从功能用C返回二维数组++ [英] return 2d array from function in C++

问题描述

我已经声明，像这样一个功能：

  unsigned char型**分类::分类（）{
      // ...
    unsigned char型**字符=新的无符号的char * [H]
的for（int i = 0; I＆LT; H，我++）
    字符由[i] =新的无符号字符[W * 3]。// ...返回＆安培;字符;
//注意：这是返回字符;我得到以下几点：不能转换'无符号字符*'到'unsigned char型**'的回报
 

这是给我的警告：

  Classifier.cpp：在成员函数'unsigned char型**分类::分类（）'：
Classifier.cpp：124：警告：​​局部变量的地址'字符'返回
 

这是确定以忽略？基本上，我的问题是你如何返回引用到在函数中定义的数组？

我希望能够做到

  unsigned char型** someData =分类（）;
 

解决方案

这是的从不没关系忽略。你返回一个局部变量的地址。当你离开区分该地址将变得无效（）的堆栈帧，的前的调用者有机会使用它。

您只需要返回该变量的值，而不是：

 返回字符;
 

I have a function declared like so:

unsigned char** Classifier::classify(){
      //...
    unsigned char **chars = new unsigned char *[H];
for(int i = 0; i < H; i++)
    chars[i] = new unsigned char[W*3];

//...

return &chars;
//note: when this is "return chars;" I get the following:  cannot convert ‘unsigned char*’ to ‘unsigned char**’ in return

This is giving me the warning:

Classifier.cpp: In member function ‘unsigned char** Classifier::classify()’:
Classifier.cpp:124: warning: address of local variable ‘chars’ returned

Is this ok to ignore? Basically, my question is how do you return a reference to an array that is defined in the function?

I want to be able to do

unsigned char** someData = classify();

解决方案

This is never okay to ignore. You're returning the address of a local variable. That address will become invalid when you leave classify()'s stack frame, before the caller has a chance to use it.

You only need to return the value of that variable instead:

return chars;

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