Reputation: 37
Pointer manipulation in C
I am new to C programming. While solving one of my class assignments, I came across the following code snippet. I did not understand what it does.
Can any one tell me what is the meaning of following C syntax,
((char *)0 +1) or ((int*)0 +1))
Upvotes: 2
Views: 596
Answers (3)
Reputation: 400109
The (char *) 0 part creates a pointer to character data, at address 0. This address is then incremented by one, triggering undefined behavior since pointers to address 0 (also known as NULL in C) cannot be used in pointer arithmetic. The second part does the same but for pointer to integer data.
If the compiler simply treats NULL as the address (which is common but, again, not required which is why this is undefined behavior) the resulting addresses, if viewed numerically, will not be the same, since pointer arithmetic in C is done in terms of the type being pointed at, and typically sizeof (int) > sizeof (char).
Upvotes: 6
Reputation: 162317
Can any one tell me what is the meaning of following C syntax,
((char *)0 +1) or ((int*)0 +1))
Nothing by the terms of the C standard, because it's not defined. This code invokes undefined behavior on part of the C compiler. Let me explain:
In C every pointer may either point to some object of the type the pointer dereferences to or it may be 0, which is then called a null pointer. Null pointers can not be used in →pointer arithmetic.
Note that the actual representation of a null pointer on the metal, i.e. the bits the variable has on the machine may be something different than all zeros. But on the C side of things the null pointer always compares equal to an integer of the value 0. Moreover null pointers of different types also compare equal by definition. However comparisons of non null pointers of different types invokes undefined behavior. Also you can cast any pointer to a void* pointer, and back. Also you can cast every pointer to an integer of type uintptr_t and back. But casting from a pointer to type A to a pointer of type B (where B is not void*) invokes undefined behavior.
The special function malloc is defined by the C language specification to return a void* pointer that can be cast to any pointer type, though. But say you use it to allocate some memory for an array of char and later you cast that to int this again invokes undefined behavior.
Now you may ask: "What is undefined behavior?". Well, it just means, that the language standard doesn't define it and an implementer may go about it in any way seen fit. On most plattforms writing something like ((char*)0 + 1) may do something naively expected (creating a pointer, pointing to address 1), but it may as well make the compiler build an artificial intelligence, that at first chases you down the street, then gains consciousness and finally takes over the world, turning humans into batteries. So be careful about what you do ;)
Upvotes: 1
Reputation: 1
In C you have to tell compiler which type you mean to use, this is called "casting". For example:
char *c; //define c as "char pointer" (pointer to char)
c = ((char *)0 + 1); //this casts "0 + 1" to "char pointer" type, in this example not strictly necessary but adds some clarification to code
Upvotes: -1