Question

When I execute the following code:

int main()
{
    char **temp;
    printf("Size of **temp %d", sizeof(**temp));
    printf("Size of *temp %d", sizeof(*temp));
    printf("Size of temp %d", sizeof(temp));
    return 0;
}

I get:

Size of **temp 1
Size of *temp 8
Size of temp 8

What I don’t understand is how does a char pointer have a size of 8? Is it machine independent?

Answer 1

In the original question you weren’t calling sizeof.
duskwuff fixed that for you.

The output produced was:

Size of **temp 1
Size of  *temp 8
Size of   temp 8

Reason:

On a 64-bit architecture, pointers are 8-bytes (regardless of what they point to)

 **temp is of type char ==> 1 byte
  *temp is of type pointer-to-char ==> 8 bytes
   temp is of type pointer-to-pointer-to-char ==> 8 bytes