Question

I have a function with the following signature:

float* Interpolate(float t, UINT iOrder, UINT iDimension, float** ppPointsArray);

When trying to call it as follows:

float ppfValues[2][1];
ppfValues[0][0] = 0.0f;
ppfValues[1][0] = 10.0f;

float* pfResult = MyMathFuncs::Interpolate(0.5f,2,1,ppfValues);

I get the following error:

Error: argument of type float(*)[1] is incompatible with parameter of type "float**"

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If I want to call it properly, I should do it like this:

float** ppfValues = new float*[2];
ppfValues[0] = new float(0.0f);
ppfValues[1] = new float(10.0f);

float* pfResult = MyMathFuncs::Interpolate(0.5f,2,1,ppfValues);

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Now the question is: I thought float[x][y] was actually the same as a float**
Why are they not? What are the technical reasons? And what are they exactly, then?

Answer 1

I thought float[x][y] was actually the same as a float**

It all boils down to the fact that arrays and pointers aren’t equivalent. Below is a list of C FAQs (even if this is a C++ question) which stress this fact in various ways.

• My compiler complained when I passed a two-dimensional array to a
  function expecting a pointer to a pointer.
• But I heard that char a[] was identical to char *a
• So what is meant by the “equivalence of pointers and arrays” in C?
• How do I write functions which accept two-dimensional arrays when the width is not known at compile time?