C Does type before malloc even matter ?

[Xenon02]

Hello !

I've been testing some stuff with pointers.
Here is one of them :

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

int main()
{
    int *matrix = (int *)malloc(1);
 
    *(matrix) = 2562323;
 
    printf("%d", *matrix);
    return 0;
}

I tried to check if the output will be an error because I've only reserved 1 byte, but when I wrote 2562323 into this one byte it saved and printed this value. How ??? What if in next byte after the malloc there is something ?

Same went when I changed malloc from int to char

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

int main()
{
    int *matrix = (char *)malloc(1);
 
    *(matrix) = 2562323;
 
    printf("%d", *matrix);
    return 0;
}

I expected that the only number that will appear is 2. Because the memory is for char, so only 1 byte, so number "2" at the beggining will be saved in this one byte and the rest will be ignored Does this "type" infront of malloc matter ?

 

[entrocraft]

It's undefined behavior to assign an object to uninitialized memory. int has a defined 2 or 4 byte width, so maybe the compiler is using sizeof(int) from the data type of the declared matrix pointer. A program compiled using this code could arbitrarily fail or produce different outputs.

The type in front of malloc() doesn't matter, it's just a redundant operation that gets implicitly converted to int * after casting explicitly to char *.

The (int *) cast to an int * pointer isn't needed. Here's a documented ANSI C example that confirms this (with the non-standard alloc() function).

extern void *alloc(size_t);
double *dp = alloc(sizeof *dp);

The following shows the correct, standards-compliant version of your code.

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

int main() {
    int *matrix = malloc(sizeof(int));

    if (matrix != NULL) {
      *matrix = (int) 2562323;
    } else {
      exit(EXIT_FAILURE);
    }

    printf("%d", *matrix);
    return 0;
}

The (int) cast is explicit because sizeof(int) can be 2 bytes on 32-bit systems and 2562323 exceeds 2 bytes. If sizeof(int) is 2 instead of 4, the output is truncated to 24910.

 

[entrocraft]

I suggest using long instead of int to avoid truncation as well.

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

int main() {
    long *matrix = malloc(sizeof(long));

    if (matrix != NULL) {
      *matrix = 2562323;
    } else {
      exit(EXIT_FAILURE);
    }

    printf("%ld", *matrix);
    return 0;
}

 

[user001]

I think type conversion before malloc is quietly doesn't matter, however position 1 must coherent to position 2.

main()
{
    int * p1 = malloc(sizeof(int));
//  ^^^ position 1           ^^^ position 2
}

It's a pedantic way to add type conversion before molloc etc.

main()
{
    int * p = (int *)malloc(sizeof(int));
    *(char *)p = 1; // No if you were a newbie to C, don't do this way.
    free(p);
    return 0;
}