1

This is related to this question. I have a function void doConfig(mappings, int& numOfMappings) and I'm not sure how to declare mappings. It is a two dimensional array whose elements are chars. The first dimension is determined at run time, and will be computed in the body of the function. The second dimension is always going to be 2. What is the code for this? I'd imagine it to be char** mappings or something like that. Also in C++ arrays are always passed by reference right? So I don't need to use & even though I intend to use the value when the function returns?

EDIT: Basically I want to return this char (*mapping)[2] = new char[numOfMappings][2];

as per 2to1mux's suggestion I still cannot get it to work. The array appears to getting the right values but something is going wrong when the doConfig() function returns.

int main()
{
    int numOfMappings = 0;
    char **mappings;
    doConfig(mappings, numOfMappings);
    cout << "This is mappings" << mappings << endl;//this address is different than the one given in doConfig(), is that wrong?
    cout << "this is numOfMappings: " << numOfMappings << endl;
    cout << mappings[0][0] << "->" << mappings[0][1] << endl;//program crashes here
    //code removed
    return EXIT_SUCCESS;
}

void doConfig(char **mappings, int& numOfMappings)
{
    //code removed, numOfMappings calculated
    for(int j = 0; j < numOfMappings; j++)
    {
        getline(settingsFile, setting);
        mappings[j] = new char[2];
        mappings[j][0] = setting.at(0);
        mappings[j][1] = setting.at(2);
    }
    for(int j = 0; j < numOfMappings; j++)
        cout << mappings[j][0] << "->" << mappings[j][1] << endl;//everything is as expected so array created ok
    cout << "This is mappings" << mappings << endl;//different address than the one give in main
}

OK I got it working now but mainly from haking around. Could people please explain there solutions as to how they known when to use * and &?

4

4 回答 4

2

（跟进我对链接问题的回答。）

直接（但相当复杂）的语法是

char (*create_mappings(size_t n))[2]
{
  // Allocate an char[n][2] array
  char (*mappings)[2] = new char[n][2];

  // Initailize `mappings[i][j]` in any way you want...

  return mappings;
}

但更好的主意是让它更具可读性typedef

typedef char Char2[2];

Char2 *create_mappings(size_t n)
{
  // Allocate an char[n][2] array
  Char2 *mappings = new Char2[n];

  // Initailize `mappings[i][j]` in any way you want...

  return mappings;
}
于 2013-05-17T17:24:23.917 回答
2

由于您标记了您的问题 C++，而不是 C，我想您可能需要一个适当的解决方案。

template<typename T>
using vectorOf2D = std::vector<std::array<T, 2>>;

vectorOf2D<char> getMappings() {
    return /* whatever you do to fill those */;
    // (most probably) using NRVO to ellide the copy
}

如果您担心访问可能会很复杂：

auto mappings = getMappings();

functionTakingAMapping(mappings[i]);
char element = mappings[0][1];
于 2013-05-16T17:59:50.950 回答
1

I'll answer your questions last-in-first-out:

  1. Correct, you won't need to use & here.

  2. The term by-reference technically doesn't apply to passing arrays, but the simple answer to your question is that you are never passing a copy of an array to a function. Any changes made to a parameter of type array will apply to the original array, not a copy.

  3. I suggest passing a double-pointer:

    void doConfig(char **mappings, int& numOfMappings)

    You will be able to access members of mappings exactly as you would a 2d array. Example:

    mappings[2][3] = 'b';

EDIT: Here is new suggestion based on your clarification

void doConfig(char** mappings, int& numOfMappings){

    /*Compute numOfMappings -- 
      this integer is passed by-reference, so it can be used outside function
      to figure out the size allocated within the function*/

    mappings = new char*[numOfMappings];
    for(int i=0; i < numOfMappings; i++){
        mappings[i] = new char[2];
    }
    /*Do whatever you need to do with mappings*/

    /*Return nothing because function is void -- since mappings is passed as
      pointer, changes are maintained after function ends*/

}
于 2013-05-14T04:49:17.310 回答
0

您可以返回指向二维数组的指针。

例如，

char **ptr;


return ptr;

在传递数组的地址时，如果要传递二维数组的起始位置的地址而不是特定元素的地址，则不必使用 & 运算符。

于 2013-05-14T04:45:43.830 回答