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Home/ Questions/Q 8870651
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T17:47:25+00:00 2026-06-14T17:47:25+00:00

I’m currently doing a University project which is marked heavily upon the speed and

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I’m currently doing a University project which is marked heavily upon the speed and efficiency of my solution. Minor changes I make to the code have massive impacts, as the particular function I am writing is called many hundreds of thousands of times.

I have written the main functionality of my project now, and am currently in the process of optimising everything I possibly can. One particular part of my code that I am questioning looks like this:

array[i] *= -1;

Which I was considering optimising to:

array[i] = 0 - array[i];

Would changing this code actually affect the speed? Is a subtraction operation faster than a multiplication operation? Or is this kind of issue a thing of the past?

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  1. Editorial Team
    Editorial Team
    2026-06-14T17:47:29+00:00Added an answer on June 14, 2026 at 5:47 pm

    Overlooking the fact that you should probably use this instead:

    array[i] = -array[i];
    

    as it’s much clearer IMO since it directly states intent, lets check what the compiler does (GCC 4.7.2 on x86-64) for this program:

    #include <stdio.h>
    #include <time.h>
    
    int main(void)
    {
        time_t t = time(NULL);
        t *= -1;
        return 0;
    }
    
    gcc -S mult.c -o 1.s

    And for this:

    #include <stdio.h>
    #include <time.h>
    
    int main(void)
    {
        time_t t = time(NULL);
        t = 0 - t;
        return 0;
    }
    
    gcc -S sub.c -o 2.s

    Now compare the two assembly outputs:

    diff 1.s 2.s

    Nothing is printed. The compiler generated the same exact code for both versions. So the answer is: it doesn’t matter what you use. The compiler will choose whatever is fastest. This is a pretty easy optimization to make (if you can even call it an optimization), so we can assume that virtually every compiler out there will pick the fastest way to do it for a given CPU architecture.

    For reference, the generated code is:

    int main()
    {
        time_t t = time(NULL);
           mov    edi,0x0
           call   12 
           mov    QWORD PTR [rbp-0x8],rax
    
        t *= -1;
           neg    QWORD PTR [rbp-0x8]
    
        t = 0 - t;
           neg    QWORD PTR [rbp-0x8]
    
        return 0;
           mov    eax,0x0
    }

    In both cases, it uses NEG to negate the value. t *= -1 and t = 0 - t both generate:

    neg QWORD PTR [rbp-0x8]
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