(*array[i]) * size2); // 2
for (int j = 0; j < size2; j++) {
array[i][j] = malloc(sizeof(*array[i][j]) * size3); // 3
}
}
// free arrays
for (int i = 0; i < size1; i++) {
for (int j = 0; j < size2; j++) {
free(array[i][j]);
}
free(array[i]);
}
free(array);
return 0;
}
Onitted in the above examples to make them simple, but you should check results of malloc() to see if the allocations are successful.
I am doing something like this;
int main()
{
int *b[2], j;
for (j = 0; j < 2; j++)
{
b[j] = (int *)malloc(12 * sizeof(int));
}
return 0;
}
Please tell me what this instruction really means? And how can I pass this array of pointers to a function to access values like *(B[0]+1),*(B[1]+1) etc?
int main(void)
{
int *b[2], j; // initialization of an array of pointers to integers (size = 2)
for (j = 0; j < 2; j++) // for each of the pointers
{
b[j] = malloc(12 * sizeof (int)); // allocate some space = 12 times size of integer in bytes (usually 4)
}
return 0;
}
If you want to pass this array to a function you can just pass b
foo(b);
I'm quite new to pointers and C in general.
void moveUpToTop(int num, int dim, int index) {
int i,j;
double *temp = w[index];
double *zero = w[0];
for(i = index; i > 0; i--) {
double *ptrA = w[i];
double *ptrB = w[i - 1];
for(j = 0; j < dim; j++) {
*(ptrA + j) = *(ptrB + j);
}
}
for(j = 0; j < dim; j++) {
*(zero + j) = *(temp + j);
}
}
Having this with a 2-dimensional array w, defined as double **w. I'd like to move some array value with index "index" up to the top of the array using pointers because that is the exercise we've got to do.
first of all, I'm saving one of the array locations, then I'm moving all array locations one level up.
What am I doing wrong?
Given to me is the following code in what I have to design the sort function.
double **w;
int main (void) {
int dim, num;
int i, j;
scanf ("%d %d", &dim, &num);
w = calloc (num, sizeof (double *));
for (i = 0; i < num; i++) {
w[i] = calloc (dim + 1, sizeof (double));
int sum = 0;
for (j = 0; j < dim; j++) {
scanf ("%le", &w[i][j]);
sum += w[i][j] * w[i][j];
}
w[i][dim] = sqrt(sum);
}
sort(num, dim);
for(i = 0; i < num; i++) {
for(j = 0; j < dim; j++) {
printf("%e ", w[i][j]);
}
printf("\n");
}
return 0;
}
Your problem is that you save a pointer to the original data but you don't save the data itself.
double *temp = w[index]; // Here you save the pointer
but in the first loop you overwrite data:
for(i = index; i > 0; i--) {
double *ptrA = w[i]; // Same as w[index] in first loop
double *ptrB = w[i - 1];
for(j = 0; j < dim; j++) {
*(ptrA + j) = *(ptrB + j); // The data pointed to by w[index] is overwritten
// in the first loop
So this code is no copying the original data at index:
for(j = 0; j < dim; j++) {
*(zero + j) = *(temp + j); // Data at index have been overwritten
// so this doesn't do what you want
}
To fix this, it is not sufficient to save a pointer to index(i.e. double *temp = w[index];). Instead you need to save all data that double *temp = w[index]; points to.
So you need to malloc some data to hold a copy. Then copy the data in a for-loop and use the copied data when restoring to zero.
BTW: Also notice that the code given to you uses a very ugly construct. It is allocating dim + 1 to save an extra double. Your move function therefore also need to use dim + 1 instead of just dim
gcc 4.6.2 c89
Allocating memory for a 2D array and filling with characters.
However, I don't seem to be filling as when I print nothing is displayed.
