Question: about C pointer variable

[scottgai]

I was confused about such a fact:
when I declare a array of char type like the following:
char name[30]="scott";
then I have a printf statement:
printf("Addr: %x\n&Addr: %x\n", name, &name);
I got the same value of these two address.
who can tell me the reason about this?
Thanks in advance

 

[Zyrenthian]

Replace the first %x with a %s and remove the & on the second name. %x specifies hex output, %s is string.

Matt

 

[Zyrenthian]

oh... disregard that, i misread it. I dont know why the pointer to a memory location is equal to a pointer to a pointer to a memory location. I would assume it has something to do with the internals of the compiler.

Matt

 

[ankan]

There are no 2 addresses. Its the same memory location where the array is stored (to be precise, the 1st element of the array is stored)... name and &name point to the same memeory location, and hence the same value. Ankan.

Please do correct me if I am wrong. s-)

 

[scottgai]

I agree with what Ankan's said. "name" is the variable name of an array. So "&name" is valued as the address of this variable, it's equivalent to "&name[0]". Things would be different if the variable is defined as:

char * name = "scott"

This time "name" is the address of the first element of "scott" and "&name" is the address of pointer variable "name".
Thanks Ankan and Matt a lot!