c++ - 错误 : argument of type "float(*)[1] " is incompatible with parameter of type "float** "

Question

我有一个具有以下签名的函数:

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

尝试按如下方式调用时:

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

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

我收到以下错误:

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

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如果我想正确地调用它，我应该这样做:

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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现在的问题是:我认为 float[x][y] 实际上与 float 相同**他们为什么不呢？技术原因是什么？那么它们到底是什么？

Answer 1

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

这一切都归结为数组和指针不等同。下面是 C 常见问题解答列表(即使这是一个 C++ 问题)，它们以各种方式强调了这一事实。

• My compiler complained when I passed a two-dimensional array to afunction 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?