String length

This is a discussion on String length within the C++ Programming forums, part of the General Programming Boards category; I have tried the following program: #include<iostream.h> #include<fstream.h> #include<string.h> class Member { private: int Age; int l,L; char Name[40]; public: ...

  1. #1
    Registered User
    Join Date
    Nov 2001
    Posts
    8

    Question String length

    I have tried the following program:

    #include<iostream.h>
    #include<fstream.h>
    #include<string.h>

    class Member
    {
    private:
    int Age;
    int l,L;
    char Name[40];

    public:

    void SetName(char* name)
    {

    l= strlen(name);
    cout<<"l="<<l<<endl;

    for(int i=0;i<l;i++)
    {
    Name[i] = name[i];
    cout<<"Name["<<i<<"] = "<<Name[i]<<endl;
    }

    cout<<Name;
    for(int j=l;j<sizeof(Name);j++)
    {
    Name[j] = ' ';
    }

    }

    void DisplayName()
    {
    L = strlen(Name);
    cout<<"The length of the Name is :"<<L<<endl;
    for(int y=0;y<l;y++)
    {
    cout<<Name[y];
    }

    }

    void SetAge(int age)
    {
    Age = age;
    }

    void DisplayAge()
    {
    cout<<"Age = "<<Age<<endl;
    }
    };

    void main()
    {
    char s[40];

    Member Kokila;

    cout<<"Please enter a name"<<endl;
    cin.getline(s,40);


    Kokila.SetAge(14);
    Kokila.DisplayAge();

    Kokila.SetName(s);
    Kokila.DisplayName();

    }

    I have entered Tom as input. In SetName() method, the value of l =3. In DisplayName() method, the value of L = 43( ie sum of the length of Name[40] from the class member and the length of the input string). I cannot understand why .

    Anybody ..can explain me...

    Thanks,
    Kokila.

  2. #2
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    162
    If I understand your question correctly, the reason your string lenght is being displayed as 43 is because you are setting all the elements of the Name array to a space, which in ASCII text is 32. If you want to clear the array try something like:

    strncpy(Name, NULL, sizeof(Name))//not sure if this will cause a stack error or not

    and if that doesn't work you can always do this:

    for(USHORT i=0; i<sizeof(Name); i++){
    Name[i] = '\0'; //'\0' is basically a NULL value
    }

    Good Luck!

  3. #3
    ATH0 quzah's Avatar
    Join Date
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    14,826
    You may also want to consider memcpy. It is called like:

    memcpy( tohere, thisCharacter, thisMany );

    Thus, 'memcpy( myBuffer, NULL, sizeof(myBuffer) );

    Quzah.
    Hope is the first step on the road to disappointment.

  4. #4
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    2,934
    I edited this to change the bottom for-loop also. Should be l+1.

    void SetName(char* name)
    {

    l= strlen(name);
    cout<<"l="<<l<<endl;

    for(int i=0;i<l;i++)
    //To get the string terminator, should be
    for(int i=0;i<=l;i++)

    {
    Name[i] = name[i];
    cout<<"Name["<<i<<"] = "<<Name[i]<<endl;
    }

    cout<<Name;
    for(int j=l;j<sizeof(Name);j++)
    //I think this should be
    for(int j=l+1;j<sizeof(Name);j++)
    {
    Name[j] = ' ';
    }

    }

    //Another option is strcpy(Name,name);
    Last edited by swoopy; 12-19-2001 at 12:41 PM.

  5. #5
    Registered User
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    Sep 2001
    Posts
    164
    Originally posted by quzah
    You may also want to consider memcpy. It is called like:

    memcpy( tohere, thisCharacter, thisMany );

    Thus, 'memcpy( myBuffer, NULL, sizeof(myBuffer) );

    Quzah.
    You mean memset, right? memcopy copies memory.
    // Gliptic

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