****New to C!****
I am running Dev-C++ 4.9.9.2 on Windows 7 (64 bit build)
My computer has 39GB of Physical Memory.
I am trying to create a large two dimensional array. I have already created code that tells me how many dimensions it has, and how many items are in each dimension.
As an example, let's say the array is two dimensional: One million items long, and 6 wide:
[1,2,3,4,5,6],
[1,2,3,4,5,6],
[1,2,3,4,5,6],
...and on to one million items.
I have tried:
float MyArray[1000000][6];
but this crashes Dev-C. It seems to fail when I try to initialize an array larger than:
float Myarray[86486][6];
I imagine I am experiencing a "stack overflow" which amuses me since that is the name of this site.
I have been digging around and it seems I need to use malloc to help C understand how much memory to reserve. I have seen good examples of how to use this to set up a 1 dimensional array, but I would very much appreciate example code of how to set this up with a 2 dimensional array.
I have seen the example here: Initializing a large two dimensional array in C
But I'm afraid I am too much of a beginner in c to understand the brief explanation.
As background: I am coming from python where you can make an array of (almost) any size or dimension by just declaring MyArray=[] and then filling it with whatever you want.
Thank you!
Yeah, you're running up against the limit of the size of an individual stack frame.
Here's one approach:
#include <stdlib.h>
...
int main(void)
{
/**
* declare myArray as a *pointer* to a 6-element array of float
*/
float (*myArray)[6];
/**
* dynamically allocate space for 1 million objects of type
* "6-element array of float"
*/
myArray = malloc(sizeof *myArray * 1000000);
...
myArray[i][j] = ...;
}
Why this works:
The subscript operation a[i] is interpreted as *(a + i); that is, we compute the address of the i'th element (not byte) after a (the base address of the array) and dereference it. Since myArray is a pointer to a 6-element array of float, myArray[i] gives us the address of the i'th 6-element array of float after myArray.
The advantage of this approach is that the memory is allocated in a contiguous chunk, and you can subscript myArray like any 2-d array.
Related
I am trying to create a struct containing 2 Variable Length Array (buffer_size is the variable parameter acquired at run time).
Here is my code:
struct data {
float *c; //2D array
float *mtdt; //1D array
};
struct data c_matrice;
c_matrice.c = malloc((90 * sizeof (float*) * buffer_size));
c_matrice.mtdt = malloc(90 * sizeof (float*));
The idea is to link the structure's members to arrays that are dynamically allocated.
Here is the compiler error
expected « = », « , », « ; », « asm » or « __attribute__ » before « . » token
c_matrice.c = malloc((90 * sizeof (float*) * buffer_size));
And when I try to access those members, I get
subscripted value is neither array nor pointer nor vector
I haven't been able to find a solution to my problem in the previous questions. Frankly as a beginner I don't get everything. What am I missing?
EDIT 1: Ok I got rid of the first error by moving the last two lines into my main.c rather than a .h file (This was a basic stupid mistake). Now I still face the
subscripted value is neither array nor pointer nor vector
when I try to access the struct with something like this
pmoy = pow(10,(c_matrice->c[i][curve2apply]/20))*pmax;
And by the way, the whole code is really big, and what I presented you was a small part of the actual code.
What you've done here:
c_matrice.c = malloc((90 * sizeof (float*) * buffer_size));
Is allocate one long buffer of size 90 * size of pointer-to-float * buffer_size.
You have a bunch of options in how you implement a 2D array in C.
One approach is to change what you have there to:
c_matrice.c = malloc((90 * sizeof (float) * buffer_size));
So you've allocated space for 90*buffer_size floats (rather than pointers to floats).
