How to write exact 1MB array in C?

Question

I want to initialize an array of size 1MB. So my goal is finally write that 1MB to a file. I am curious every time i use this formula it is giving less than 1mb.

int len = (1048576)/sizeof(int);
data = (int *) malloc(len);

What is correct way ? Thank you

Edit - As per the comments I have changed the code .

#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>
#include <string.h>

int main(int argc, char *argv[]) {
    int *data;
    int bytes = (1024*1024);
    data = (int *) malloc(bytes);
    for(int i=0;i<bytes;i++){
     data[i] = (int)rand();
     printf("%d",data[i]);
    }
    return 0;

}

After compiling it and I tried dumping the data like below

 mpicc -o a mpiFileSize.c

 ./a > dump.dat

Now I see the file size of dump.dat. Why its 2.5MB ?

`malloc()` takes the number of bytes to allocate. There is no need to divide by `sizeof(int)`. — kichik, Nov 03 '16 at 16:17
If your goal is to write one MB to a file; why do you think you need 1 MB in memory too? You could write 1024 x 1024 bytes to your file for example (so a simply loop would do)?! — GhostCat, Nov 03 '16 at 16:17
programming 101: read the manual before writing code. 102: read the manual before asking a question on SO. — user3528438, Nov 03 '16 at 16:19
What is `size(int)` supposed to do? Do you mean `sizeof(int)`? If so, that means you're trying to store 1024^3 / 4 bytes, since each integer is typically 4 bytes. — Random Davis, Nov 03 '16 at 16:20
@mouviciel That's not dynamic. The compiler will turn that into `int data[4194304];` (on a machine with 4-byte ints) I do declarations like that all the time because it makes your logic for choosing the size much clearer than a magic number like 4194304. — Carey Gregory, Nov 03 '16 at 16:39
@CareyGregory: I know that's not dynamic: it's my point. Why does OP choose malloc() over a static declaration? — mouviciel, Nov 03 '16 at 18:16
you get a file larger than 1mb because possibly because of the `include` statements, which literally include (copy and paste) the header files to the file you `include` them in — Fabulous, Nov 03 '16 at 18:32

score 4 · Accepted Answer · answered Nov 03 '16 at 17:31

4

Try this

#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>
#include <string.h>

int main(int argc, char *argv[]) {
char *data;
int bytes = (1024*1024);
data = (char *) malloc(bytes);
for(int i=0;i<bytes;i++){
 data[i] = (char) rand();
 printf("%c",data[i]);
}
return 0;

}

You shoul use character instead of integer.

answered Nov 03 '16 at 17:31

Muhammad Mohsin Khan

146
7

Yes It generates 1 MB data. – ajayramesh Nov 03 '16 at 17:34
why not integers ? any thoughts on that – ajayramesh Nov 03 '16 at 17:35
1

Actually character is of 1 byte and integer is 16 or 32 depending upon the underlying architecture. – Muhammad Mohsin Khan Nov 03 '16 at 17:36

Erick Souza · Answer 2 · 2022-06-18T04:45:07.633

Although it was already properly answered. Just a plus to the answer, if one wants to choose the amount of MBs to allocate would make something like:

#include <malloc.h>

#define Mebabyte (1024 * 1024)

int main(int argc, char** argv)
{
    void* data = malloc(2 * Megabyte);
    
    // Do your work here...
    
    free(data);
    return 0;
}

If you wanted to allocate more than 2 MBs just change the 2. As already stated before do not use integers as it's going to have more than 1 byte of size. Instead use char or unsigned char. And as stated by another post, there's no need to cast the result of malloc since void* can be turned to a pointer to any type (and in fact it's done implicitly by the compiler).

see: Why does this code segfault on 64-bit architecture but work fine on 32-bit?

How to write exact 1MB array in C?

2 Answers2