错误：类型“float（*）[1]”的参数与“float **”类型的参数不兼容

Question

我有一个带有以下签名的函数：

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

尝试按以下方式调用时：

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**"

---

如果我想正确地打电话，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);

---

现在问题是：我认为float [x] [y]实际上与浮动**相同 他们为什么不呢？技术原因是什么？那么它们究竟是什么呢？

Answer 1

  

我认为float [x] [y]实际上和float **

相同

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

• 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?