Why can't (or doesn't) the compiler optimize a predictable addition loop into a multiplication?

Solution 1:

The compiler can't generally transform

for (int c = 0; c < arraySize; ++c)
    if (data[c] >= 128)
        for (int i = 0; i < 100000; ++i)
            sum += data[c];

into

for (int c = 0; c < arraySize; ++c)
    if (data[c] >= 128)
        sum += 100000 * data[c];

because the latter could lead to overflow of signed integers where the former doesn't. Even with guaranteed wrap-around behaviour for overflow of signed two's complement integers, it would change the result (if data[c] is 30000, the product would become -1294967296 for the typical 32-bit ints with wrap around, while 100000 times adding 30000 to sum would, if that doesn't overflow, increase sum by 3000000000). Note that the same holds for unsigned quantities, with different numbers, overflow of 100000 * data[c] would typically introduce a reduction modulo 2^32 that must not appear in the final result.

It could transform it into

for (int c = 0; c < arraySize; ++c)
    if (data[c] >= 128)
        sum += 100000LL * data[c];  // resp. 100000ull

though, if, as usual, long long is sufficiently larger than int.

Why it doesn't do that, I can't tell, I guess it's what Mysticial said, "apparently, it does not run a loop-collapsing pass after loop-interchange".

Note that the loop-interchange itself is not generally valid (for signed integers), since

for (int c = 0; c < arraySize; ++c)
    if (condition(data[c]))
        for (int i = 0; i < 100000; ++i)
            sum += data[c];

can lead to overflow where

for (int i = 0; i < 100000; ++i)
    for (int c = 0; c < arraySize; ++c)
        if (condition(data[c]))
            sum += data[c];

wouldn't. It's kosher here, since the condition ensures all data[c] that are added have the same sign, so if one overflows, both do.

I wouldn't be too sure that the compiler took that into account, though (@Mysticial, could you try with a condition like data[c] & 0x80 or so that can be true for positive and negative values?). I had compilers make invalid optimisations (for example, a couple of years ago, I had an ICC (11.0, iirc) use signed-32-bit-int-to-double conversion in 1.0/n where n was an unsigned int. Was about twice as fast as gcc's output. But wrong, a lot of values were larger than 2^31, oops.).

Solution 2:

This answer does not apply to the specific case linked, but it does apply to the question title and may be interesting to future readers:

Due to finite precision, repeated floating-point addition is not equivalent to multiplication. Consider:

float const step = 1e-15;
float const init = 1;
long int const count = 1000000000;

float result1 = init;
for( int i = 0; i < count; ++i ) result1 += step;

float result2 = init;
result2 += step * count;

cout << (result1 - result2);

Demo

Solution 3:

The compiler contains various passes which does the optimization. Usually in each pass either an optimization on statements or loop optimizations are done. At present there is no model which does an optimization of loop body based on the loop headers. This is hard to detect and less common.

The optimization which was done was loop invariant code motion. This can be done using a set of techniques.

Solution 4:

Well, I'd guess that some compilers might do this sort of optimization, assuming that we are talking about Integer Arithmetics.

At the same time, some compilers might refuse to do it because replacing repetitive addition with multiplication might change the overflow behavior of the code. For unsigned integer types, it shouldn't make a difference since their overflow behavior is fully specified by the language. But for signed ones, it might (probably not on 2's complement platform though). It is true that signed overflow actually leads to undefined behavior in C, meaning that it should be perfectly OK to ignore that overflow semantics altogether, but not all compilers are brave enough to do that. It often draws a lot of criticism from the "C is just a higher-level assembly language" crowd. (Remember what happened when GCC introduced optimizations based on strict-aliasing semantics?)

Historically, GCC has shown itself as a compiler that has what it takes to take such drastic steps, but other compilers might prefer to stick with the perceived "user-intended" behavior even if it is undefined by the language.

Solution 5:

It does now -- at least, clang does:

long long add_100k_signed(int *data, int arraySize)
{
    long long sum = 0;

    for (int c = 0; c < arraySize; ++c)
        if (data[c] >= 128)
            for (int i = 0; i < 100000; ++i)
                sum += data[c];
    return sum;
}

compiles with -O1 to

add_100k_signed:                        # @add_100k_signed
        test    esi, esi
        jle     .LBB0_1
        mov     r9d, esi
        xor     r8d, r8d
        xor     esi, esi
        xor     eax, eax
.LBB0_4:                                # =>This Inner Loop Header: Depth=1
        movsxd  rdx, dword ptr [rdi + 4*rsi]
        imul    rcx, rdx, 100000
        cmp     rdx, 127
        cmovle  rcx, r8
        add     rax, rcx
        add     rsi, 1
        cmp     r9, rsi
        jne     .LBB0_4
        ret
.LBB0_1:
        xor     eax, eax
        ret

Integer overflow has nothing to do with it; if there is integer overflow that causes undefined behavior, it can happen in either case. Here's the same kind of function using int instead of long:

int add_100k_signed(int *data, int arraySize)
{
    int sum = 0;

    for (int c = 0; c < arraySize; ++c)
        if (data[c] >= 128)
            for (int i = 0; i < 100000; ++i)
                sum += data[c];
    return sum;
}