Is it possible to dereference a void pointer without type-casting in the C programming language?
Also, is there any way of generalizing a function which can receive a pointer and store it in a void pointer and by using that void pointer, can we make a generalized function?
for e.g.:
void abc(void *a, int b)
{
if(b==1)
printf("%d",*(int*)a); // If integer pointer is received
else if(b==2)
printf("%c",*(char*)a); // If character pointer is received
else if(b==3)
printf("%f",*(float*)a); // If float pointer is received
}
I want to make this function generic without using if-else statements - is this possible?
Also if there are good internet articles which explain the concept of a void pointer, then it would be beneficial if you could provide the URLs.
Also, is pointer arithmetic with void pointers possible?
Is it possible to dereference the void pointer without type-casting in C programming language...
No, void
indicates the absence of type, it is not something you can dereference or assign to.
is there is any way of generalizing a function which can receive pointer and store it in void pointer and by using that void pointer we can make a generalized function..
You cannot just dereference it in a portable way, as it may not be properly aligned. It may be an issue on some architectures like ARM, where pointer to a data type must be aligned at boundary of the size of data type (e.g. pointer to 32-bit integer must be aligned at 4-byte boundary to be dereferenced).
For example, reading uint16_t
from void*
:
/* may receive wrong value if ptr is not 2-byte aligned */
uint16_t value = *(uint16_t*)ptr;
/* portable way of reading a little-endian value */
uint16_t value = *(uint8_t*)ptr
| ((*((uint8_t*)ptr+1))<<8);
Also, is pointer arithmetic with void pointers possible...
Pointer arithmetic is not possible on pointers of void
due to lack of concrete value underneath the pointer and hence the size.
void* p = ...
void *p2 = p + 1; /* what exactly is the size of void?? */