赋值运算符不起作用

Question

I have a constructor to represent big numbers and store them into a vector: my class is composed of a vector, a boolean and an int and is supposed to represent big numbers.

    CBigInt(const char* str2)
{
    string str=string(str2);
    num.resize(200000);
    if (str[0]=='-') 
    {
        signe=true; 
        pos=str.length()-1;
        for (int i=pos;i>0;i--)
        {
            num[pos-i]=str[i]-48;
        }
    }
    else
    {
        signe=false; 
        pos=str.length();
        for (int i=pos;i>=0;i--)
        {
            num[pos-i-1]=str[i]-48;
        }
    }       
}

and I get an error: http://pastebin.com/cy82XaLF

Answer 1

A conversion sequence can only contain at most one implicit user-defined conversion. Yours contains two: const char[] to string , then string to CBigInt .

A simple fix is to make one conversion explicit:

CBigInt autre("123456789012345678901111");

Answer 2

在已经拥有的构造函数之上声明并实现一个附加的构造函数：

CBigInt::CBigInt(const char* str):CBigInt(string(str)){}

Answer 3

Probabily they doesn't work because the compile do not provide two implicit convertion at same operation, one from Const Char* to String , and String to CBigInt . And only when the type was a CBigInt is possible call operator = .

This could be repare at this peace of code that compile correctily with your class:

   CBigInt a("99") //justo one implicit convertion, to const char * to string
   string std;
   a = std; //just one implicit convertion, to String to CBigInt
   a = string("99"); //just one implicit convertion, to String to CBigInt

My sugestion is copy your code from CBigInt(string) to CBigInt(const char*) and adpatad to looking for '\\0' at the end of string.

Another suggestion that I have to you is exchange your signature from CBigInt(string) to CBigInt(const string&) , it's faster because doesn't copy the string and no code have to be changed.