The Often Misunderstood GEP Instruction

1. Introduction
2. The Questions
   1. Why is the extra 0 index required?
   2. What is dereferenced by GEP?
   3. Why can you index through the first pointer but not subsequent ones?
   4. Why don't GEP x,0,0,1 and GEP x,1 alias?
   5. Why do GEP x,1,0,0 and GEP x,1 alias?
3. Summary

Written by: Reid Spencer.

Introduction

This document seeks to dispel the mystery and confusion surrounding LLVM's GetElementPtr (GEP) instruction. Questions about the wiley GEP instruction are probably the most frequently occuring questions once a developer gets down to coding with LLVM. Here we lay out the sources of confusion and show that the GEP instruction is really quite simple.

The Questions

When people are first confronted with the GEP instruction, they tend to relate it to known concepts from other programming paradigms, most notably C array indexing and field selection. However, GEP is a little different and this leads to the following questions, all of which are answered in the following sections.

1. What is the first index of the GEP instruction?
2. Why is the extra 0 index required?
3. What is dereferenced by GEP?
4. Why don't GEP x,0,0,1 and GEP x,1 alias?
5. Why do GEP x,1,0,0 and GEP x,1 alias?

What is the first index of the GEP instruction?

Quick answer: The index stepping through the first operand.

The confusion with the first index usually arises from thinking about the GetElementPtr instruction as if it was a C index operator. They aren't the same. For example, when we write, in "C":

      AType* Foo;
      ...
      X = &Foo->F;

it is natural to think that there is only one index, the selection of the field `F`. However, in this example, `Foo` is a pointer. That pointer must be indexed explicitly in LLVM. C, on the other hand, indexs through it transparently. To arrive at the same address location as the C code, you would provide the GEP instruction with two index operands. The first operand indexes through the pointer; the second operand indexes the field `F` of the structure, just as if you wrote:

      X = &Foo[0].F;

Sometimes this question gets rephrased as:

> Why is it okay to index through the first pointer, but subsequent pointers won't be dereferenced?

The answer is simply because memory does not have to be accessed to perform the computation. The first operand to the GEP instruction must be a value of a pointer type. The value of the pointer is provided directly to the GEP instruction as an operand without any need for accessing memory. It must, therefore be indexed and requires an index operand. Consider this example:

      struct munger_struct {
        int f1;
        int f2;
      };
      void munge(struct munger_struct *P)
      {
        P[0].f1 = P[1].f1 + P[2].f2;
      }
      ...
      munger_struct Array[3];
      ...
      munge(Array);

In this "C" example, the front end compiler (llvm-gcc) will generate three GEP instructions for the three indices through "P" in the assignment statement. The function argument `P` will be the first operand of each of these GEP instructions. The second operand indexes through that pointer. The third operand will be the field offset into the `struct munger_struct` type, for either the `f1` or `f2` field. So, in LLVM assembly the `munge` function looks like:

      void %munge(%struct.munger_struct* %P) {
      entry:
        %tmp = getelementptr %struct.munger_struct* %P, int 1, uint 0
        %tmp = load int* %tmp
        %tmp6 = getelementptr %struct.munger_struct* %P, int 2, uint 1
        %tmp7 = load int* %tmp6
        %tmp8 = add int %tmp7, %tmp
        %tmp9 = getelementptr %struct.munger_struct* %P, int 0, uint 0
        store int %tmp8, int* %tmp9
        ret void
      }

In each case the first operand is the pointer through which the GEP instruction starts. The same is true whether the first operand is an argument, allocated memory, or a global variable.

To make this clear, let's consider a more obtuse example:

      %MyVar = unintialized global int
      ...
      %idx1 = getelementptr int* %MyVar, long 0
      %idx2 = getelementptr int* %MyVar, long 1
      %idx3 = getelementptr int* %MyVar, long 2

These GEP instructions are simply making address computations from the base address of `MyVar`. They compute, as follows (using C syntax):

- idx1 = (char*) &MyVar + 0
- idx2 = (char*) &MyVar + 4
- idx3 = (char*) &MyVar + 8

Since the type `int` is known to be four bytes long, the indices 0, 1 and 2 translate into memory offsets of 0, 4, and 8, respectively. No memory is accessed to make these computations because the address of `%MyVar` is passed directly to the GEP instructions.

The obtuse part of this example is in the cases of `%idx2` and `%idx3`. They result in the computation of addresses that point to memory past the end of the `%MyVar` global, which is only one `int` long, not three `int`s long. While this is legal in LLVM, it is inadvisable because any load or store with the pointer that results from these GEP instructions would produce undefined results.

Why is the extra 0 index required?

