sizeof array

[Iam3R]

how the sizeof array is the total number of elements multiped by its base data type size, when the name of the array gives only the base address.

int array [5];
       sizeof array ;

gives 20 bytes.

why not 4 bytes.

because

array = &array[0]
.

[Ancient Dragon]

>>why not 4 bytes.

You answered your own question in the first sentence you posted. Because 5 * sizeof(int) = 5 * 4 = 20. Simple grade-school math.

[Iam3R]

>>why not 4 bytes.

You answered your own question in the first sentence you posted. Because 5 * sizeof(int) = 5 * 4 = 20. Simple grade-school math.

why sizeof does not consider name of the array as some address and give pointer size bytes.

[gerard4143]

check out this link:
http://en.wikipedia.org/wiki/Sizeof

[Narue]

>why sizeof does not consider name of the array
>as some address and give pointer size bytes.
Because the conversion from array to pointer only occurs in a value context. The operand of sizeof is being used in an object context, so the conversion doesn't happen.

There are three object contexts for an array:

1. As the operand to sizeof ( sizeof a ).
2. As the operand to address-of ( &a ).
3. As a string literal initializer ( char a[] = "booga"; )

In all other cases the name of an array is converted to a pointer to the first element.