I can initialize a one dimensional array in c with or without initializing its size:
int x[] = {1,2,3,4,5};
int y[5] = {1,2,3,4,5};
But, when I try to do the same for a two dimensional array such as
int x[][] = {{1,2,3},
{4,5,6}};
I get an error: array type has incomplete element type. The same error occurs if I declare and initialize the array on different lines.
However, I can initialize it while stating the size:
int x[2][3] = {{1,2,3},
{4,5,6}};
There is no error with this one.
My question is, is it possible to initialize a multi dimensional array without first initializing its size? I ask this because for an eventual project, I need to be able to declare arrays and initialize them later, and their size will not be known when compiling.
is it possible to initialize a multi dimensional array without first initializing its size?
No, not in the way you are proposing or anything similar to that.
I need to be able to declare arrays and initialize them later, and they will have dynamic sizes.
If the size is unknown at compile time, you should allocate the array using a = malloc(x * y * sizeof(value_t)). Then index into it like a[i + j*y].
Is it possible to initialize a multi dimensional array without first initializing its size?
=> No, it is not possible.
The thing possible is that you take the size of the array and then allocate memory using calloc.
I'm asking you to use calloc because this way all the array elements will be initially initialized as 0.
Now, use of calloc will be applied as:
Assuming you want 2D array, so you take variable row and column as input from the user.
Then use,
int *x;
x = calloc( ( row * column), sizeof(datatype) );
In this case the datatype would be int.
In short code would appear as :
int row, column, *x;
/* TAKING INPUT FROM THE USER */
printf("\n\t Enter the number of rows : ");
scanf("%d", &row);
printf("\n\t Enter the number of columns : ");
scanf("%d", &column);
/* DYNAMICALLY ALLOCATING MEMORY TO x USING calloc */
x = calloc( (row * column), sizeof(int));
I hope this code solves your problem.
I have one more thing to share with you.
In your this line of code :
int x[][] = {{1,2,3},
{4,5,6}};
This initialization is just missing one thing and that thing is mention of column size and otherwise code is correct.
So, Correction to your code :
int x[][3] = {{1,2,3},
{4,5,6}};
This would work fine.
Keep on trying! These type of things can only be known while practicing.
Happy Coding!
Yes, but you're missing a comma:
int x[2][3] = {{1,2,3},
{4,5,6}};
All array elements (even inner arrays) need to be comma-separated
Related
I came across an old code where I need to allocate memory for an array of matrices. Currently it is done by creating a structure matrix and allocating memory by making an arrray of structures.
Like so.
struct matrix
{
int x[13][13];
};
int main()
{
matrix *push = (matrix*)malloc(sizeof(matrix) * 1000);
//do stuff
free(push);
return 0;
}
But now the question.. Do I really need to do this? I know I am allowed to declare a variable of type int *matrix[13][13];
But I cant seem to figure out a way to allocate memory to this variable.
I tried matrix = (int***)malloc(sizeof(int[13][13])*1000); which results in
E0137 expression must be a modifiable lvalue and honestly in doesnt seem correct either
I know there are many better and creative ways of allocating an array like this, but I am curious
Questions
int *matrix[13][13] what will this kind of variable even mean or represent? Is it a pointer to a 2D matrix or s it an pointer to an array of 2D matrices?
How will I use the above mentioned variable?
How will I allocate memory for the above variable without using any more variables?
Clarifications
From comments it seems int *matrix[13][13] represents a matrix of 13x13 int*
What I am asking is actually, is there a way to have a variable with a dynamically allocated first dimension and static 2nd and 3rd dimensions without using structures typedefs etc
EDIT
As pointed out by felix
what I am looking for is int (*matrix)[13][13] = (int(*)[13][13])malloc(1000*sizeof(matrix));
I have noticed that sometimes beginner programmers are finding typedef-ing the multidimensional matrices which have to be dynamically allocated easier as it reduces the problem to the single star pointer and the sizeof of the defined type and pointer to it is much easier to understand.
#include <stdio.h>
typedef int my13x13matrix[13][13];
int main(void) {
my13x13matrix *mylargematrix = malloc(1000 * sizeof(*mylargematrix));
mylargematrix[999][12][12] = 5;
printf("%d \n", mylargematrix[999][12][12]);
return 0;
}
you can also use the single star pointer arithmetic.
my13x13matrix *anotherpointer = &mylargematrix[10];
(*anotherpointer)[5][5] = 1;
anotherpointer++;
(*anotherpointer)[5][5] = 2;
printf("%d %d\n", mylargematrix[10][5][5], mylargematrix[11][5][5]);
No, you don't want to use
int *matrix[13][13];
That will declare a [13][13] array of int*s.