Quick answer: there are no superfluous indices.

This question arises most often when the GEP instruction is applied to a global variable which is always a pointer type. For example, consider this:

      %MyStruct = uninitialized global { float*, int }
      ...
      %idx = getelementptr { float*, int }* %MyStruct, long 0, ubyte 1

The GEP above yields an `int*` by indexing the `int` typed field of the structure `%MyStruct`. When people first look at it, they wonder why the `long 0` index is needed. However, a closer inspection of how globals and GEPs work reveals the need. Becoming aware of the following facts will dispell the confusion:

1. The type of `%MyStruct` is not `{ float*, int }` but rather `{ float*, int }*`. That is, `%MyStruct` is a pointer to a structure containing a pointer to a `float` and an `int`.
2. Point #1 is evidenced by noticing the type of the first operand of the GEP instruction (`%MyStruct`) which is `{ float*, int }*`.
3. The first index, `long 0` is required to step over the global variable `%MyStruct`. Since the first argument to the GEP instruction must always be a value of pointer type, the first index steps through that pointer. A value of 0 means 0 elements offset from that pointer.
4. The second index, `ubyte 1` selects the second field of the structure (the `int`).

What is dereferenced by GEP?

Quick answer: nothing.

The GetElementPtr instruction dereferences nothing. That is, it doesn't access memory in any way. That's what the Load and Store instructions are for. GEP is only involved in the computation of addresses. For example, consider this:

      %MyVar = uninitialized global { [40 x int ]* }
      ...
      %idx = getelementptr { [40 x int]* }* %MyVar, long 0, ubyte 0, long 0, long 17

In this example, we have a global variable, `%MyVar` that is a pointer to a structure containing a pointer to an array of 40 ints. The GEP instruction seems to be accessing the 18th integer of the structure's array of ints. However, this is actually an illegal GEP instruction. It won't compile. The reason is that the pointer in the structure must be dereferenced in order to index into the array of 40 ints. Since the GEP instruction never accesses memory, it is illegal.

In order to access the 18th integer in the array, you would need to do the following:

      %idx = getelementptr { [40 x int]* }* %, long 0, ubyte 0
      %arr = load [40 x int]** %idx
      %idx = getelementptr [40 x int]* %arr, long 0, long 17

In this case, we have to load the pointer in the structure with a load instruction before we can index into the array. If the example was changed to:

      %MyVar = uninitialized global { [40 x int ] }
      ...
      %idx = getelementptr { [40 x int] }*, long 0, ubyte 0, long 17

then everything works fine. In this case, the structure does not contain a pointer and the GEP instruction can index through the global variable, into the first field of the structure and access the 18th `int` in the array there.

Why don't GEP x,0,0,1 and GEP x,1 alias?

Quick Answer: They compute different address locations.

If you look at the first indices in these GEP instructions you find that they are different (0 and 1), therefore the address computation diverges with that index. Consider this example:

      %MyVar = global { [10 x int ] }
      %idx1 = getlementptr { [10 x int ] }* %MyVar, long 0, ubyte 0, long 1
      %idx2 = getlementptr { [10 x int ] }* %MyVar, long 1

In this example, `idx1` computes the address of the second integer in the array that is in the structure in %MyVar, that is `MyVar+4`. The type of `idx1` is `int*`. However, `idx2` computes the address of the next structure after `%MyVar`. The type of `idx2` is `{ [10 x int] }*` and its value is equivalent to `MyVar + 40` because it indexes past the ten 4-byte integers in `MyVar`. Obviously, in such a situation, the pointers don't alias.

Why do GEP x,1,0,0 and GEP x,1 alias?

Quick Answer: They compute the same address location.

These two GEP instructions will compute the same address because indexing through the 0th element does not change the address. However, it does change the type. Consider this example:

      %MyVar = global { [10 x int ] }
      %idx1 = getlementptr { [10 x int ] }* %MyVar, long 1, ubyte 0, long 0
      %idx2 = getlementptr { [10 x int ] }* %MyVar, long 1

In this example, the value of `%idx1` is `%MyVar+40` and its type is `int*`. The value of `%idx2` is also `MyVar+40` but its type is `{ [10 x int] }*`.

Summary

In summary, here's some things to always remember about the GetElementPtr instruction:

1. The GEP instruction never accesses memory, it only provides pointer computations.
2. The first operand to the GEP instruction is always a pointer and it must be indexed.
3. There are no superfluous indices for the GEP instruction.
4. Trailing zero indices are superfluous for pointer aliasing, but not for the types of the pointers.
5. Leading zero indices are not superfluous for pointer aliasing nor the types of the pointers.