Playing with pointers, dazed and confused by char * strings

Question

So i was playing with pointers to understand different use cases, I'm not so experienced at this and unable to wrap my head around some ideas. I'll mark the lines where i have a problem. Please help me understand pointers.

on line 6:

Why can't I refer a string as an array, since that's how it's stored in memory i think. How to access individual characters of a string?

on line 7-8:

why is there a difference of 8 bytes between &str and &str + 1 when int datatype takes 4-bytes

on line 9,10 and 12:

Why does using *str to refer to the string take int as input? and why does it cast a single byte char to int instead of casting 4 bytes?

char *str = "Ninechars";
printf("Start\n");
printf("%s\n", str);
printf("%p\n", str);    // address of first char of str
printf("%p\n", str+1);  // address of second char of str
// printf("%s\n", str[1]);  // [segfault] warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [TODO]
printf("%p\n", &str);   // address to head of the string? NO, address of the pointer var str
printf("%p\n", &str+1); // [TODO], difference of 8bytes b/w this and previous address value? [WHY]?
printf("%d\n", *str);   // what integer is this? -> whatever can be made from 'N'=78
printf("%d\n", *(str+1));   // 'i'=105, but why is it taking single byte ints [TODO]
printf("%ld\n", sizeof(1)); // 4 bytes, as expected
// printf("%s\n", *str);    // [?]Wrong, [HOW][WHY], format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’[TODO]
printf("End\n");

Output:
Start
Ninechars
0x40088f
0x400890
0x7fff95a594e8
0x7fff95a594f0
78
105
4
End

The full code file is here: http://ideone.com/ryag80

Answer 1

Let's examine it line by line:

---

char *str = "Ninechars";
printf("Start\n");
printf("%s\n", str);
printf("%p\n", str);    // address of first char of str
printf("%p\n", str+1);  // address of second char of str

The above addresses are part of the data segment of process' virtual memory.

---

printf("%s\n", str[1]);

Actually, str[1] == *(str + 1) . What %s expects though is a char * , thus (str + 1) or &str[1] .

---

printf("%p\n", &str);   
printf("%p\n", &str+1);

&str is the address of str , a pointer to char . It's on the stack.

The difference of 8 bytes means that, on your system , the size of a pointer to char is 8 bytes. Try printf("\\nsizeof(char*) == %zd\\n", sizeof(char*)) to verify it.

---

printf("%d\n", *str);

It's ASCII code for N .

---

printf("%d\n", *(str+1));

Indeed 'i' == 105 ( ASCII code again), and you instructed printf() to print it as an int using %d .

---

printf("%ld\n", sizeof(1)); // 4 bytes, as expected
printf("%s\n", *str);    // [?]Wrong, [HOW][WHY], format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’[TODO]

Again, using %s , printf() expects a string ( char * in C), but *str is actually a char (an 8-byte int ).