What is a void pointer?Consider the following code snippet
What happens when the line void *k is encountered?Code:#include<stdio.h> void main() { float i=10,*j; void *k; k=&i; j=k; printf("%f",*j); }
How is the assignment done?Does that mean k holds the address of i?Also,why doesn't the following code work?
Thanks.Code:#include<stdio.h> void main() { float i=10,*j; void *k; k=&i; printf("%d",*k); }
Yes, k points to i.Originally Posted by Sylar Persie
The type information is lost, hence you cannot dereference a pointer to void.
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I'm sorry,what does that mean?Could you elaborate?
A void pointer is just taken to mean a raw generic address, not an address pointing to a specific data type. When you dereference a pointer you need to know what type it is supposed to be pointing at. Given a generic pointer to some address, the compiler rightly complains when you attempt to dereference it since it cannot determine what it should be pointing to. You - the programmer - can logically follow the source and see that it is pointing to an int or a float or whatever but the compiler does not have that luxury, its simply an address without any associated type information.
Consider this, if you dereference an integer pointer, the compiler knows to take sizeof(int) bytes at the address the pointer points to, and treat those sizeof(int) bytes as an integer. If you have a double pointer that for whatever reason points to the same address as the above integer pointer, when you dereference the double pointer, it is going to take sizeof(double) bytes and interpret as a double value. If you have a generic void pointer to that same address, how does the compiler know how many bytes to grab when you try to dereference the pointer if there is no type associated with the pointer.
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Thanks hk_mp5kpdw....that made things clear
