Accessing array values via pointer arithmetic vs. subscripting in C

Mecki has a great explanation. From my experience, one of the things that often matters with indexing vs. pointers is what other code sits in the loop. Example:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <iostream>

using namespace std;

typedef int64_t int64;
static int64 nsTime() {
  struct timespec tp;
  clock_gettime(CLOCK_REALTIME, &tp);
  return tp.tv_sec*(int64)1000000000 + tp.tv_nsec;
}

typedef int T;
size_t const N = 1024*1024*128;
T data[N];

int main(int, char**) {
  cout << "starting\n";

  {
    int64 const a = nsTime();
    int sum = 0;
    for (size_t i=0; i<N; i++) {
      sum += data[i];
    }
    int64 const b = nsTime();
    cout << "Simple loop (indexed): " << (b-a)/1e9 << "\n";
  }

  {
    int64 const a = nsTime();
    int sum = 0;
    T *d = data;
    for (size_t i=0; i<N; i++) {
      sum += *d++;
    }
    int64 const b = nsTime();
    cout << "Simple loop (pointer): " << (b-a)/1e9 << "\n";
  }

  {
    int64 const a = nsTime();
    int sum = 0;
    for (size_t i=0; i<N; i++) {
      int a = sum+3;
      int b = 4-sum;
      int c = sum+5;
      sum += data[i] + a - b + c;
    }
    int64 const b = nsTime();
    cout << "Loop that uses more ALUs (indexed): " << (b-a)/1e9 << "\n";
  }

  {
    int64 const a = nsTime();
    int sum = 0;
    T *d = data;
    for (size_t i=0; i<N; i++) {
      int a = sum+3;
      int b = 4-sum;
      int c = sum+5;
      sum += *d++ + a - b + c;
    }
    int64 const b = nsTime();
    cout << "Loop that uses more ALUs (pointer): " << (b-a)/1e9 << "\n";
  }
}

On a fast Core 2-based system (g++ 4.1.2, x64), here's the timing:

    Simple loop (indexed): 0.400842
    Simple loop (pointer): 0.380633
    Loop that uses more ALUs (indexed): 0.768398
    Loop that uses more ALUs (pointer): 0.777886

Sometimes indexing is faster, sometimes pointer arithmetic is. It depends on the how the CPU and compiler are able to pipeline the loop execution.

This might be a bit off topic (sorry) because it doesn't answer your question regarding execution speed, but you should consider that premature optimization is the root of all evil (Knuth). In my opinion, specially when still (re)learning the language, by all means write it the way it is easiest to read first. Then, if your program runs correct, consider optimizing speed. Most of the time you code will be fast enough anyway.

You need to understand the reason behind this claim. Have you ever questioned yourself why it is faster? Let's compare some code:

int i;
int a[20];

// Init all values to zero
memset(a, 0, sizeof(a));
for (i = 0; i < 20; i++) {
    printf("Value of %d is %d\n", i, a[i]);
}

They are all zero, what a surprise :-P The question is, what means a[i] actually in low level machine code? It means

1. Take the address of a in memory.

2. Add i times the size of a single item of a to that address (int usually is four bytes).

3. Fetch the value from that address.

So each time you fetch a value from a, the base address of a is added to the result of the multiplication of i by four. If you just dereference a pointer, step 1. and 2. don't need to be performed, only step 3.

Consider the code below.

int i;
int a[20];
int * b;

memset(a, 0, sizeof(a));
b = a;
for (i = 0; i < 20; i++) {
    printf("Value of %d is %d\n", i, *b);
    b++;
}

This code might be faster... but even if it is, the difference is tiny. Why might it be faster? "*b" is the same as step 3. of above. However, "b++" is not the same as step 1. and step 2. "b++" will increase the pointer by 4.

(important for newbies: running ++ on a pointer will not increase the pointer one byte in memory! It will increase the pointer by as many bytes in memory as the data it points to is in size. It points to an int and the int is four bytes on my machine, so b++ increases b by four!)

Okay, but why might it be faster? Because adding four to a pointer is faster than multiplying i by four and adding that to a pointer. You have an addition in either case, but in the second one, you have no multiplication (you avoid the CPU time needed for one multiplication). Considering the speed of modern CPUs, even if the array was 1 mio elements, I wonder if you could really benchmark a difference, though.

That a modern compiler can optimize either one to be equally fast is something you can check by looking at the assembly output it produces. You do so by passing the "-S" option (capital S) to GCC.

