Difference between char * and char (*)[25]

Question

#include<stdio.h>

int main()
{
    char str[25] = "Hello World";
    printf("%s\n", &str+2);
    printf("%p\n",&str);
    printf("%p\n",&str+2);
    return 0;
}

The program above flashes an error for line 6 saying:

warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char (*)[25]’ [-Wformat=]

I wonder what's the difference between these two. And the output always shows two addresses having a difference of 32 bytes. I was expecting an output with a difference of 2 as I'm trying to print the address of str and address of str+2. Can you explain?

some insight: http://stackoverflow.com/questions/15177420/what-does-sizeofarray-return/15177499#15177499 — Karoly Horvath, Jun 19 '16 at 11:58

Sergio · Answer 1 · 2016-06-19T12:12:38.453

The difference for this two cases is quite simple. Assume that we have an array of 25 characters: char arr[25];. By convention expression arr evaluated to a pointer to first array element, i.e. arr is the same as &arr[0]. Note, that here unary operator & is applied to lvalue arr[0] which has type of char. And as result & gives an expression with type char * (i.e. pointer to char)

When we say &arr - the lvalue is arr, that is char [25] (i.e. array of 25 characters). And as result the whole expression has type char (*)[25] (i.e. pointer to array of 25 characters). So, we've got expression with completely another type, that is not compatible with char *.

Few notes about indexing. Difference between &arr + 2 and &arr is 50 bytes (0x32 in hexadecimal), because sizeof(arr) == 25.

score 1 · Answer 2 · answered Jun 19 '16 at 12:09

char * is a pointer to char while char (*)[25] is a pointer to an array of 25 char's so these types are different and the compiler rightfully issues a warning.

Moreover, since &str is the address of the str array and is of type char (*)[25], the value of &str+2 will be equal to:

&str + 2*sizeof(char (*)[25]) = &str + 2*25 = &str + 50 = &str + 0x32

Since a pointer is incremented according to the type it points to.

Weather Vane · Answer 3 · 2016-06-19T12:15:28.497

0

EDIT: after OP added a new line 6.

This is the correct syntax to print a string and pointer values. In all cases there was a & where it should not be, because a reference to the array identifier decays it to a pointer.

#include <stdio.h>

int main()
{
    char str[25] = "Hello World";
    printf("%s\n", str+2);
    printf("%p\n", (void*)str);
    printf("%p\n", (void*)(str+2));
    return 0;
}

Program output:

llo World
0018FF20
0018FF22

EDIT 2: This shows the effect of passing the pointer of the array. The compiler thinks you are addressing the third element of an array of strings each size 25. I have contrived to show how this undefined behaviour actually manifested.

#include <stdio.h>

int main()
{
    char str[25] = "Hello World";
    char abc[] = "abcdefghijklmnopqrstuvwxyz";
    printf("%s\n", &str+2);
    return 0;
}

Program output:

wxyz

edited Jun 19 '16 at 12:15

answered Jun 19 '16 at 11:32

Weather Vane

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36
56

The question asks for an explanation. But you'd better search for a dup... – Karoly Horvath Jun 19 '16 at 11:33
You want a syntax error explaining? I did that by giving the correct syntax. – Weather Vane Jun 19 '16 at 11:35
@WeatherVane : What was the problem with my program? I want to know the reason to why the compiler raised the error. Thanks for the correct answer though. – 7_R3X Jun 19 '16 at 11:48
1

You have no idea. That's fine. Just be honest. – Karoly Horvath Jun 19 '16 at 11:58
MSVC compiler does not give a warning or an error, but does not give a sensible result. – Weather Vane Jun 19 '16 at 11:58
@KarolyHorvath I have been trying to contrive some code that shows how `&str[25]` is treated as an array of that size, but did not yet make anything to demonstrate. – Weather Vane Jun 19 '16 at 12:02
@KarolyHorvath I have added a bit more. – Weather Vane Jun 19 '16 at 12:15
@WeatherVane, are you sure it is `printf("%s\n", &str+2);` and not `printf("%p\n", &str+2);`? [Cpp.sh](http://cpp.sh/6rhu) throws a warning. – abhishek_naik Jun 19 '16 at 12:19
@BatCoder OP asked for an explanation of the warning from line 6 `printf("%s\n", &str+2);` that is using the `%s` format, and also asked why the other two lines don't give addresses that differ by 2. Although it is the same error in all three lines. – Weather Vane Jun 19 '16 at 12:21

score 0 · Accepted Answer · answered Jun 19 '16 at 12:15

After a little research I came up with this explanation. &str+2 is treated as a pointer to an array. char (*)[25] says that it's a pointer to a character array of size 25. To find the explain to why the output shows 32 byte difference I ran the program again and notices the output was displayed in hex ie

0x7ffd865b2710
0x7ffd865b2742

When converted to base 10, it becomes a difference of 50 bytes which is a pointer to the array that lies at second place to the current array(remember &str+2). TO make it more clear here's a little memory diagram.

&str   -----> current array ie str. array is 25 bytes long.
&str+1 -----> array lies after 1 position to current array. This address is 25 bytes away from &str.
&str+2-----> array lies after 2 position to current array. This address is 50 bytes away from &str.

Difference between char * and char (*)[25]

4 Answers4