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Converting Arrays: float[][] to float**

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Started by jdub January 09, 2013 11:07 PM

2 comments, last by wintertime 11 years, 4 months ago

jdub

459

Author

January 09, 2013 11:07 PM

Hi I am trying to compile this code:


void mult_arrays(const float **src1, const float **src2, float **output)
{
	//for every element in the output array, multiply 
	for (int y = 0; y < 10; y++) {

		for (int x = 0; x < 10; x++) {

			output[y][x] = 0.0f;

			for (int i = 0; i < 10; i++) 
				output[y][x] += src1[y][i]*src2[i][x];
		}
	}
}

int main(void)
{    
	//part 2
	float arr1[10][10] = 
	{
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
	};

	float arr2[10][10] = 
	{
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
		{1.0f,2.0f,3.0f,4.0f,5.0f,6.0f,7.0f,8.0f,9.0f,},
	};

	float output[10][10];

	mult_arrays(arr1, arr2, output);
	return 0;
}

The error occurs when I call mult_arrays(): "float (*)[10]" is incompatible with parameter of type "float **"

1) Why does this error occur? 2) How can it be fixed?

Thankyou

J.W.

Brother Bob

10,350

January 09, 2013 11:26 PM

The core of the problem is that arrays are not pointers, nor are pointers arrays. Therefore, if you have a float[10][10], then the array itself consists of ten arrays of ten elements each. However, the parameters to the function expects a pointer that points to an array of pointers, which in turn points to individual arrays. In short, in main(), arr1[0] is an array of ten floats, but in mult_arrays(), src1[0] is a pointer, and the two are incompatible.

The solution depends on what you expect. Seeing that your code assumes arrays of fixed size, you could make the parameters of fixed-size arrays too.

void mult_arrays(const float src1[10][], const float src2[10][], float output[10][]) { ... }

All but the last dimension of multi-dimensional arrays must be static, so it's OK to leave out the last dimension since it doesn't do anything anyway in the parameter list. Alternatively, if you use C++, you can use references to arrays.

void mult_arrays(const float (&src1)[10][10], const float (&src2)[10][10], float (&output)[10][10])

The last dimension is required now since the complete type has to be encoded into the reference type.

ultramailman

1,720

January 10, 2013 12:22 AM

Try float (*src)[] instead. src will be a pointer to arrays, instead of pointer to pointers.

This will work if you know the size of the 2nd dimension at compile time, so something like float (*src)[10], and not
float (*src)[aVariableSize]

wintertime

4,154

January 10, 2013 11:58 AM

Theres also the possibility to prepare a pointer array before calling the function, which would give you the freedom to not only use it on size 10.

Though it looks like you are coding a matrix multiply and there are probably some premade matrix libs if you google for them.

Converting Arrays: float[][] to float**

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