You need
// pointer to "an array of 13 x 13 ints"
int (*matrix)[13][13];
or simplify it by using a type alias.
typedef int matrix_type[13][13];
matrix_type* matrix_ptr;
You can allocate memory for such a variable using
matrix_type* matrix_ptr = malloc(1000*sizeof(*matrix_ptr));
and assign values to it using
matrix_ptr[0][0][0] = 0;
...
matrix_ptr[999][12][12] = 0;
I have seen the following declaration of two dimensional array.
int arr[][3] = { {1,2,3}, {4,5,6}};
My question is how can I allocate following multidimensional array in run time based on user input of first dimension?
#define M 10
#define N 15
int arr[][M][N]
Start by declaring a pointer suitable for accessing the array:
int (*array)[M][N];
Then allocate memory for the array based on the user input:
array = malloc(P * sizeof(*array)); // P is the value obtained from the user
Then use the pointer as if it was a 3D array:
array[x][y][z] = 42;
Don't forget to free the memory when you're done with it.
C allows variable-length arrays. So after reading the first dimension from the user, you can declare the array with that size.
int n;
printf("How big is it? ");
scanf("%d", &n);
int arr[n][M][N];
i want to create a matrix in C, but size of the matrix must be determine by user.There is my code.
int row1,column1;
printf("Please enter number of rows in first matrix: ");
scanf("%d",&row1);
printf("Please enter number of columns in first matrix: ");
scanf("%d",&column1);
int M1[row1][column1];
i get errors with row1 and column1(in the last line).What is the correct way of taking size of a matrix from an user?
You have to dynamically allocate the array as you don't know the size at compile time.
My hint: use a single array it's much simpler:
int M1[] = new int[row1 * column1];
Then address it as
M1[column + line * row1];
If you absolutely need a 2D matrix, refer to this question: How do I declare a 2d array in C++ using new?
And don't forget to delete[] properly your array.
Related: dynamic allocating array of arrays in C
First allocate an array of pointers:
M1 = (int**)malloc(row1 * sizeof(int*));
And then point each to another array.
for(i = 0; i < row1; i++)
M1[i] = (int*)malloc(column1 * sizeof(int));
In c to create matrix of user defined size you need to use dynamic allocation using malloc or alloca(). you can read this link to get information on creating arrays of user defined size in c
This is because you cannot initialize an array with array length as a variable. Declare a pointer and use malloc to dynamically allocate the array.
int row1,column1;
printf("Please enter number of rows in first matrix: ");
scanf("%d",&row1);
printf("Please enter number of columns in first matrix: ");
scanf("%d",&column1);
int **arr;
//allocate the memory
arr=malloc(sizeof(int*)*row1);
int i;
for (i=0;i<row1;i++)
*(arr+i)=malloc(sizeof(int)*column1);
//do what you want to do
//free the memory
for (i=0;i<row1;i++)
free(*(arr+i));
free(arr);
I am doing an assignment for class and I thought I would bug you all with a question:
So the purpose of the program is for the user to enter the size of the array and then initialize it with some data. Max size is 20. My issue is that the array crashes when the sets the size beyond 14 and tries to initialize it. So forexample, if it sets the size as 20, it accepts it, but when I try to initialize it, it crashes. I have no idea what I am doing. your help is much appreciated.
Thank you very much,
Essi
int main ()
{
int sizeOfArray = 0; //size Of Array
float myArray[sizeOfArray];
//I have a piece a line or two of code that asks the user for the size of the
array
printf("Let's Initialize an array of size %d!!\n", sizeOfArray);
do
{
printf("Enter array element %d : ", initCounter+1);
myArray[initCounter] = userInitArray();
initCounter++;
}while (initCounter < sizeOfArray);
}
float userInitArray()
{
float num;
scanf("%f", &num);
return num;
}
These two lines
int sizeOfArray = 0; //size Of Array
float myArray[sizeOfArray];
Create an empty array. So whatever you try to store in this array later on is access out of bounds and invokes undefined behavior. The fact that your program crashes on 14th call is simply luck. It could have crashed on the first one just as well.
int sizeOfArray = 0; //size Of Array
float myArray[sizeOfArray];
Your array is created here with a size of zero. It doesn't magically expand when you later increase sizeOfArray. You need to get the size variable set first (from your 'line or two of code' user input) then create the array based on that.