You then need to calculate indexes yourself:
float get_matrix_element(struct data *c_matrix, size_t row, size_t column) {
return c_matrix->c[row*buffer_size+column];
}
That's a very popular and very efficient way of storing the data because it's stored as one block of memory and you can do useful things like allocate it in a single block and iterate through it without concern for structure:
float get_matrix_sum(struct data *c_matrix) {
size_t sz=buffer_size*90;
float sum=0.0f;
for(size_t i=0;i<sz;++i){
sum+=c_matrix->c[i];
}
return sum;
}
There are other ways of doing this including:
Declare a 90 long 1D array of pointers to float and then allocate rows of floats.
The downside is 91 malloc()/free() operations instead of 1.
The upside is you could allocate a ragged array with different length rows.
Declare a static (compile time) sized array float c[90][buffer_size];
Where buffer_size is a compile time constant.
The downside is it's compile time fixed (and if large and a local variable may break the stack).
The upside is managing the internal r*90+c row calculation is taken off you.
If c is a member of the struct, then you must use c_matrice.c, not c_matrice->c. And carefully note everything people tell you about c not being a two dimensional array. To allocate these, there's a ton of question/answers on SO and you must not ask that question once again. :-)
It is more than one questions. I need to deal with an NxN matrix A of integers in C. How can I allocate the memory in the heap? Is this correct?
int **A=malloc(N*sizeof(int*));
for(int i=0;i<N;i++) *(A+i)= malloc(N*sizeof(int));
I am not absolutely sure if the second line of the above code should be there to initiate the memory.
Next, suppose I want to access the element A[i, j] where i and j are the row and column indices starting from zero. It it possible to do it via dereferencing the pointer **A somehow? For example, something like (A+ni+j)? I know I have some conceptual gap here and some help will be appreciated.
not absolutely sure if the second line of the above code should be there to initiate the memory.
It needs to be there, as it actually allocates the space for the N rows carrying the N ints each you needs.
The 1st allocation only allocates the row-indexing pointers.
to access the element A[i, j] where i and j are the row and column indices starting from zero. It it possible to do it via dereferencing the pointer **
Sure, just do
A[1][1]
to access the element the 2nd element of the 2nd row.
This is identical to
*(*(A + 1) + 1)
Unrelated to you question:
Although the code you show is correct, a more robust way to code this would be:
int ** A = malloc(N * sizeof *A);
for (size_t i = 0; i < N; i++)
{
A[i] = malloc(N * sizeof *A[i]);
}
size_t is the type of choice for indexing, as it guaranteed to be large enough to hold any index value possible for the system the code is compiled for.
Also you want to add error checking to the two calls of malloc(), as it might return NULL in case of failure to allocate the amount of memory requested.
The declaration is correct, but the matrix won't occupy continuous memory space. It is array of pointers, where each pointer can point to whatever location, that was returned by malloc. For that reason addressing like (A+ni+j) does not make sense.
Assuming that compiler has support for VLA (which became optional in C11), the idiomatic way to define continuous matrix would be:
int (*matrixA)[N] = malloc(N * sizeof *matrixA);
In general, the syntax of matrix with N rows and M columns is as follows:
int (*matrix)[M] = malloc(N * sizeof *matrixA);
Notice that both M and N does not have to be given as constant expressions (thanks to VLA pointers). That is, they can be ordinary (e.g. automatic) variables.
Then, to access elements, you can use ordinary indice syntax like:
matrixA[0][0] = 100;
Finally, to relase memory for such matrices use single free, e.g.:
free(matrixA);
free(matrix);
You need to understand that 2D and higher arrays do not work well in C 89. Beginner books usually introduce 2D arrays in a very early chapter, just after 1D arrays, which leads people to assume that the natural way to represent 2-dimensional data is via a 2D array. In fact they have many tricky characteristics and should be considered an advanced feature.
If you don't know array dimensions at compile time, or if the array is large, it's almost always easier to allocate a 1D array and access via the logic
array[y*width+x];
so in your case, just call
int *A;
A = malloc(N * N * sizeof(int))
A[3*N+2] = 123; // set element A[3][2] to 123, but you can't use this syntax
It's important to note that the suggestion to use a flat array is just a suggestion, not everyone will agree with it, and 2D array handling is better in later versions of C. However I think you'll find that this method works best.
