指向char的指针的地址与指向char的指针的指针之间的区别

Question

I have following sample of code:

#include <stdio.h>

int main(void)
{
   char *text[2];
   text[0] = "Hello";
   text[1] = "World";

   printf("Address of text[0]: %p\n", text[0]);
   printf("Address of text   : %p\n", text);

   return 0;
}

The output of this program is:

Address of text[0]: 0x400694
Address of text   : 0x7ffcac41b000

I am wondering why these addresses differ in length. Why is address of first pointer to char only 6 digits length?

Answer 1

There is no difference in "address length", the difference is in the value .

In this particular case, I assume the text[0] pointer points to a data segment that holds the compile time value "Hello" , and the text pointer points to a runtime address.

Since these are at different memory segments, they can be "very far" and so you get the output you see.

Answer 2

The first address is the address of "Hello", which is stored in your data segment.

The second address is the address of text[] , which is stored in your stack.

These two areas of memory are far from each other, so one has many digits, the other has few digits.

Apparently, %p renders only as many digits as necessary to represent the address.

Answer 3

printf("Address of text[0]: %p\n", text[0]);

prints the address of C string (the address first element of array points to), while:

printf("Address of text   : %p\n", text);

prints the address of array's first element.

Answer 4

To get the address of the first item of the array text , you'll have to use:

printf("Address of text[0]: %p\n", (void *)&text[0]);

In your question, you are printing the address of the first character of the string "Hello" (in other words, of the content of text[0]), not the address of the first element of the array text , ie not the address of text[0] , just the address stored in text[0] .