Exception thrown by C++ CopyConstructor! !
过去多啦不再A梦
过去多啦不再A梦 2017-05-31 10:39:26
0
3
846

code show as below:

void CopyStr(char *&destination, char *&source) {
    int sz = strlen(source) + 1;//此处引发异常!!
    destination = new char[sz];
    for (unsigned i = 0; source[i] != 'rrreee'; i++)
        destination[i] = source[i];
    destination[sz - 1] = 'rrreee';
    return;
}

Data::Data(Data &adata) {
    CopyStr(adata.P_name, P_name);
    CopyStr(adata.address, address);
    CopyStr(adata.number, number);
}

Code Description: The Data class now has three char* members, namely P_name, address and number.

Compiled picture:

Please give me the answer, thank you! !

过去多啦不再A梦
过去多啦不再A梦

reply all(3)
伊谢尔伦

First, the parameter char *&source is changed to char *const &source, and secondly, CopyStr(adata.P_name, P_name) is changed to CopyStr(P_name, adata.P_name).
This is the code I tested, it can be run directly:

#include <iostream>
#include <cstring>
using std::cout;
using std::endl;
using std::strlen;
class Data
{
public:
  char *a = "";
  char *b = "";
  char *c = "";
  Data(char *a, char *b, char *c):a(a),b(b),c(c) {};
  Data(Data const &data);
};


int CopyStr(char * &des,  char *  const &source)
{
  int sz = strlen(source) + 1;
  des = new char(sz);
  for (unsigned i = 0; source[i] != 'rrreee'; i++)
    *(des +i)  = source[i];
  *(des+sz - 1) = 'rrreee';
  return 0;
}

Data::Data(Data const  &data)
{ 
  CopyStr(a, data.a);
  CopyStr(b, data.b);
  CopyStr(c, data.c);
}

int main()
{
  Data data = Data("abcd", "b", "c");
  char * p = "232323";
  Data data2 =  Data(data);
  char * p2 = "1232";
  CopyStr(p2, p);
  cout << p2 << endl;
  cout << data.a << endl;
  cout << data2.a << endl;
  return 0;
}
我想大声告诉你

Where does the second parameter (such as P_name) come from when calling CopyStr?

phpcn_u1582

vs will fill the uninitialized memory area with 0xCCCCCCCC (this is also the origin of scalper);
Considering that you accessed 0xCCCCCCCC, either you passed in an illegal pointer, or this char* The string corresponding to the pointer does not end with '0';
You can try to output the value of source, and then try to output the data pointed to by the pointer byte by byte

Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template