Is there a way we can copy every element from one multidimensional array to another multidimensional array by just doing one memcpy operation?
int array1[4][4][4][4];
int array2[4][4][4][4];
int main()
{
memset(&array1,1,sizeof(array1));
memset(&array2,0,sizeof(array2));
printf_all("value in array2 %d \n",array2[1][1][1][1]);
memcpy(&array2,&array1,sizeof(array2));
printf("memcopied in array2 from array1 \n");
printf("value in array2 %d \n",array2[1][1][1][1]); //not printing 1
}
Your code is correct. You should not expect the output to show you a value of 1. You should expect it to show you a value of 16843009, assuming a 4 byte int.
The reason is: you are filling array1 with bytes of value 1, not with ints of value 1. i.e. binary 00000001000000010000000100000001 (0x01010101) is being filled into all the int elements with your memset operation.
So regardless of the size of int on your machine (unless it's a single byte!) you should not expect to see the value 1.
I hope this helps.
Yes, your code should already be correct.
You have to consider memory layout when doing this. The arrays are all in one block, multi dimensional is essentially a math trick done by the compiler.
Your code says copy this memory content to the other memory block. since both share the same layout they will contain the same values.
The following code also just copies the values, but access is handled differently so you would have to think about how to get the order of elements correct.
int array1[4][4][4][4]; //elements 256
int array2[256];
int main()
{
memcpy(&array2,&array1,sizeof(array1)); //will also copy
// original access via: a + 4 * b + 16 * c + 64 * d
}
Multidimensional array in C is a flat block of memory with no internal structure. Memory layout of a multidimensional array is exactly the same as that of a 1-dimensional array of the same total size. The multidimensional interface is implemented through simple index recalculation. You can always memcpy the whole multidimensional array exactly as you do it in your code.
This, of course, only applies to built-in multidimensional arrays, explicitly declared as such (as in your code sample). If you implement a hand-made multidimensional array as an array of pointers to sub-arrays, that data structure will not be copyable in one shot with memcpy.
However, apparently you have some misconceptions about how memset works. Your memset(&array1,1,sizeof(array1)); will not fill the array with 1s, meaning that your code is not supposed to print 1 regardless of which array you print. memset interprets target memory as an array of chars, not as an array of ints.
memset can be used to set memory to zero. As for non-zero values, memset is generally unsuitable for initializing arrays of any type other than char.
I know there is several questions about that which gives good (and working) solutions, but none IMHO which says clearly what is the best way to achieve this.
So, suppose we have some 2D array :
int tab1[100][280];
We want to make a pointer that points to this 2D array.
To achieve this, we can do :
int (*pointer)[280]; // pointer creation
pointer = tab1; //assignation
pointer[5][12] = 517; // use
int myint = pointer[5][12]; // use
or, alternatively :
int (*pointer)[100][280]; // pointer creation
pointer = &tab1; //assignation
(*pointer)[5][12] = 517; // use
int myint = (*pointer)[5][12]; // use
OK, both seems to work well. Now I would like to know :
what is the best way, the 1st or the 2nd ?
are both equals for the compiler ? (speed, perf...)
is one of these solutions eating more memory than the other ?
what is the more frequently used by developers ?
//defines an array of 280 pointers (1120 or 2240 bytes)
int *pointer1 [280];
//defines a pointer (4 or 8 bytes depending on 32/64 bits platform)
int (*pointer2)[280]; //pointer to an array of 280 integers
int (*pointer3)[100][280]; //pointer to an 2D array of 100*280 integers
Using pointer2 or pointer3 produce the same binary except manipulations as ++pointer2 as pointed out by WhozCraig.