You may also want to impose some sensible upper limit on your array size so you don't blow up your stack when trying, for example, to create a one-billion-entry array :-)
You have a (variable-length) array of size zero. You need to first ask for the size, and then allocate the array. Otherwise any attempts to assign to array elements would result in undefined behaviour.
You could do :
int sizeOfArray; //size Of Array
printf("tell me the size of the array");
scanf("%d",&sizeOfArray);
float myArray[sizeOfArray]; // not a good practice
The right way to do it would be:
int sizeOfArray; //size Of Array
float *myArray;
printf("tell me the size of the array");
scanf("%d",&sizeOfArray);
myArray=malloc(sizeof(float)*sizeOfArray);
You may use the pointer as a common array then.
and call like this: myArray[3] = doSomething();
EDIT Note that since you already know the max size you could avoid doing dynamic allocations listed above:
#Define MAXSIZE 20
int main ()
{
int sizeOfArray; //size Of Array
float myArray[MAXSIZE];
printf("tell me the size of the array\n");
scanf("%d",&sizeOfArray);
printf("\nLet's Initialize an array of size %d!!\n", sizeOfArray);
do
{
printf("Enter the element at myArray[%d] : ", initCounter+1);
myArray[initCounter] = userInitArray();
initCounter++;
}while (initCounter < sizeOfArray);
}
float userInitArray()
{
float num;
scanf("%f", &num);
return num;
}
Probably this last option is what your teacher is actually looking for.
You need to read sizeOfArray before you allocate myArray dynamically like this:
float * myArray = malloc(sizeOfArray * sizeof(float));
This is allocating sizeof(float) * sizeOfArray bytes of memory on heap and assigning address of allocated memory to myArray.
This is maybe hard to understand about arrays in C, they are really just pointers into memory - in your program the array myArray is allocated statically on stack and is of size 0. You cannot add any elements to it or assign to any index, it will not grow, its forewer 0 float elements long. Best thing that can happen to your program in this case is that it will crash. Worst case, it will not crash and strange things will happen;
You really should read something about memory allocation/management in C.
I think you forget to add function prototype in the beginning of your program (before main).
And also
int sizeOfArray = 0; //size Of Array
float myArray[sizeOfArray];
is wrong.
As you are using variable length array (valid in C99), you can do it as
int sizeOfArray; //size Of Array
printf("Let's Initialize an array of size %d!!\n", sizeOfArray);
float myArray[sizeOfArray];
dynamic arrays are not supported in C or C++.
Change your array to:
float* myAraray;
//later when you have the size , allocate it:
myArray = (float*)malloc(arrSize * sizeof(float));
I am working with a 2-dimensional array of structs which is a part of another struct. It's not something I've done a lot with so I'm having a problem. This function ends up failing after getting to the "test" for-loop near the end. It prints out one line correctly before it seg faults.
The parts of my code which read data into a dummy 2-d array of structs works just fine, so it must be my assigning array to be part of another struct (the imageStruct).
Any help would be greatly appreciated!
/*the structure of each pixel*/
typedef struct
{
int R,G,B;
}pixelStruct;
/*data for each image*/
typedef struct
{
int height;
int width;
pixelStruct *arr; /*pointer to 2-d array of pixels*/
} imageStruct;
imageStruct ReadImage(char * filename)
{
FILE *image=fopen(filename,"r");
imageStruct thisImage;
/*get header data from image*/
/*make a 2-d array of of pixels*/
pixelStruct imageArr[thisImage.height][thisImage.width];
/*Read in the image. */
/*I know this works because I after storing the image data in the
imageArr array, I printed each element from the array to the
screen.*/
/*so now I want to take the array called imageArr and put it in the
imageStruct called thisImage*/
thisImage.arr = malloc(sizeof(imageArr));
//allocate enough space in struct for the image array.
