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Old 03-14-2005, 07:45 AM   #1
pablowablo
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pointer notation vs array notation?


According to some text I'm reading *(array + i) is much faster than array[i] since the following has to be done on the latter: &array[0] + i * sizeof(TYPE)

I was wondering if pointer notation is really faster than array notation. Could anyone verify this for me? The reason that I'm having some doubt about this is because the compiler is smart enought to know the size of the element in pointer arithmetic but why not in array notation?


Thanks!
 
Old 03-14-2005, 09:02 AM   #2
perfect_circle
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Re: pointer notation vs array notation?

Quote:
Originally posted by pablowablo
According to some text I'm reading *(array + i) is much faster than array[i] since the following has to be done on the latter: &array[0] + i * sizeof(TYPE)

I was wondering if pointer notation is really faster than array notation. Could anyone verify this for me? The reason that I'm having some doubt about this is because the compiler is smart enought to know the size of the element in pointer arithmetic but why not in array notation?
Thanks!

You can test it by yourself if you want, by creating loops, but I wouldn't bother with those kind of things, as long as you use -O2 both will run in the same time.
 
Old 03-14-2005, 09:14 AM   #3
zeropash
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hmm lets see
int arr[20];

int func1(int i) {
*(arr+i) = 10;
}

int func2(int i) {
arr[i] = 10;
}

int main() {
func1(9);
func2(9);
}
compile this and run it using gdb
disassemble func1 gives
Dump of assembler code for function func1:
0x08048334 <func1+0>: push %ebp
0x08048335 <func1+1>: mov %esp,%ebp
0x08048337 <func1+3>: mov 0x8(%ebp),%eax
0x0804833a <func1+6>: movl $0xa,0x8049580(,%eax,4)
0x08048345 <func1+17>: leave
0x08048346 <func1+18>: ret
0x08048347 <func1+19>: nop
disassemble func2 gives
Dump of assembler code for function func2:
0x08048348 <func2+0>: push %ebp
0x08048349 <func2+1>: mov %esp,%ebp
0x0804834b <func2+3>: mov 0x8(%ebp),%eax
0x0804834e <func2+6>: movl $0xa,0x8049580(,%eax,4)
0x08048359 <func2+17>: leave
0x0804835a <func2+18>: ret
0x0804835b <func2+19>: nop
End of assembler dump.

Now spot the difference (I cannot)
you can see that the compiler produces the exact same code. so dont expect any change in performance .
you can check on other arch too.

Last edited by zeropash; 03-14-2005 at 09:16 AM.
 
Old 03-14-2005, 09:24 AM   #4
Hivemind
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zeropash's idea is a good one. If you want to know if two different ways produces the exact same code, simply study the assembler output!
 
Old 03-14-2005, 09:38 AM   #5
perfect_circle
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Of course all those things depend from the compiler. Maybe older compilers used to output different code for those 2 things but not any more.
 
Old 03-14-2005, 12:34 PM   #6
aluser
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I think your text is simply broken because when you do pointer arithmetic it multiplies by the sizeof the thing pointed to by the pointer anyway.

Code:
int *foo;
...
*(foo + i) = blah;
is equivalent to
Code:
int *foo;
...
*(int*)(((char*)foo)+sizeof(*foo)*i) = blah;
which makes it exactly the same as the other syntax.

On the other hand, something along these lines which might make your code faster, particularly without optimizations turned on, is doing

Code:
char *p;
for (p = s; *p != '\0'; ++p) {
    ...
}
instead of
Code:
int i;
for (i = 0; s[i] != '\0'; ++i) {
    ...
}
I'd bet that optimizations turns the second form into the first.
 
  


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