This question already has answers here:
Segmentation fault on large array sizes
(7 answers)
Closed 9 years ago.
If I keep the value of rows to 100000, the program works fine but, if I make rows one million as 1000000, the program gives me segmentation fault. What is the reason? I am running below on Linux 2.6x RHEL kernel.
#include<stdio.h>
#define ROWS 1000000
#define COLS 4
int main(int args, char ** argv)
{
int matrix[ROWS][COLS];
for(int col=0;col<COLS;col++)
for(int row=0;row < ROWS; row++)
matrix[row][col] = row*col;
return 0;
}
The matrix is a local variable inside your main function. So it is "allocated" on the machine call stack.
This stack has some limits.
You should make your matrix a global or static variable or make it a pointer and heap-allocate (with e.g. calloc or malloc) the memory zone. Don't forget that calloc or malloc may fail (by returning NULL).
A better reason to heap-allocate such a thing is that the dimensions of the matrix should really be a variable or some input. There are few reasons to wire-in the dimensions in the source code.
Heuristic: don't have a local frame (cumulated sum of local variables' sizes) bigger than a kilobyte or two.
[of course, there are valid exceptions to that heuristic]
You are allocating a stack variable, the stack of each program is limited.
When you try to allocate too much stack memory, your kernel will kill your program by sending it a SEGV signal, aka segmentation fault.
If you want to allocate bigger chunks of memory, use malloc, this function will get memory from the heap.
Your system must not allow you to make a stack allocation that large. Make matrix global or use dynamic allocation (via malloc and free) and you should be ok.
Related
This question already has answers here:
Array index out of bound behavior
(10 answers)
Why don't I get a segmentation fault when I write beyond the end of an array?
(4 answers)
Closed 6 years ago.
I am trying to implement a simple sieve and to my help I found the following code:
int main(int argc, char *argv[])
{
int *array, n=10;
array =(int *)malloc(sizeof(int));
sieve(array,n);
return 0;
}
void sieve(int *a, int n)
{
int i=0, j=0;
for(i=2; i<=n; i++) {
a[i] = 1;
}
...
For some reason this works, but I think it should not! The space that is allocated for the variable array is only enough to support one integer, but a[i] for i = 2...10 are called in the function sieve. Shouldn't this cause problems?
I tried to change the implementation to
int array[10], n = 10;
which caused "Abort trap: 6" on runtime. However, this I understand since array[10] will be outside of the space allocated. But shouldn't the same be true also for the code where malloc i used?
Truly confusing.
You are correct in some ways. For example this line:
array =(int *)malloc(sizeof(int));
only allocates space for one integer, not 10. It should be:
array =(int *)malloc(sizeof(int) * 10);
However, that does not mean the code will fail. At least not immediately. When you get back a pointer from malloc and then start writing beyond the bounds of what you have allocated, you might be corrupting something in memory.
Perhaps you are writing over a structure that malloc uses to keep track of what it was asked for, perhaps you are writing over someone else's memory allocation. Perhaps nothing at all - malloc usually allocates more than it is asked for in order to keep the chunks it gives out manageable.
If you want something to crash, you usually have to scribble beyond an operating system page boundary. If you are using Windows or Linux or whatever, the OS will give you (or malloc in this case) some memory in a set of block, usually 4096 bytes in size. If you scribble within that block, the operating system will not care. If you go outside it, you will cause a page fault and the operating system will usually destroy your process.
It was much more fun in the days of MS-DOS. This was not a "protected mode" operating system - it did not have hardware enforced page boundaries like Windows or Linux. Scribbling beyond your area could do anything!
This question already has answers here:
Segmentation Fault, large arrays
(1 answer)
Getting a stack overflow exception when declaring a large array
(8 answers)
Closed 6 years ago.
This is my part of the code that occurred and segmentation fault:
int main (int argc, char *argv[]) {
printf ("====================================================\n");
double pointArray[MAX_NUM_OF_POINTS][DIMENSION];
double range;
int num_of_nearest;
double queryPoint[DIMENSION];
int counter;
int dist;
int num;
printf ("====================================================\n");
}
where MAX_NUM_OF_POINTS was defined to be 100,000,000.
However, when I changed this number to be smaller like 100,000, the segmentation fault disappeared.
Could anyone tell me the reason?
Local variables are created on the stack, which has a limited amount of space. By attempting to create an array of at least 100000000 doubles, each of which is probably 8 bytes, it is too large for the stack and causes a segfault.
