char c[] = {'a','b','c'}; int* p = &c[0]; printf("%i\n", sizeof(*p)); //Prints out 4 printf("%i\n", sizeof(*c)); //Prints out 1 I am extremely confused about this section of code. Both p and c represent the address of the array c at the 0th index. But why does sizeof(*p) print out 4? Shouldn't it be 1?
Because p is of type int *, so *p is of type int, which is apparently 4 bytes wide on your implementation.
And use %zu for printing size_t (what sizeof yields) if you don't want your program to invoke undefined behavior.
int and use "%d" or "%i" if your implementation (coughMicrosoftcough) doesn't support "%zu".unsigned long and use "%lu"
int p pointer accept address of char .this code int* p = &c[0]. i though only void* pointers could take any typechar c[10], gcc warns initialization from incompatible pointer typesizeof(*p) is the size of the int object to which p points.
sizeof(*p) will print size of p which is 4 because of int but c is of char that's why it is 1
In C (not C99) the sizeof operator is strictly a compile time calculation. so when sizeof (*p) [a dereferenced ptr to an integer] is evaluated then the size of int is four.
Note. the (*p) portion of that statement is a "cast" operator. So, sizeof is NOT a function call as in sizeof(xyz), rather it is like sizeof var_name or sizeof (int *).
When your program runs it changes the object pointed to by p, but the value of sizeof (*p) was already computed and hard-coded into your executable load module.
I think your confusion is that you were thinking that C would figure out what data type p was pointing to when your program runs.