*thisImage.arr = *imageArr; /*put imageArr into the thisImage imagestruct*/
//test to see if assignment worked: (this is where it fails)
for (i = 0; i < thisImage.height; i++)
{
for (j = 0; j < thisImage.width; j++)
{
printf("\n%d: R: %d G: %d B: %d\n", i ,thisImage.arr[i][j].R,
thisImage.arr[i][j].G, thisImage.arr[i][j].B);
}
}
return thisImage;
}
(In case you are wondering why I am using a dummy array in the first place, well it's because when I started writing this code, I couldn't figure out how to do what I am trying to do now.)
EDIT: One person suggested that I didn't initialize my 2-d array correctly in the typedef for the imageStruct. Can anyone help me correct this if it is indeed the problem?
You seem to be able to create variable-length-arrays, so you're on a C99 system, or on a system that supports it. But not all compilers support those. If you want to use those, you don't need the arr pointer declaration in your struct. Assuming no variable-length-arrays, let's look at the relevant parts of your code:
/*data for each image*/
typedef struct
{
int height;
int width;
pixelStruct *arr; /*pointer to 2-d array of pixels*/
} imageStruct;
arr is a pointer to pixelStruct, and not to a 2-d array of pixels. Sure, you can use arr to access such an array, but the comment is misleading, and it hints at a misunderstanding. If you really wish to declare such a variable, you would do something like:
pixelStruct (*arr)[2][3];
and arr would be a pointer to an "array 2 of array 3 of pixelStruct", which means that arr points to a 2-d array. This isn't really what you want. To be fair, this isn't what you declare, so all is good. But your comment suggests a misunderstanding of pointers in C, and that is manifested later in your code.
At this point, you will do well to read a good introduction to arrays and pointers in C, and a really nice one is C For Smarties: Arrays and Pointers by Chris Torek. In particular, please make sure you understand the first diagram on the page and everything in the definition of the function f there.
Since you want to be able to index arr in a natural way using "column" and "row" indices, I suggest you declare arr as a pointer to pointer. So your structure becomes:
/* data for each image */
typedef struct
{
int height;
int width;
pixelStruct **arr; /* Image data of height*width dimensions */
} imageStruct;
Then in your ReadImage function, you allocate memory you need:
int i;
thisImage.arr = malloc(thisImage.height * sizeof *thisImage.arr);
for (i=0; i < thisImage.height; ++i)
thisImage.arr[i] = malloc(thisImage.width * sizeof *thisImage.arr[i]);
Note that for clarity, I haven't done any error-checking on malloc. In practice, you should check if malloc returned NULL and take appropriate measures.
Assuming all the memory allocation succeeded, you can now read your image in thisImage.arr (just like you were doing for imageArr in your original function).
Once you're done with thisImage.arr, make sure to free it:
for (i=0; i < thisImage.height; ++i)
free(thisImage.arr[i]);
free(thisImage.arr);
In practice, you will want to wrap the allocation and deallocation parts above in their respective functions that allocate and free the arr object, and take care of error-checking.
I don't think sizeof imageArr works as you expect it to when you're using runtime-sized arrays. Which, btw, are a sort of "niche" C99 feature. You should add some printouts of crucial values, such as that sizeof to see if it does what you think.
Clearer would be to use explicit allocation of the array:
thisImage.arr = malloc(thisImage.width * thisImage.height * sizeof *thisImage.arr);
I also think that it's hard (if even possible) to implement a "true" 2D array like this. I would recommend just doing the address computation yourself, i.e. accessing a pixel like this:
unsigned int x = 3, y = 1; // Assume image is larger.
print("pixel at (%d,%d) is r=%d g=%d b=%d\n", x, y, thisImage.arr[y * thisImage.width + x]);
I don't see how the required dimension information can be associated with an array at run-time; I don't think that's possible.
height and width are undefined; you might want to initialise them first, as in
thisImage.height = 10; thisImage.width = 20;
also,
what is colorRGB?
*thisImage.arr = *imageArr; /*put imageArr into the thisImage imagestruct*
This won't work. You have to declare arr as colorRGB **, allocate it accordingly, etc.
it looks like you are trying to copy array by assignment.
You cannot use simple assignment operator to do that, you have to use some function to copy things, for example memcpy.
*thisImage.arr = *imageArr;
thisimage.arr[0] = imagearr[0];
The above statements are doing the same thing.
However this is not most likely what causes the memory corruption
since you are working with two dimensional arrays, do make sure you initialize them correctly.
Looking at the code, should not even compile: the array is declared as one-dimensional in your image structure but you refer to as two-dimensional?