Here's the code of first C code (optimization level -Os has been used, which means optimize for code size and speed, but don't do speed optimizations that will increase code size noticeably, unlike -O2 and much unlike -O3):

_main:
    pushl   %ebp
    movl    %esp, %ebp
    pushl   %edi
    pushl   %esi
    pushl   %ebx
    subl    $108, %esp
    call    ___i686.get_pc_thunk.bx
"L00000000001$pb":
    leal    -104(%ebp), %eax
    movl    $80, 8(%esp)
    movl    $0, 4(%esp)
    movl    %eax, (%esp)
    call    L_memset$stub
    xorl    %esi, %esi
    leal    LC0-"L00000000001$pb"(%ebx), %edi
L2:
    movl    -104(%ebp,%esi,4), %eax
    movl    %eax, 8(%esp)
    movl    %esi, 4(%esp)
    movl    %edi, (%esp)
    call    L_printf$stub
    addl    $1, %esi
    cmpl    $20, %esi
    jne L2
    addl    $108, %esp
    popl    %ebx
    popl    %esi
    popl    %edi
    popl    %ebp
    ret

Same with the second code:

_main:
    pushl   %ebp
    movl    %esp, %ebp
    pushl   %edi
    pushl   %esi
    pushl   %ebx
    subl    $124, %esp
    call    ___i686.get_pc_thunk.bx
"L00000000001$pb":
    leal    -104(%ebp), %eax
    movl    %eax, -108(%ebp)
    movl    $80, 8(%esp)
    movl    $0, 4(%esp)
    movl    %eax, (%esp)
    call    L_memset$stub
    xorl    %esi, %esi
    leal    LC0-"L00000000001$pb"(%ebx), %edi
L2:
    movl    -108(%ebp), %edx
    movl    (%edx,%esi,4), %eax
    movl    %eax, 8(%esp)
    movl    %esi, 4(%esp)
    movl    %edi, (%esp)
    call    L_printf$stub
    addl    $1, %esi
    cmpl    $20, %esi
    jne L2
    addl    $124, %esp
    popl    %ebx
    popl    %esi
    popl    %edi
    popl    %ebp
    ret

Well, it's different, that's for sure. The 104 and 108 number difference comes of the variable b (in the first code there was one variable less on stack, now we have one more, changing stack addresses). The real code difference in the for loop is

movl    -104(%ebp,%esi,4), %eax

compared to

movl    -108(%ebp), %edx
movl    (%edx,%esi,4), %eax

Actually to me it rather looks like the first approach is faster(!), since it issues one CPU machine code to perform all the work (the CPU does it all for us), instead of having two machine codes. On the other hand, the two assembly commands below might have a lower runtime altogether than the one above.

As a closing word, I'd say depending on your compiler and the CPU capabilities (what commands CPUs offer to access memory in what way), the result might be either way. Either one might be faster/slower. You cannot say for sure unless you limit yourself exactly to one compiler (meaning also one version) and one specific CPU. As CPUs can do more and more in a single assembly command (ages ago, a compiler really had to manually fetch the address, multiply i by four and add both together before fetching the value), statements that used to be an absolute truth ages ago are nowadays more and more questionable. Also who knows how CPUs work internally? Above I compare one assembly instructions to two other ones.

I can see that the number of instructions is different and the time such an instruction needs can be different as well. Also how much memory these instructions needs in their machine presentation (they need to be transferred from memory to CPU cache after all) is different. However modern CPUs don't execute instructions the way you feed them. They split big instructions (often referred to as CISC) into small sub-instructions (often referred to as RISC), which also allows them to better optimize program flow for speed internally. In fact, the first, single instruction and the two other instructions below might result in the same set of sub-instructions, in which case there is no measurable speed difference whatsoever.

Regarding Objective-C, it is just C with extensions. So everything that holds true for C will hold true for Objective-C as well in terms of pointers and arrays. If you use Objects on the other hand (for example, an NSArray or NSMutableArray), this is a completely different beast. However in that case you must access these arrays with methods anyway, there is no pointer/array access to choose from.

"using pointer arithmetic is generally faster than subscripting for array access"

Nah. It's the same operation either way. Subscripting is syntactic sugar for adding (element size * index) to the array's start address.

That said, when iterating over the elements in an array, taking a pointer to the first element and increasing it each time through the loop will usually be slightly faster than calculating the current element's position from the loop variable each time. (Though it is unusual for this to matter much in a real life application. Examine your algorithm first, premature optimisation is the root of all evil, etc etc)