CODEWITHSUNDEEP
c++arraysstringcharacter
Expert Novice
Expert Novice

Reputation: 1963

Strcpy does not copy an allocated character array

Why does this not work:

SomeClass::SomeClass(char *lit) //Ctor
{
    str = new char[strlen(lit)+1]; // str is a pointer to char in SomeClass
    strcpy(str,"have");
    cout << str << " " << "In Ctor" << " +Size=" << strlen(str)<< endl;
}

The above code shows a string with length 0. But this code works:

SomeClass::SomeClass(char *lit)
{
    char newstr[strlen(lit)+1];
    strcpy(newstr,"have");
    cout << newstr << " " << "In Ctor" << " +Size=" << strlen(newstr)<< endl;
}

Here is the complete code.

EDIT:
Added the link to Ideone which OP removed after I answered the Question.
Without the link to source code, this Q & answer to it is useless.

Upvotes: 1

Views: 1160

Answers (2)

Mike DeSimone
Mike DeSimone

Reputation: 42825

Another way to write String::leng():

int String::leng()
{
    char *endPtr = str;
    while(*endPtr)
        endPtr++;
    return endPtr - str;
}

Upvotes: 0

Alok Save
Alok Save

Reputation: 206636

There is no problem with the strcpy, You are just messing your pointer.

The problem is here: 

 str = new char[strlen(lit)+1];
 strcpy(str,lit);
 length=leng();    <------------- str points to \0 after this call
 cout << str << " " << "In Ctor" << " +Size=" << strlen(lit)<< endl;

str is your class member and You move the pointer str to point to the \0 in the function leng(), Naturally, You don't see any output in the next statement.

Solution is to hold the starting address in a separate pointer inside the function. 

int String :: leng()
{
      int length=0;
      char *tempPtr= str;       <----------- Store the address in a temporary pointer
      while(*str)
      {
                 length++;
                 str++;
      }
      str = tempPtr;            <---------- Point the Pointer member to right address again
      return length;
}

Upvotes: 4

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