I recommend using typedef (producing same binary code as above pointer3)
typedef int myType[100][280];
myType *pointer3;
Note: Since C++11, you can also use keyword using instead of typedef
using myType = int[100][280];
myType *pointer3;
in your example:
myType *pointer; // pointer creation
pointer = &tab1; // assignation
(*pointer)[5][12] = 517; // set (write)
int myint = (*pointer)[5][12]; // get (read)
Note: If the array tab1 is used within a function body => this array will be placed within the call stack memory. But the stack size is limited. Using arrays bigger than the free memory stack produces a stack overflow crash.
The full snippet is online-compilable at gcc.godbolt.org
int main()
{
//defines an array of 280 pointers (1120 or 2240 bytes)
int *pointer1 [280];
static_assert( sizeof(pointer1) == 2240, "" );
//defines a pointer (4 or 8 bytes depending on 32/64 bits platform)
int (*pointer2)[280]; //pointer to an array of 280 integers
int (*pointer3)[100][280]; //pointer to an 2D array of 100*280 integers
static_assert( sizeof(pointer2) == 8, "" );
static_assert( sizeof(pointer3) == 8, "" );
// Use 'typedef' (or 'using' if you use a modern C++ compiler)
typedef int myType[100][280];
//using myType = int[100][280];
int tab1[100][280];
myType *pointer; // pointer creation
pointer = &tab1; // assignation
(*pointer)[5][12] = 517; // set (write)
int myint = (*pointer)[5][12]; // get (read)
return myint;
}
Both your examples are equivalent. However, the first one is less obvious and more "hacky", while the second one clearly states your intention.
int (*pointer)[280];
pointer = tab1;
pointer points to an 1D array of 280 integers. In your assignment, you actually assign the first row of tab1. This works since you can implicitly cast arrays to pointers (to the first element).
When you are using pointer[5][12], C treats pointer as an array of arrays (pointer[5] is of type int[280]), so there is another implicit cast here (at least semantically).
In your second example, you explicitly create a pointer to a 2D array:
int (*pointer)[100][280];
pointer = &tab1;
The semantics are clearer here: *pointer is a 2D array, so you need to access it using (*pointer)[i][j].
Both solutions use the same amount of memory (1 pointer) and will most likely run equally fast. Under the hood, both pointers will even point to the same memory location (the first element of the tab1 array), and it is possible that your compiler will even generate the same code.
The first solution is "more advanced" since one needs quite a deep understanding on how arrays and pointers work in C to understand what is going on. The second one is more explicit.
int *pointer[280]; //Creates 280 pointers of type int.
In 32 bit os, 4 bytes for each pointer. so 4 * 280 = 1120 bytes.
int (*pointer)[100][280]; // Creates only one pointer which is used to point an array of [100][280] ints.
Here only 4 bytes.
Coming to your question, int (*pointer)[280]; and int (*pointer)[100][280]; are different though it points to same 2D array of [100][280].
Because if int (*pointer)[280]; is incremented, then it will points to next 1D array, but where as int (*pointer)[100][280]; crosses the whole 2D array and points to next byte. Accessing that byte may cause problem if that memory doen't belongs to your process.
Ok, this is actually four different question. I'll address them one by one:
are both equals for the compiler? (speed, perf...)
Yes. The pointer dereferenciation and decay from type int (*)[100][280] to int (*)[280] is always a noop to your CPU. I wouldn't put it past a bad compiler to generate bogus code anyways, but a good optimizing compiler should compile both examples to the exact same code.
is one of these solutions eating more memory than the other?
As a corollary to my first answer, no.
what is the more frequently used by developers?
Definitely the variant without the extra (*pointer) dereferenciation. For C programmers it is second nature to assume that any pointer may actually be a pointer to the first element of an array.
what is the best way, the 1st or the 2nd?
That depends on what you optimize for:
Idiomatic code uses variant 1. The declaration is missing the outer dimension, but all uses are exactly as a C programmer expects them to be.
