<pre>
AType* Foo;
...
- X = Foo[1];</pre>
+ X = &Foo->F;</pre>
<p>it is natural to think that there is only one index, the constant value
<tt>1</tt>. 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.</p>
+ pointer, the second index the element of the structure just as if it was:</p>
+ <pre>
+ X = &Foo[0].F;</pre>
<p>Sometimes this question gets rephrased as:</p>
<blockquote><i>Why is it okay to index through the first pointer, but
subsequent pointers won't be dereferenced?</i></blockquote>
<p>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:</p>
+ 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:</p>
<pre>
struct munger_struct {
int f1;