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I have this simple program:

#include <stdio.h>

struct time
{
    int hours ;
    int minutes ;
    int seconds ;
} t[2] = {1,2,3,4,5,6};

int main() {    
    struct time *tt = t;
    printf("%d\n", (*tt));
    printf("%d", *(tt+1));
}

Now the expected output should be:

1
4

But the output I am getting is 2 exactly same memory addresses. I sent this program to other people and some are having the expected output but some aren't. I reckon this is an issue with version of C or GCC. Is it resolvable ? Is there any explanation as to why this is happening in my version of C?

The most weird thing is that even after I have dereferenced the pointer, it still printing some address. How is this possible?

On the other hand if I change the program to this:

$include <stdio.h>

struct time
{
    int hours ;
    int minutes ;
    int seconds ;
} t[2] = {1,2,3,4,5,6};

int main() {    
    struct time *tt = t;
    printf("%d\n", (*tt).hours);
    printf("%d", (*(tt+1)).hours);
}

It prints the expected output.

What is the explanation for this behaviour? Thanks.

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1

3 Answers 3

Reset to default
2

The problem is not in the compiler but your understanding of C. Let's compile with GCC version 7.3.0 (Ubuntu 7.3.0-27ubuntu1~18.04):

% gcc struct_time.c      
struct_time.c: In function ‘main’:
struct_time.c:12:14: warning: format ‘%d’ expects argument of type ‘int’, 
          but argument 2 has type ‘struct time’ [-Wformat=]
     printf("%d\n", (*tt));
             ~^     ~~~~~
struct_time.c:13:14: warning: format ‘%d’ expects argument of type ‘int’,
          but argument 2 has type ‘struct time’ [-Wformat=]
     printf("%d", *(tt+1));
             ~^   ~~~~~~~

Your format is %d which expects an int as the corresponding argument, and you're passing in struct time. The behaviour of the program is undefined, because the conversion specifier does not match the type of the actual argument.

It doesn't matter where the values reside in memory - their type is also important. Even though the address of tt[0] is the same as tt[0].hours, they have a distinct type. In one you want the whole struct, in another you want the value contained in the first member.

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5 Comments

Ohhh the last sentence answers one of my question! But why do types matter? I mean if we are just dereferencing an integer location, shouldnt we get an integer in return?
@MohammadAreebSiddiqui: Supposing that tt is given the address 0x100, then, starting at 0x100, there is both an int (the first thing in the structure) and struct time. They are both at the same location. If you wrote *0x100, how would the compiler know whether you meant just the int or the entire struct time? The way C handles this is to associate types with expressions. The type of tt is “array of 2 struct time”. When used in *tt, it is automatically converted to “pointer to struct time”, and then **tt has the type struct time. So it is a struct time, not an int.
@MohammadAreebSiddiqui: To refer to the int, you must use a C expression that refers to the member of the structure, not to the structure, such as tt[0].hours or (*tt).hours.
You cannot be referencing an integer location because there is no such thing as an integer location. A given location, assuming proper alignment, can be any type. How is printf() supposed to know how much data to read at that location? The only type information you send it is in the conversion specifier. That's why it's important to use the right one.
@EricPostpischil I tried doing **tt but its giving me an error saying *tt is already a struct time type. This tells me that the compiler is assuming *tt to be of struct time type. Also, when i changed the type of tt to be of int* and then printed *tt then it printed 1 which is the expected behaviour.
1

Your code is invalid. Here's the error your compiler should be producing (maybe you are not enabling warnings?):

error: format specifies type 'int' but the argument has type 'struct time' [-Werror,-Wformat]
    printf("%d\n", (*tt));

You simply cannot write code like this and then complain about its behavior. You're telling printf() you're passing an integer but then you pass a struct. Anything could happen.

For bonus points, you have a style problem which a reasonable compiler will also warn about:

error: suggest braces around initialization of subobject [-Werror,-Wmissing-braces]
} t[2] = {1,2,3,4,5,6};

That's because you're initializing 2 structs with a single list of numbers. It should be {{1,2,3},{4,5,6}}.

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3 Comments

When I print tt and &(tt->hours), I get the same address which is the location of an integer, so deferencing that location should give an integer value right?
No. Not right. It should not give an integer value. That memory address is not "the location of an integer." It is just a memory address. There is no type associated with it. You are the one who put an int at the beginning of the struct
@MohammadAreebSiddiqui: The address of the first member of the struct is the same as the address of the struct. However, printf() is a "variadic function" which receives its arguments via a special mechanism which depends on their types. That is, passing an int to printf() has a different effect compared with passing a struct. So telling printf() you've passed an int when you have actually passed a struct (not the address of a struct, but the actual struct value) is invalid.
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It behaves works ok because the array is used as a pointer (which is 100% ok)

so the tt points to the first array element (this with index 0) and tt + 1 to the second (index 1).

When you add one to this pointer (and this pointer is of the type pointer to the struct) the resulting pointer references the next structure of this type.

If you dereference pointers to the structs or unions use ->

(*tt).hours == tt -> hours and (tt + 1).hours == (tt + 1) -> hours

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