If you want to make it explicit that you are pointing to an array, you can use variant 2. However, many seasoned C programmers will think that there's a third dimension hidden behind the innermost *. Having no array dimension there will feel weird to most programmers.
given the following function signature:
void readFileData(FILE* fp, double inputMatrix[][], int parameters[])
this doesn't compile.
and the corrected one:
void readFileData(FILE* fp, double inputMatrix[][NUM], int parameters[])
my question is, why does the compiler demands that number of columns will be defined when handling a 2D array in C? Is there a way to pass a 2D array to a function with an unknown dimensions?
thank you
Built-in multi-deminsional arrays in C (and in C++) are implemented using the "index-translation" approach. That means that 2D (3D, 4D etc.) array is laid out in memory as an ordinary 1D array of sufficient size, and the access to the elements of such array is implemented through recalculating the multi-dimensional indices onto a corresponding 1D index. For example, if you define a 2D array of size M x N
double inputMatrix[M][N]
in reality, under the hood the compiler creates an array of size M * N
double inputMatrix_[M * N];
Every time you access the element of your array
inputMatrix[i][j]
the compiler translates it into
inputMatrix_[i * N + j]
As you can see, in order to perform the translation the compiler has to know N, but doesn't really need to know M. This translation formula can easily be generalized for arrays with any number of dimensions. It will involve all sizes of the multi-dimensional array except the first one. This is why every time you declare an array, you are required to specify all sizes except the first one.
As the array in C is purely memory without any meta information about dimensions, the compiler need to know how to apply the row and column index when addressing an element of your matrix.
inputMatrix[i][j] is internally translated to something equivalent to *(inputMatrix + i * NUM + j)
and here you see that NUM is needed.
C doesn't have any specific support for multidimensional arrays. A two-dimensional array such as double inputMatrix[N][M] is just an array of length N whose elements are arrays of length M of doubles.
There are circumstances where you can leave off the number of elements in an array type. This results in an incomplete type — a type whose storage requirements are not known. So you can declare double vector[], which is an array of unspecified size of doubles. However, you can't put objects of incomplete types in an array, because the compiler needs to know the element size when you access elements.
For example, you can write double inputMatrix[][M], which declares an array of unspecified length whose elements are arrays of length M of doubles. The compiler then knows that the address of inputMatrix[i] is i*sizeof(double[M]) bytes beyond the address of inputMatrix[0] (and therefore the address of inputMatrix[i][j] is i*sizeof(double[M])+j*sizeof(double) bytes). Note that it needs to know the value of M; this is why you can't leave off M in the declaration of inputMatrix.
A theoretical consequence of how arrays are laid out is that inputMatrix[i][j] denotes the same address as inputMatrix + M * i + j.¹
A practical consequence of this layout is that for efficient code, you should arrange your arrays so that the dimension that varies most often comes last. For example, if you have a pair of nested loops, you will make better use of the cache with for (i=0; i<N; i++) for (j=0; j<M; j++) ... than with loops nested the other way round. If you need to switch between row access and column access mid-program, it can be beneficial to transpose the matrix (which is better done block by block rather than in columns or in lines).
C89 references: §3.5.4.2 (array types), §3.3.2.1 (array subscript expressions)
C99 references: §6.7.5.2 (array types), §6.5.2.1-3 (array subscript expressions).
¹ Proving that this expression is well-defined is left as an exercise for the reader. Whether inputMatrix[0][M] is a valid way of accessing inputMatrix[1][0] is not so clear, though it would be extremely hard for an implementation to make a difference.
This is because in memory, this is just a contiguous area, a single-dimension array if you will. And to get the real offset of inputMatrix[x][y] the compiler has to calculate (x * elementsPerColumn) + y. So it needs to know elementsPerColumn and that in turn means you need to tell it.
No, there's not. The situation's pretty simple really: what the function receives is really just a single, linear block of memory. Telling it the number of columns tells it how to translate something like block[x][y] into a linear address in the block (i.e., it needs to do something like address = row * column_count + column).
Other people have explained why, but the way to pass a 2D array with unknown dimensions is to pass a pointer. The compiler demotes array parameters to pointers anyway. Just make sure it's clear what you expect in your API docs.