I'm doing dynamic memory allocation testing, but when I try to get the size of the array I always have the same result.
int main()
{
int *teste=malloc(sizeof(int) * 10);
int len=sizeof(teste) / sizeof(int);
print("%i\n", len);
return 0;
}
Compiling with gcc:
bash-4.2$ ./teste
2
It does not matter if I put sizeof(int) * 100 or 10 , it always returns 2 :(
What am I doing wrong?
The idiomatic sizeof(teste) / sizeof(int) is only able to calculate the in bytes size of statically allocated buffers:
int teste[ 123 ];
printf("%d\n", sizeof(teste) / sizeof(int) ); /* 492 / 4 = 123 */
In your case, teste is a pointer to a dynamically allocated memory region and sizeof(teste) is the size in bytes that that pointer occupies, not the size of the memory it is pointing to.
int * teste = NULL;
printf("%d\n", sizeof(teste) / sizeof(int) ); /* 8 / 4 = 2 */
What about:
int main()
{
int len = sizeof(int) * 10;
int *teste=malloc(len);
print("%d\n", len); /* 40 */
free(teste);
return 0;
}
Thanks for the collaborations, in your example Lacobus in my project it would not work because I do not know the size of the array when creating and I always reallocate but I remembered something that can be done there of the data structure classes.
I made a structure
struct __buf {
int *ptr;
int len;
};
As I adjust the size of the array I update the structure, like this:
__buf buffer_serial;
...
buffer_serial.ptr = malloc ( sizeof(int) * len);
buffer_seria.len = len;
...