If you declare the array as a global, it will not reside on the stack but in the data segment instead, which can handle larger variables. Alternately, you can create the array dynamically using malloc in which case it lives on the stack.
This however raises the question as to why you need an array that large. You may need to rethink your design to see if there is a more memory efficient way of doing what you want.
This question already has answers here:
Segmentation fault on large array sizes
(7 answers)
Closed 3 years ago.
Program with large global array:
int ar[2000000];
int main()
{
}
Program with large local array:
int main()
{
int ar[2000000];
}
When I declare an array with large size in the main function, the program crashes with "SIGSEGV (Segmentation fault)".
However, when I declare it as global, everything works fine. Why is that?
Declaring the array globally causes the compiler to include the space for the array in the data section of the compiled binary. In this case you have increased the binary size by 8 MB (2000000 * 4 bytes per int). However, this does mean that the memory is available at all times and does not need to be allocated on the stack or heap.
EDIT: #Blue Moon rightly points out that an uninitialized array will most likely be allocated in the bss data segment and may, in fact, take up no additional disk space. An initialized array will be allocated statically.
When you declare an array that large in your program you have probably exceeded the stack size of the program (and ironically caused a stack overflow).
A better way to allocate a large array dynamically is to use a pointer and allocate the memory on the heap like this:
using namespace std;
int main() {
int *ar;
ar = malloc(2000000 * sizeof(int));
if (ar != null) {
// Do something
free(ar);
}
return 0;
}
A good tutorial on the Memory Layout of C Programs can be found here.
This question already has answers here:
Memory allocation for global and local variables
(3 answers)
Segmentation fault on large array sizes
(7 answers)
Closed 7 years ago.
I'm not too experienced in C, but I've been recently writing some program in the language to speed it up a little bit (it was originally written in python). I don't really have a problem anymore since I already managed to solve my original problem. However, I would like to know why this solution works.
I have a data structure representing complex numbers defined as
typedef struct _fcomplex { float re, im; } fcomplex;
Then I want to create an array of complex numbers as:
fcomplex M[N];
Here N is a large number (something like ~10^6). Then I initialize the array with zeros in a function that essentially runs through all the indices and sets the values in the array. It's something like:
fcomplex num = {0.0, 0.0};
int i;
for (i=0 ; i < N ; i++) {
M[i] = num;
}
However, when I run the code, it results in a segmentation fault. However, if I use malloc() to allocate space for the array instead as
fcomplex* M = malloc(N*sizeof(fcomplex));
and then do everything as before, the code works fine. Also, for smaller values of N, the code runs fine either way.
As I said, using malloc() already solved the problem, but I would like to know why?
It depends where you allocated the array. If it's inside a function, then the variable is allocated on the stack, and by default, (I assume you're running linux) the stack size is 8Mb.
You can find it out using ulimit -s and also modify this value, for instance ulimit -s 1000000.
You may want to have a look on those questions:
Memory allocation for global and local variables
Segmentation fault on large array sizes (suggested by #Ed)
This question already has answers here:
Segmentation fault on large array sizes
(7 answers)
Closed 8 years ago.
When I want to #define for SIZE 1.000.000 , my program crashed before it start main function, but when i #define for SIZE 100.000 it work. I have two arrays initialization in my program.
#define SIZE 1000000
char *baza_vocka[SIZE];
char b_vocka[SIZE];
EDIT: They are local variables.
In case of 1M you're trying to allocate an array on the stack, which is bigger than the stack size. For such sizes you need to allocate the memory on heap.
For example:
int main()
{
// allocation on the heap
char **baza_vocka = malloc(sizeof(char*)*SIZE);
char *b_vocka = malloc(sizeof(char)*SIZE);
// working code
baza_vocka[1] = 0x0;
b_vocka[1] = 'a';
// cleaning the heap
free(b_vocka);
free(baza_vocka);
}
I guess that baza_vocka is a local variable, perhaps inside main
You are experimenting some stack overflow, .... Local call frames should usually be small (a few kilobytes) in your call stack
You should not have so big local variables. Allocate such big arrays in the heap using malloc or calloc. Don't forget to test that they are not failing. Read documentation of malloc(3), and don't forget to free such a heap allocated array. Beware of memory leaks & buffer overflows. Use valgrind if available.
On current desktops, stack space is a few megabytes, so you usually should limit each call frame to a few kilobytes (or a few dozens of kilobytes).