Question

HeJ小鼠！我想学习使用C中函数的指针。所以我写了一个小计算器，包含私有操作的函数和一个指针，根据输入的操作符号切换它们。我不知道它是否会比调用单个函数更有效，但它只是一个测试。

因此，在Calculator类公共部分内部，此函数假设将POper指针设置为适当的操作：

 void Calculate() {
    switch (m_s) {
    default:
      printf("Error, wrong operation sign\n");
      break;
    case '+':
      POper = &Calculator::Add;
      break;
    case '-':
      POper = &Calculator::Sub;
      break;
    case '*':
      POper = &Calculator::Multi;
      break;
    case '/':
      POper = &Calculator::Divide;
      break;
    }
  }

当我尝试编译fallowing代码时，gcc给我这种类型的错误交战：

../src/main.cpp: In member function ‘void Calculator::Calculate()’:
../src/main.cpp:16:13: error: cannot convert ‘float (Calculator::*)()’ to ‘float (*)()’ in assignment
       POper = &Calculator::Add;
             ^
../src/main.cpp:19:13: error: cannot convert ‘float (Calculator::*)()’ to ‘float (*)()’ in assignment
       POper = &Calculator::Sub;
             ^
../src/main.cpp:22:13: error: cannot convert ‘float (Calculator::*)()’ to ‘float (*)()’ in assignment
       POper = &Calculator::Multi;
             ^
../src/main.cpp:25:13: error: cannot convert ‘float (Calculator::*)()’ to ‘float (*)()’ in assignment
       POper = &Calculator::Divide;

我真的不明白，POper和使用成员都是Calculator类的一部分。这对cannot convert ‘float (Calculator::*)()’ to ‘float (*)()’ in assignment意味着什么？差异是什么？

指针功能错误转换

0 个答案: