Array-syntax vs pointer-syntax and code generation

Question

In the book, "Understanding and Using C Pointers" by Richard Reese it says on page 85,

int vector[5] = {1, 2, 3, 4, 5};

The code generated by vector[i] is different from the code generated by *(vector+i) . The notation vector[i] generates machine code that starts at location vector , moves i positions from this location, and uses its content. The notation *(vector+i) generates machine code that starts at location vector , adds i to the address, and then uses the contents at that address. While the result is the same, the generated machine code is different. This difference is rarely of significance to most programmers.

You can see the excerpt here. What does this passage mean? In what context would any compiler generate different code for those two? Is there a difference between "move" from base, and "add" to base? I was unable to get this to work on GCC -- generating different machine code.

Answer 1

The quote is just wrong. Pretty tragic that such garbage is still published in this decade. In fact, the C Standard defines x[y] as *(x+y).

The part about lvalues later on the page is completely and utterly wrong too.

IMHO, the best way to use this book is to put it in a recycling bin or burn it.

Answer 2

I think what the original text may be referring to is some optimizations which some compiler may or may not perform.

Example:

for ( int i = 0; i < 5; i++ ) {
  vector[i] = something;
}

vs.

for ( int i = 0; i < 5; i++ ) {
  *(vector+i) = something;
}

In the first case, an optimizing compiler may detect that the array vector is iterated over element by element and thus generate something like

void* tempPtr = vector;
for ( int i = 0; i < 5; i++ ) {
  *((int*)tempPtr) = something;
  tempPtr += sizeof(int); // _move_ the pointer; simple addition of a constant.
}

It might even be able to use the target CPU's pointer post-increment instructions where available.

For the second case, it is "harder" for the compiler to see that the address that's calculated via some "arbitrary" pointer arithmetics expression shows the same property of monotonically advancing a fixed amount in each iteration. It might thus not find the optimization and calculate ((void*)vector+i*sizeof(int)) in each iteration which uses an additional multiplication. In this case there's no (temporary) pointer that gets "moved" but only a temporary address re-calculated.

However, the statement probably does not universally hold for all C compilers in all versions.

Update:

I checked the above example. It appears that without optimizations enabled at least gcc-8.1 x86-64 generates more code (2 extra instructions) for the second (pointer-arithmethics) form than the first (array index).

See: https://godbolt.org/g/7DaPHG

However, with any optimizations turned on (-O...-O3) the generated code is the same (length) for both.

Answer 3

I've got 2 C files: ex1.c

% cat ex1.c
#include <stdio.h>

int main (void) {
    int vector[5] = { 1, 2, 3, 4, 5 };
    printf("%d\n", vector[3]);
}

and ex2.c,

% cat ex2.c
#include <stdio.h>

int main (void) {
    int vector[5] = { 1, 2, 3, 4, 5 };
    printf("%d\n", *(vector + 3));
}

And I compile both into assembly, and show the difference in generated assembly code

% gcc -S ex1.c; gcc -S ex2.c; diff -u ex1.s ex2.s
--- ex1.s       2018-07-17 08:19:25.425826813 +0300
+++ ex2.s       2018-07-17 08:19:25.441826756 +0300
@@ -1,4 +1,4 @@
-       .file   "ex1.c"
+       .file   "ex2.c"
        .text
        .section        .rodata
 .LC0:

Q.E.D.

---

The C standard very explicitly states (C11 n1570 6.5.2.1p2):

2. A postfix expression followed by an expression in square brackets [] is a subscripted designation of an element of an array object. The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2))). Because of the conversion rules that apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the initial element of an array object) and E2 is an integer, E1[E2] designates the E2-th element of E1 (counting from zero).

---

Additionally, the as-if rule applies here - if the behaviour of the program is the same, the compiler can generate the same code even though the semantics weren't the same.

Answer 4

Let me try to answer this "in the narrow" (others have already described why the description "as-is" is somewhat lacking/incomplete/misleading):

In what context would any compiler generate different code for those two?

A "not-very-optimizing" compiler might generate different code in just about any context, because, while parsing, there's a difference: x[y] is one expression (index into an array), while *(x+y) are two expressions (add an integer to a pointer, then dereference it). Sure, it's not very hard to recognize this (even while parsing) and treat it the same, but, if you're writing a simple/fast compiler, then you avoid putting "too much smarts into it". As an example:

char vector[] = ...;
char f(int i) {
    return vector[i];
}
char g(int i) {
    return *(vector + i);
}

Compiler, while parsing f(), sees the "indexing" and may generate something like (for some 68000-like CPU):

MOVE D0, [A0 + D1] ; A0/vector, D1/i, D0/result of function

OTOH, for g(), compiler sees two things: first a dereference (of "something yet to come") and then the adding of integer to pointer/array, so being not-very-optimizing, it could end up with:

MOVE A1, A0   ; A1/t = A0/vector
ADD A1, D1    ; t += i/D1
MOVE D0, [A1] ; D0/result = *t

Obviously, this is very implementation dependent, some compiler might also dislike using complex instructions as used for f() (using complex instructions makes it harder to debug the compiler), the CPU might not have such complex instructions, etc.

Is there a difference between "move" from base, and "add" to base?

The description in the book is arguably not well-worded. But, I think the author wanted to describe the distinction shown above - indexing ("move" from base) is one expression, while "add and then dereference" are two expressions.

This is about compiler implementation, not language definition, the distinction which should have also been explicitly indicated in the book.

Answer 5

I tested the Code for some compiler variations, most of them give me the same assembly code for both instructions (tested for x86 with no optimization). Interesting is, that the gcc 4.4.7 does exactly, what you mentioned: Example:

Other langauges like ARM or MIPS are doing sometimes the same, but I didn`t tested it all. So it seems their was a difference, but later versions of gcc "fixed" this bug.

Answer 6

The Standard specifies the behavior of arr[i] when arr is an array object as being equivalent to decomposing arr to a pointer, adding i, and dereferencing the result. Although the behaviors would be equivalent in all Standard-defined cases, there are some cases where compilers process actions usefully even though the Standard does require it, and the handling of arrayLvalue[i] and *(arrayLvalue+i) may differ as a consequence.

For example, given

char arr[5][5];
union { unsigned short h[4]; unsigned int w[2]; } u;

int atest1(int i, int j)
{
if (arr[1][i])
    arr[0][j]++;
return arr[1][i];
}
int atest2(int i, int j)
{
if (*(arr[1]+i))
    *((arr[0])+j)+=1;
return *(arr[1]+i);
}
int utest1(int i, int j)
{
    if (u.h[i])
        u.w[j]=1;
    return u.h[i];
}
int utest2(int i, int j)
{
    if (*(u.h+i))
        *(u.w+j)=1;
    return *(u.h+i);
}

GCC's generated code for test1 will assume that arr[1][i] and arr[0][j] cannot alias, but the generated code for test2 will allow for pointer arithmetic to access the entire array, On the flip side, gcc will recognize that in utest1, lvalue expressions u.h[i] and u.w[j] both access the same union, but it's not sophisticated enough to notice the same about *(u.h+i) and *(u.w+j) in utest2.