pointer

Question

int main(){
    int a[5]={1,2,3,4,5};
    char *str="hello";
    printf("%p %p",a,&a);
    printf("%p %p",str,&str);
}

Why in $1$st printf , both the outputs are same($a$,&$a$)

And in $2$nd printf ,both the outputs are different(str,&str)

please help!

Comments

because array name is itself address of the array i.e it is a pointer,so a and &a is same.

wheras str is not a character array,it prints the address of the string i.e 'hello' wheras &str givess the address of the 'str' which is a character pointer.hence they are diff

Answer 1

• $a$ is an array .
• $a$ decays to the address of the first element of a[] , i.e. a decays to a pointer of type int*. (pointer to an integer)
• &$a$ is also a pointer having the same numerical value of &$a$[$0$], but &$a$ has different type , int (*) [5]
• This difference comes into picture when we do the pointer arithmetic as shown in the above image.

• str and string literal "hello" has different storage location in memory.
• str is a pointer of char* type and stores the starting address of "hello"
• But str has its own address ,i.e.  &str , type is pointer to a character pointer.
• Some pointer arithmetic is also shown in above image.