C++ Passing array to functions error: incompatible types in assignment

Question

I know the question was already asked elsewhere, but I keep not getting what's wrong here. Maybe I'm doing something wrong because the two arrays are inside a structure.

(EDIT: I am getting the structure from some other code, and I cannot change it)

I'm trying to pass two arrays of float to a function, and then saving back the result of manipulation in the first array.

core.h:

typedef struct{
    //other stuff
    float m_vector[16];
} structure_t;

class CoreClass{
    private:
        structure_t s1;
        structure_t s2;

       float *MyFunction(const float *vDest, const float *vNew);
}

core.cpp:

#include "core.h"
#include "another_file.h"

void anotherFunction(){
    //....
    s1.m_vector = MyFunction(s1.m_vector, s2.m_vector); //error here
    //....    
}

float *CoreClass::MyFunction(const float *vDest, const float *vNew){
    return yet_another_function(vDest, vNew);
}

When I'm calling the function, however, I'm getting this error:

error: incompatible types in assignment of ‘float*’ to ‘float [16]’

For completeness, here is the function I'm calling, though it does not seem to do any trouble at compile time:

another_file.h

static __inline float *yet_another_function(const float *vDest, const float *vNew){
    float *tmp = new float[16];
    //tmp = matrix multiplication (vDest * vNew)
    for(int i=0; i<4; i++)
        for(int j = 0; j<4;j++)
            for(int k = 0; 4; k++)
                tmp[i + j*4] += vDest[i + k*4] * vNew[k + j*4];

    return tmp;

}

billz · Accepted Answer · 2013-02-06 09:17:28Z

2

The problem is because you are assigning pointer to array. In C++ you can't assign pointer to array.

s1.m_vector = MyFunction(s1.m_vector, s2.m_vector);
  ^^ array   ^^ return pointer

You could use copy the return valute from MyFunction to s1.m_vector. But why do you need to reassign value to s1.m_vector anyway? You could just make MyFunction function take a ref to structure_t and modify m_vector internally.

void MyFunction(structure_t& vDest, const structure_t& vNew)
{
    vDest.m_vector[0] = vNew.m_vector[0];
    //...
    vDest.m_vector[15] = vNew.m_vector[15];
}

Edit

yet_another_function(structure_t* t, structure_t& vDest, const structure_t& vNew)
{
    // blah blah
    t->m_vector[i + j*4] += vDest.m_vector[i + k*4] * vNew.m_vector[k + j*4];
}

edited Feb 6, 2013 at 9:17

answered Feb 6, 2013 at 8:35

billz

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6 Comments

ocramot Over a year ago

Sorry, I edited the question. I cannot use a ref it basically because I'm performing matrix manipulation and I need to update the old matrix values with the new ones.

billz Over a year ago

use a ref to vDest is perfect for that solution! Note, first parameter is not const, so you can modify its matrix.

ocramot Over a year ago

I wrote matrix manipulation, I meant matrix multiplication. See my updated question. How can I perform matrix manipulation without corrupting the initial matrix values?

billz Over a year ago

Then you need to copy the return value of MyFunction and delete the memory allocated by MyFunction, if you want to do it that way.

ocramot Over a year ago

I don't really like that solution, but I think I'll do like that, just like PorkyBrain suggested too, or find some other workaround.

|

odinthenerd · Accepted Answer · 2013-02-06 08:41:36Z

0

Looks like your returning a pointer to float and trying to save it to your array. structure_t.m_vector is an array not a pointer.

you could fix it like this:

float * temp = MyFunction(s1.m_vector, s2.m_vector);
for(int i=0; i<16;i++)
    s1.m_vector[i] = temp[i];
delete[] temp;

this is still quite error prone, if you new a different size other than 16 in yet_another_function you got a bug. As a general rule using std::array or std::vector and returning by value would be suggested. If you compiler supports c++ 11 move semantics returning by value will not be a performance penalty.

edited Feb 6, 2013 at 8:41

answered Feb 6, 2013 at 8:34

odinthenerd

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2 Comments

ocramot Over a year ago

So, how should I write that? Anyway I seemed to understand that in C++ writing float[], float[16] and float * is basically the same.

odinthenerd Over a year ago

when passed as arguments to a function they all turn into float *, but on the left side of operator = they are different.

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