c++ - 将数组作为参数传递给printf

有一种方法可以执行以下操作，但是只将bounds传递给printf吗？

double *bounds = getBounds();
printf("%f-%f, %f-%f, %f-%f",
    bounds[0], bounds[1],
    bounds[2], bounds[3],
    bounds[4], bounds[5] );

// what I'd like to write instead:
xxxprintf("%f-%f, %f-%f, %f-%f", bounds);

最佳答案

我认为您要优化此设置的原因是，您需要在程序中打印很多边界，并且这样编写很麻烦，容易出错等。

在C语言中，您可以使用如下宏:

#define BOUNDS_FORMAT "%f-%f, %f-%f, %f-%f"
#define BOUNDS_ARG(b) b[0], b[1], b[2], b[3], b[4], b[5]

然后像这样写:

printf(BOUNDS_FORMAT, BOUNDS_ARG(bounds));
// ... some other code, then another call, with more text around this time:
printf("Output of pass #%d: " BOUNDS_FORMAT "\n", passNumber, BOUNDS_ARG(bounds));

在C++中，更期望您使用std::cout或类似的流。然后，您可以编写一个自定义对象来为您执行此操作:

class PrintBounds {
  protected:
    const double* m_bounds;

  public:
    PrintBounds(const double* bounds)
      : m_bounds(bounds)
    {
    }

    friend std::ostream& operator<<(std::ostream& os, const PrintBounds& self)
    {
        os << self.m_bounds[0] << "-" << self.m_bounds[1] << ", "
           << self.m_bounds[2] << "-" << self.m_bounds[3] << ", "
           << self.m_bounds[3] << "-" << self.m_bounds[5];
        return os;
    }
};

然后，您将像这样使用它:

std::cout << "Some other text: " << PrintBounds(bounds) << " ...\n";

关于c++ - 将数组作为参数传递给printf，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/23564715/