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[1];

it is natural to think that there is only one index, the constant value 1. This results from C allowing you to treat pointers and arrays as equivalent. LLVM doesn't. In this example, Foo is a pointer. That pointer must be indexed. To arrive at the same address location as the C code, you would provide the GEP instruction with two indices. The first indexes through the pointer, the second index the second element of the array.

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 without any need for accessing memory. It must, therefore be indexed like any other 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;
  }
  ...
  complex 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 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):

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 ints 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.
Appendix: Discussion

The following is a real discussion from the #llvm IRC channel about the GEP instruction. You may find this instructive as it was the basis for this document.

UserComment
YorionIf x & y must alias, are [ getelementptr x,0,0,1,2 ] and [ getelementptr x,1,2 ] aliased? (they obviously have different types, but they should alias...)
Yorionoops, for the second one I meant [ getelementptr y,1,2 ]
ReidI don't see how that could be, Yorion but I'm not the authority on this
Yorionhmm..
Reidthe two geps, by definition, are going to produce different pointers which are not aliased
Yorionwould [ GEP x,1,0 ] and [ GEP y,1 ] be aliased?
Reidif the second gep was [gep y,0,0,1,2] then they should be aliased as well
Reidno, I wouldn't expect that to work either :)
Reidyou can't just arbitrarily drop leading or trailing indices :)
Reid(.. leading or trailing 0 indices, I mean)
Reidthis instruction walks through a data structure and generates a pointer to the resulting thing
Reidif the number of indices are different, you're ending up at a different place and by definition they'll have different addresses
Yorionoh, I see, because of different types, [ GEP x,0,1 ] & [ GEP x,1 ] actually might refer to different fields, but might also refer to the same ones...
Reidor, at least, that's my crude understanding of it :)
Reidno, they'll definitely refer to different fields
nicholasGEP x,0,1 ==> &((*(x+0))+1)? vs. GEP x,1 ==> &(*(x+1))?
Reidlemme grok that for a sec
Reidthat might be true in some limited definition of x, but it wouldn't be generally
nicholasoh. fields of different sizes in a structure.
Reidyup
Yorionis perhaps the type unification the reason why [ GEP x,0,1 ] and [ GEP x,1 ] cannot alias?
Reidno
Reidthey may or may not have the same type, but they are definitely different pointers
Reidlets use a concrete example for "x"
Reidsuppose x is "struct {int a, float b} *"
ReidGEP X,0,1 is going to return the address of b
ReidGEP X,1 is going to return the address of the *second* "a" (after the first b)
Yorionah, I see...
Yoriontrailing zeros are still a bit confusing...
Reidsame thing .. you're just selecting the 0th member of an array or structure
Yorionyou don't move away from the pointer, only the type is changed
Reidno, you still move away from the pointer .. the type might change, or not
Reidthe pointer definitely changes
Reidlets look at an example for trailing zero
Reidsuppose x is "int x[10][10][10][10]" (in C)
ReidGEP X,0,0 will yield you a 3 dimensional array
ReidGEP X,0,0,0,0,0 will yield you an "int"
Reidmake sense?
Yorionyes
Reidso, I think there's a law here: if the number of indices in two GEP instructions are not equivalent, there is no way the resulting pointers can alias
Reid(assuming the x and y alias)
YorionI was confused with some code in BasicAliasAnalysis that says that two pointers are equal if they differ only in trailing zeros
YorionBasicAliasAnalysis.cpp:504-518
Reidlemme look
nicholasif y1 = GEP X, 0, 0 and y2 = GEP X, 0, 0, 0, 0, 0 (from Reid's example)
nicholasthen doesn't *y1 and *y2 both refer to the same "int"?
Reidthey shouldn't
Reidhmm .. actually, maybe you're right :)
Reidthey definitely have different *types*
Yoriontrue
nicholasdifferent types just doesn't cut it. :)
Reid.. thinking on this :)
nicholassimilarly, i could create a yucky with a struct that has a char *, then have you GEP right through the pointer into the pointed-to data. That could mean that the resulting point might alias anything.
Yorionmy theory (after reading BAA) is that all zeros can be omitted, and that the pointers alias if they have the same sequence of indices
Yorionhowever, this screws the typing, so that's why zeros are for
Yorionnicholas, does that match your hunch?
nicholasI have to admit, I've had much grief with GEPIs already. I wish the semantics were plainly documented as part of their own language, instead of just relying on C aliasing rules and C semantics...
nicholasYorion: leading zeroes can't be omitted.
Reidokay, if you have two GEPs and their leading indices are an exact match, if the one with more indices only has trailing 0s then they should alias
nicholasmust alias, i think.
Reidyes, must alias, sorry
Yorionokay
YorionI'm glad we cleared this up
Reidso, if y1 = GEP X, 1,2,0 and if y2 = GEP X, 1,2,0,0,0 then y1 "must alias" y2 :)
Reidbut that doesn't work for leading 0s :)
Yorionyes, true... I was wrong
ReidI too have been having fun with GEP recently :)
Yorionbut, there're cases like [a = GEP x,1,0; b = GEP a,1,0; c = GEP b,1,0], and that should be equivalent to GEP x,1,0,1,0,1
Reidnot quite
nicholasI'm sure another rule can be written for GEPIs, but they would only apply to type-safe code.
nicholasanother *tautology
YorionReid: why not, only the type should be different...
Reidits should be equivalent to GEP x,1,0,1,0,1,0
Yorionand that must alias GEP x,1,0,1,0,1
Reidhmm, by the previous rule, I guess you're right :)
YorionI'm a bit scared that even you're a bit confused about GEP
ReidI'm glad I'm not the only one that gets a little confused wrapping my head around this stuff :)
ReidGEP has always confused me .. partly because I think its wrong :)
Reidwell, actually, not so much that GEP is wrong, but that gvars being pointers without storage
Reidi.e. when you say "%x = global int" in LLVM, the type of X is int*
Reidyet, there is no storage for that pointer
Reidits magically deduced
nicholaswell, it makes no sense to have globals be SSA...
Reidheh
Reidyeah, well .. practicalities :)
Yoriontrue
Yorionsabre gave me a reference on how globals are handled in SSA
Reidanyway, gotta run
Yorionthe paper was crappy, but I do understand now why is it implemented that way in LLVM
Yorionthx for the interesting discussion Reid
Reidheh .. its one that Chris and I keep having .. he just tells me that C has rotted my brain :)
nicholaslol
Yorionhehehe
Reidhe might be right :)
Yorionsabre: have you seen the discussion on GEP?
sabreno
sabreI'll read the backlog, j/s
sabreok, there's a lot
sabrewhat's the executive summary?
sabredo you have a q?
Yorionis it possible that GEP x,0,0,1 and GEP x,1 alias?
sabreno
Yorionand b) GEP x,1,0,0 and GEP x,1 should alias, right?
sabreI assume you mean for size = 1 ?
sabreb) yes
Yorionalthough they have different types
sabreright
Yorionokay
YorionI'm still not 100% convinced that: a=GEP x,1,0; b=GEP a,1,0; c=GEP b,1,0 cannot alias Z=GEP x,1,1,1
Yorion(that c and z cannot alias)
sabrethat's becuse they do alias
sabremustalias in fact
Yorionbut then: GEP x,1,0,1,0,1,0 = GEP x,1,1,1
sabreYorion: no
sabrec != GEP x,1,0,1,0,1,0
sabrethe first index doesn't work like that
Yorionhow does then the first index work? c and z must alias, but GEP x,1,0,1,0 != GEP x,1,1 ??
sabre*sigh*
Reid:)
Reidwe need an FAQ on this
sabreYorion: how did you get
sabre"GEP x,1,0,1,0"?
Yorionlook
sabreyou can't just concatenate subscripts
Yorionwhy?
sabrebecause... it doesn't work that way?
sabreconsider C
Yorionhow does it work?
sabreif I have blah* P
sabreP[0][1][2][3][4]
sabrethis is *not* the same as:
sabret = &P[0][1][2] ... t[3][4]
sabreYorion: Consider: struct *P
sabreP->X == P[0].X
sabreYou're losing the 0.
sabreP->X->Y == P[0].X[0].Y
sabreNot P.X.Y
sabreactually that's a bad analogy
sabrebecause C dereferences in this case
sabreTry: (&(P->X))->Y
Yorionso, a=GEP x,1,0; b=GEP a,1,0; c=GEP b,1,0, can you construct the definition of c in terms of x?
sabreyes, but you're going out of bounds :)
sabreconsider this:
sabre{ float, { double , { int } } } *P
sabreint *X = gep P, 0, 1, 1, 0
sabredo you understand the leading zero?
sabrealternatively:
sabret = gep P, 0, 1
sabret2 = gep t, 0, 1
sabreX = gep t, 0, 0
Yorionwhat's t2 for?
sabreoh
sabresorry :)
sabreX = gep t2, 0, 0
Yoriongive me a minute please
sabreok
Yorionsabre: shouldn't the type be: { float, { double, { int }* } }* P ?
sabrenope
sabrewhy the extra * ?
sabreif it helps, the type of t is { double, {int}}* and t2 is {int}* and X is int*
Yorionwait... 0 represents dereference, natural number i represents &A[i] ?
sabregep does no dereferences, ever
sabregep P, 0, 1 is equivalent to &P[0].X
sabreaka &P->X
sabregep P, 1 == &P[1] aka P+1
sabreso gep P, 0, 1 can't alias gep P, 1 just like &P->Y can't alias P+1
sabreassuming P is a pointer to struct {X, Y }
Yorionsabre: is it possible to come up with a general rule for "arithmetic of GEP indices"?
sabreYorion: of course, it's very simple
sabrejust not what you're expecting :)
sabreSee langref.html
Yorionfor example, a=GEP x,0,0,1 b=GEP a,0,0,1, c=GEP b,0,0,1, that should be equal to GEP x,0,1,1,0, right?
YorionI did
Yorionoops, equal to GEP x,0,1,1,1,0
sabrethat would be:
sabreGEP X, 0, 0, 1, 0, 1, 0, 1
Yorionyou're killing me
sabreThe basic rule when turning: A = GEP B, C D = GEP A, 0, E
sabreis that you drop the 0, turning it into
sabreGEP B, C, E
Yorionokay, that's good. any other rules?
nicholaswhat if it isn't a 0?
sabremore generally: A = GEP Ptr, B, C, ... D = GEP A, 0, E, F, ...
sabreD = GEP Ptr, B, C, ... E, F, ...
sabreif it's not zero, you generally cannot concatenate them
sabreunless the first gep has one subscript
sabrein which case you drop the zero
sabreif you look in InstCombiner::visitGetElementPtrInst, it should have this logic
Yorionwhat if there is no zero? how can I compute the offset from the base pointer in that case?
Yorionlike A=GEP B,C and D=GEP A,E,F
sabreyou don't have to combine them to compute an offset
sabreare you *just* trying to get a byte offset from the pointer?
YorionI'm trying to get offset of D from B
sabrewhy didn't you say so? :)
sabrewith all constant subscripts, it's trivial
sabrelook at SelectionDAGLowering::visitGetElementPtr
sabrein CodeGen/SelectionDAG/SelectionDAGISel.cpp
sabrebasically the rule is that you multiply the index by the size of the thing indexed
sabrethere is also a Support/GetElementPtrIterator or something
sabrethat makes it trivial to see what type is indexed by which subscript
sabrefor each subscript it gives you a type
sabreFor an array subscript you multiply the index by the indexed type
sabrefor a struct subscript, you add the field offset
sabres/array/sequentialtype/ if you're in a pedantic mood
Yorionlet's focus on structs: in that case, the above given example would be: D = GEP B,C,E,F?
sabreno
sabreyou drop the E if it's zero
sabreif it's not you can't concat
sabreare you trying to trick me into saying "yes, just append the indices"? :)
Yorionokay, let's assume E is not zero, how do I compute offset from B for D for a struct?
sabreWhy are you framing this in terms of concatenation?
Yorionno, I'm trying to understand it
sabrecomputing an offset and concatenating are entirely different
sabreLets consider a specific example
Yorionbecause I want to express certain properties in the terms of base pointers either globals or parameters
YorionI want to eliminate locals from my analysis
sabreyou realize that parmeters can point into the middle of structs?
Yorionyes
sabreyou realize invalid access paths can be constructed with geps/
sabre?
Yorionwhat do you mean by invalid access paths?
Yorionlike offseting out of the struct which is passed to the function?
sabreThe case where the subscript isn't zero is invalid code
sabrefrom a type-safety perspective
DannyBhe means untypesafe things that seem valid :)
DannyBIE they point somewhere in the struct, but not to any particular field
DannyB(or whatever)
sabreright
Yorionokay
sabreor they might point in some other struct :)
sabreIt's the equivalent to saying:
sabrestruct Foo { int A, int B; }
sabreFoo* P =
sabreT = &P->B;
sabreS = T+1
sabrethat is:
sabreT = gep 0, 1
sabreS = gep T, 1
sabreyou can't concat those and get a type-safe access path
sabreremember T = &P->B === T = &P[0].B
sabreunderstand?
Yorionlet me think for a minute
sabreConsider what the C case does, it will be most clear if you're used to C
sabre:)
sabreConsider the byte offset and why it doesn't point into P-> anything
sabreit points into P[1] not P[0]
Yorionit's perfectly fine if GEP offsets out of the type. I'd still need to express GEP in the terms of base pointers. Take the example above: T=GEP P,0,1; S=GEP T,1
Yoriontype safety is not crucial in my case
sabreThat specific example is GEP P, 1, 0
sabrehowever, you can form geps that are NOT equivalent to anything else
sabrefor example, consider:
sabrestruct X { int, char}
Yorionthat is fine. they're equivalent to something in the calling context
sabrethe same sequence points into padding
sabreand there is no gep that can do that
Yorionthe bottom line is: if the program is valid, interprocedural analysis will match that offset with something in one of the functions on the call stack
Yorionand that's all I care about
Yorioncan you give me a formula for structs for computing offsets that takes into account the case GEP T,&lt:non_zeros> and the size of the structs/fields?
sabreyes, I did above
sabreTwo things:
sabreGEP Ptr, A, X, Y, Z
sabreThe result is Ptr + A * sizeof(struct) + fieldoffs(X) + fieldoffs(Y) + fieldoffs(Z)
sabresimple enough?
sabreyou see why "A" is special?
Yoriongive me a min, I'm constructing an example
Reidsabre: I think I finally understand
Reidyour comment that GEP *never* dereferences makes a lot of sense
Reidit is only doing address calculation, so the first one is taking the address of the var
sabreIf C didn't conflate lvalues and rvalues, GEP would be much easier to understand for people
Reidyeah
Yorionso, for example: M=GEP A,B,C; N=GEP M,D,E; N = [ A + B*sizeof(struct) + fieldoffs(C) ]:(of type T) + D*sizeof(T) + fieldoffs(E)
ReidI just remember learning a hard lesson about the difference between char* A and char A[] .. long time ago when I was learning C
sabreof type T*
sabreotherwise fine
Yorionokay, I think I finally understand it
sabrewithout the T* your D sizeof will be wrong
Yoriona suggestion: the formula you gave above explains it all
YorionI'd suggest explaining it that way in documentation
sabreThat's not right though
sabreit doesn't include arrays or packed types
sabreso it is, at best, a half truth
Yoriontell me, how to compute the fieldoffs for an index?
sabrearrays can be in structs :)
Yorionin bytes
sabreidx * sizeof(arrayelt)
sabrejust like for pointers (the first index)
sabreThere are two cases: structs and sequentials
sabrefor sequentials you use idx*sizeof(sequenced type)
sabrefor structs you add their offset
sabreit's really very simple :)
sabrethe first index of a gep is always over the pointer
Yorionno I meant in LLVM, how do I convert the field offset into bytes?
sabrewhich is why it's strange
sabreif you only think about structs
sabreTargetData::getFieldOffset
sabreor something
sabrelook in SelectionDAGISel.cpp (visitGEP) as I suggested.
Yoriondo you still have the energy to go over sequential types? :-)
Yorionwhat is the offset formula for sequential types?
Reidwe just went over that: idx * sizeof(elementType)
Yorionso, if there's an array hidden somewhere in the struct, essentially I need to compute idx*sizeof() instead of fieldoffs() and that's it?
sabreyes
Reidyes
Yorionexcellent.
sabreThere are two cases: structs and sequentials
sabre[9:17pm] sabre: for sequentials you use idx*sizeof(sequenced type)
sabre[9:17pm] sabre: for structs you add their offset
sabre[9:17pm] sabre: it's really very simple :)
Yorionnow when I understand it, it is simple...
Yorionthx

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