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Question

I have a structure that I pass to a function as constant pointer, my question is the following: There is a difference between those two implementations of function updatedFields:

typedef struct
{
    int spec[100];
    int spec1[200];
    int spec2[200];
    int spec3[500];
    int spec4[100];
    int spec5[700];
    float value[100];
    char desc[1000]:
}t_product;



void updateFields_1(t_product const* context)
{

    int i,buffer[1500];
    int * pt_int;

    pt_int = (int*)context->spec1;      
    for(i = 0; i < 200; i++)
    {
        buffer[i] = pt_int[i];
    }

    pt_int = (int*)context->spec3;        
    for(i = 0; i < 500; i++)
    {
        buffer[i] = pt_int[i];
    }

    ...
}

void updateFields_2(t_product const* context)
{

    int i,buffer[1500];

    for(i = 0; i < 200; i++)
    {
        buffer[i] = context->spec1[i];
    }

    for(i = 0; i < 500; i++)
    {
        buffer[i] = context->spec3[i];
    }

    ...
}

int main(void)
{
    t_product prod;

    /* Initialisation of the structure */
    ...

    updateField(&prod);

}

I mean, there is any advantages to use pointer to member of a struct (pointer to the arrays) instead of accessing directly to the member of struture.

It's probably a dumb question but I don't know if the access of a struct member "costs" more operations.

Answer 1

It won't ever cost more in your case. Even without optimization. Actually your pt_int example is likely to be slightly worse if you don't enable optimizations.

This is because context->spec3[i] isn't dereferencing more pointers than pt_int[i] . pt_int[i] is just a pointer plus an offset, so the access can be written as @(ptr_int + 4*i) . In context->spec3[i] , it could look like there is one more pointer dereferenced, but it isn't the case. spec3 isn't a value in context , it's just an offset from context. The address you access will therefore be @(context + 2000 + 4*i) . There is only one pointer access.

Now you can wonder if @(context + 2000 + 4*i) costs more than @(ptr_int + 4*i) . It doesn't, because most architectures, including x86, AMD64 and ARM (that is, 100% of personal devices), have instructions to do accesses with constant offsets. Also, the difference can disappear at soon as you enable trivial optimizations, because context + 2000 can be converted to a single context_2000 (but compilers won't actually do that, since it can only worsen performances).

Answer 2

它确实花费更多（它必须在每次迭代时取消引用原始指针），但成本可能很小，并且中途不错的编译器将为您进行优化。