Why sizeof(array) and sizeof(&array[0]) gives different results?

Question

#include 
int main(void){
    char array[20];

    printf( \"\\nSize of array is %d\\n\", sizeof(array) );  //outputs 20
    printf(\"\\nSize of &         


        

Answer 1

sizeof(&array[0]) is returning the size of a pointer to a char. array[0] yields a char, the & returns a pointer to the char. Your compiler used 4 bytes for a pointer (32 bit).

Answer 2

In the expression sizeof array, array is not converted to a pointer type.

C11, § 6.3.2.1 Lvalues, arrays, and function designators

Except when it is the operand of the sizeof operator, the _Alignof operator, or the unary & operator, or is a string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object and is not an lvalue.

Therefore, its type is char[20], not char *. The size of this type is sizeof(char) * 20 = 20 bytes.

C11, § 6.5.3.4 The sizeof and _Alignof operators

The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand.

&array[0] type is char *. That's why the sizeof(&array[0]) gives the same result as sizeof(char *) (4 bytes on your machine).

Answer 3

The "array" itself is constant pointer, you can not address it to different location. Yes, array address is the same as &array[0], which is derived from &(array+0) and it is equal to &(array). We know that &* is eliminated so we are back to "array" again. So "array" is the same as "&*(array+0)" But in regards to sizeof operator, the operand must be in array notation, not in pointer notation. To clarify:

char *data = {"Today is Sunday"}; ... Pointer notation

char data[] = {"Today is Sunday"}; ... Array notation

In the first example with pointer notation the sizeof(data) will be the size of bytes required to store an address, which is 8 on my system. In the second example the sizeof(data) will give the number of bytes occupied by the array, which would be 15 chars in the string. But if you use pointer notation like sizeof(data+0) the result will be the number of bytes required to store an address, which is 8 on my system.

sizeof operator result is the same for pointer and for array with pointer notation.

Answer 4

The expression &array[0] is a pointer so the sizeof operator returns the size of the pointer.

Answer 5

&array[0] returns a pointer to the first index of array. Using the sizeof operator on a pointer will yield the size of a pointer (depends on ABI), in your case, it's 4.

Answer 6

I think that "array" itself is constant pointer, you can not address it to different location. It is true that we can write: array is the same as &array[0], which is derived from &*(array+0) and it is equal to &*(array). We know that &* is eliminated so we are back to "array" again. So "array" is the same as "&*(array+0)" The terms should be the same in some circumstances, but it is not the same for the sizeof operator.