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Home/ Questions/Q 7783395
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T19:49:37+00:00 2026-06-01T19:49:37+00:00

Why does this not work: SomeClass::SomeClass(char *lit) //Ctor { str = new char[strlen(lit)+1]; //

  • 0

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.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-01T19:49:38+00:00Added an answer on June 1, 2026 at 7:49 pm

    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;
    }
    
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