arrays - C - SizeOf Pointers

Question

char c[] = {'a','b','c'};
int* p = &c[0];
printf("%i
", sizeof(*p)); //Prints out 4
printf("%i
", 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?

Answer 1

Because p is of type int *, so *p is of type int, which is apparently 4 bytes wide on your implementation.

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And use %zu for printing size_t (what sizeof yields) if you don't want your program to invoke undefined behavior.