[c] What are the rules for casting pointers in C?

K&R doesn't go over it, but they use it. I tried seeing how it'd work by writing an example program, but it didn't go so well:

#include <stdio.h> 
int bleh (int *); 

int main(){
    char c = '5'; 
    char *d = &c;

    bleh((int *)d); 
    return 0;  
}

int bleh(int *n){
    printf("%d bleh\n", *n); 
    return *n; 
}

It compiles, but my print statement spits out garbage variables (they're different every time I call the program). Any ideas?

I suspect you need a more general answer:

There are no rules on casting pointers in C! The language lets you cast any pointer to any other pointer without comment.

But the thing is: There is no data conversion or whatever done! Its solely your own responsibilty that the system does not misinterpret the data after the cast - which would generally be the case, leading to runtime error.

So when casting its totally up to you to take care that if data is used from a casted pointer the data is compatible!

C is optimized for performance, so it lacks runtime reflexivity of pointers/references. But that has a price - you as a programmer have to take better care of what you are doing. You have to know on your self if what you want to do is "legal"

When thinking about pointers, it helps to draw diagrams. A pointer is an arrow that points to an address in memory, with a label indicating the type of the value. The address indicates where to look and the type indicates what to take. Casting the pointer changes the label on the arrow but not where the arrow points.

d in main is a pointer to c which is of type char. A char is one byte of memory, so when d is dereferenced, you get the value in that one byte of memory. In the diagram below, each cell represents one byte.

-+----+----+----+----+----+----+-
 |    | c  |    |    |    |    | 
-+----+----+----+----+----+----+-
       ^~~~
       | char
       d

When you cast d to int*, you're saying that d really points to an int value. On most systems today, an int occupies 4 bytes.

-+----+----+----+----+----+----+-
 |    | c  | ?1 | ?2 | ?3 |    | 
-+----+----+----+----+----+----+-
       ^~~~~~~~~~~~~~~~~~~
       | int
       (int*)d

When you dereference (int*)d, you get a value that is determined from these four bytes of memory. The value you get depends on what is in these cells marked ?, and on how an int is represented in memory.

A PC is little-endian, which means that the value of an int is calculated this way (assuming that it spans 4 bytes): * ((int*)d) == c + ?1 * 28 + ?2 * 2¹6 + ?3 * 2²4. So you'll see that while the value is garbage, if you print in in hexadecimal (printf("%x\n", *n)), the last two digits will always be 35 (that's the value of the character '5').

Some other systems are big-endian and arrange the bytes in the other direction: * ((int*)d) == c * 2²4 + ?1 * 2¹6 + ?2 * 28 + ?3. On these systems, you'd find that the value always starts with 35 when printed in hexadecimal. Some systems have a size of int that's different from 4 bytes. A rare few systems arrange int in different ways but you're extremely unlikely to encounter them.

Depending on your compiler and operating system, you may find that the value is different every time you run the program, or that it's always the same but changes when you make even minor tweaks to the source code.

On some systems, an int value must be stored in an address that's a multiple of 4 (or 2, or 8). This is called an alignment requirement. Depending on whether the address of c happens to be properly aligned or not, the program may crash.

In contrast with your program, here's what happens when you have an int value and take a pointer to it.

int x = 42;
int *p = &x;

-+----+----+----+----+----+----+-
 |    |         x         |    | 
-+----+----+----+----+----+----+-
       ^~~~~~~~~~~~~~~~~~~
       | int
       p

The pointer p points to an int value. The label on the arrow correctly describes what's in the memory cell, so there are no surprises when dereferencing it.

You have a pointer to a char. So as your system knows, on that memory address there is a char value on sizeof(char) space. When you cast it up to int*, you will work with data of sizeof(int), so you will print your char and some memory-garbage after it as an integer.

Casting pointers is usually invalid in C. There are several reasons:

1. Alignment. It's possible that, due to alignment considerations, the destination pointer type is not able to represent the value of the source pointer type. For example, if int * were inherently 4-byte aligned, casting char * to int * would lose the lower bits.

2. Aliasing. In general it's forbidden to access an object except via an lvalue of the correct type for the object. There are some exceptions, but unless you understand them very well you don't want to do it. Note that aliasing is only a problem if you actually dereference the pointer (apply the * or -> operators to it, or pass it to a function that will dereference it).

The main notable cases where casting pointers is okay are:

1. When the destination pointer type points to character type. Pointers to character types are guaranteed to be able to represent any pointer to any type, and successfully round-trip it back to the original type if desired. Pointer to void (void *) is exactly the same as a pointer to a character type except that you're not allowed to dereference it or do arithmetic on it, and it automatically converts to and from other pointer types without needing a cast, so pointers to void are usually preferable over pointers to character types for this purpose.

2. When the destination pointer type is a pointer to structure type whose members exactly match the initial members of the originally-pointed-to structure type. This is useful for various object-oriented programming techniques in C.

Some other obscure cases are technically okay in terms of the language requirements, but problematic and best avoided.

char c = '5'

A char (1 byte) is allocated on stack at address 0x12345678.

char *d = &c;

You obtain the address of c and store it in d, so d = 0x12345678.

int *e = (int*)d;

You force the compiler to assume that 0x12345678 points to an int, but an int is not just one byte (sizeof(char) != sizeof(int)). It may be 4 or 8 bytes according to the architecture or even other values.

So when you print the value of the pointer, the integer is considered by taking the first byte (that was c) and other consecutive bytes which are on stack and that are just garbage for your intent.