为什么数组保存C中第一个元素的地址？

Question

Some says that array in c is not a pointer to the first element of the array why is array name a pointer to the first element of the array? So when you print the array in c using "printf" for example why it is showing the address of the first element instead of the array elements?

Update:

const char  h[10]="Hello";
printf("%p", h);

Output: 00AFFE0C

well as one answer said said that this happens because of the %type I specified and that made sense because when I write

printf("%s", h);
OR
printf(h);

Output: Hello

here two questions rises:

1) In printf(h); why it is not decaying the array and printing the pointer value because in decaying array will be converted to pointer

2) how can I print an array of int in the same way ie what is the %type in printf for array of int

Answer 1

The difference between an array and a pointer is that the compiler knows about the array size and dimensions so it can perform some static checks address computations that are not possible with just pointers.

Additionally in the case of printf() it doesn't matter what you pass it because it's a variadic function. It doesn't know anything about the type of the arguments that you pass it except from the format string so if you pass a %s argument you're in effect casting it to char* . The only reason the compiler may give you a warning if you pass an argument of the wrong type is because it's an intrinsic function and the compiler knows about it.

Most functions that take arrays take them as pointer arguments so you're also implicitly casting the array to a pointer when calling a function unless the function specifies an array type as argument.