Am I doing something wrong here?
char **attributes = NULL;
/* TODO: Check for memory being allocated */
attributes = malloc(3 * sizeof(char*));
int i = 0;
int k = 0;
for(i = 0; i < 3; i++) {
for(k = 0; k < 5; k++) {
sdp_attributes[i] = malloc(5 * sizeof(char));
sdp_attributes[i][k] = k;
}
}
for(i = 0; i < 3; i++) {
for(k = 0; k < 5; k++) {
printf("attributes[i][k] [ %c ]\n", attributes[i][k]);
}
}
Many thanks for any advice,
Two major issues:
First Issue:
for(i = 0; i < 3; i++) {
for(k = 0; k < 5; k++) {
sdp_attributes[i] = malloc(5 * sizeof(char));
You are reallocating sdp_attributes[i] at each iteration of the inner loop - thereby overwriting it each time. You probably wanted this instead:
for(i = 0; i < 3; i++) {
sdp_attributes[i] = malloc(5 * sizeof(char));
for(k = 0; k < 5; k++) {
Second Issue:
sdp_attributes[i][k] = k;
You are basically writing the lower ascii characters. Most of them are not printable.
Something like this might do what you want:
sdp_attributes[i][k] = k + '0';
You probably want:
for (i = 0; i < 3; i++)
{
attributes[i] = malloc(5 * sizeof(char));
for (k = 0; k < 5; k++)
{
attributes[i][k] = k;
}
}
This ignores error checking on the allocation.
It also fixes the name of the array to match the declaration, but your code either wasn't compiling (don't post non-compiling code unless your question is about why it doesn't compile!) or you have another variable called sdp_attributes declared somewhere which you weren't showing us.
Your code was leaking a lot of memory. Each time around the k-loop, you allocated a new array of 5 characters and stored the pointer in attributes[i] (or sdp_attributes[i]), storing the new pointer over what was there before, so you overwrote the value of the first 4 pointers. You could not possibly free the first four items - they were lost irretrievably. Also, on the last iteration, you initialized the 5th element of the final array, but the previous 4 were not initialized and therefore contained indeterminate garbage.
Also, in your printing loop, the values in the array are control characters ^#, ^A, ^B, ^C and ^D; these do not necessarily print well with %c (especially not ^#, which is also known as NUL or '\0'). The printf() statement might be better written as:
printf("attributes[%d][%d] [ %d ]\n", i, k, attributes[i][k]);
This prints the array indexes (rather than simply the characters [i][k] for each entry), and prints the control characters as integers (since the char values are promoted to int when passed to printf()) rather than as control characters.
(It's also more conventional to use i and j for a pair of nested loops, and i, j, and k for triply nested loops, etc. However, that's a very minor issue.)
for(i = 0; i < 3; i++) {
for(k = 0; k < 5; k++) {
sdp_attributes[i] = malloc(5 * sizeof(char));
sdp_attributes[i][k] = k;
}
}
Your erasing the allocated memory every time you loop in the inner most loop.
Here is a correct version.
for(i = 0; i < 3; i++) {
sdp_attributes[i] = malloc(5 * sizeof(char));
for(k = 0; k < 5; k++) {
sdp_attributes[i][k] = k;
}
}
And you should fix your declaration:
attributes = malloc(3 * sizeof(char*));
to
sdp_attributes = malloc(3 * sizeof(char*));
Don't forget to free up all the memory allocated
for(i = 0; i < 3; i++)
{
free(sdp_attributes[i]);
}
free(sdp_attributes);
The correct way to allocate and assign elements to 2d array is as follows (but this is a int array, you can try and change it for char array):
One thing to note: As mentioned by #Mysticial, you should add/subtract '0' to your int value to get the char value when using ASCII character set (remember our itoa() functions!).
#include <stdio.h>
#include <stdlib.h>
int main()
{
int row, column;
int **matrix;
int i, j, val;
printf("Enter rows: ");
scanf("%d", &row);
printf("Enter columns: ");
scanf("%d", &column);
matrix = (int **) malloc (sizeof(int *) * row);
for (i=0 ; i<row ; i++)
matrix[i] = (int *) malloc (sizeof(int) * column);
val=1;
for (i=0 ; i<row ; i++) {
for (j=0 ; j<column; j++) {
matrix[i][j] = val++;
}
}
for (i=0 ; i<row ; i++) {
for (j=0 ; j<column; j++) {
printf("%3d ", matrix[i][j]);
}
printf("\n");
}
for (i=0 ; i<row ; i++)
free(matrix[i]);
free(matrix);
return 0;
}
Few points to note:
error handling should be added for malloc()
malloc()'ed memory must be free()'ed