Simple C++ pointer confusion

Question

void changeStr(char *str)
{
    str = "D";
}

void changeStr(char **str)
{
    *str = "S";
}

    char str[] = "Good";
    changeStr(str); 
    cout<<str<<endl;
    char *p = str;
    //*p = 'j';
    changeStr(&p);
    cout<<str<<endl;

I just try to change the value of str[] that array. WITHOUT RETURN!

I think the first changeStr just pass in pointer of str , and change that value, but actually it did not change it.

The second I use pointer of pointer but also cannot work.

Answer 1

Let's take it step by step.

char str[] = "Good";

You are creating an array of characters, 5 characters long with the following content:

{ 'G', 'o', 'o', 'd', '\\0' }

changeStr(str);

Here, you are passing that array to a function. Since arrays decay into pointers, this call is perfectly OK.

void changeStr(char *str)
{
    str = "D";
}

Now, here comes the first issue. You are probably confusing "D" and 'D' . If you want to change the first character in the array you need to do it the following way:

str[0] = 'D';

This would work fine. Changing the pointer won't do anything, because it's a local variable that holds a pointer to the beginning of the array, not the array itself. If you want to replace the entire content of the array with just { 'D', '\\0' } , you would need to use strcpy .

strcpy(str,"D");

Now, let's check the last part. Here you mix things up a bit.

char *p = str;
changeStr(&p);

You are creating a new variable that points to the beginning of the array and you pass pointer to that variable into the next function.

void changeStr(char **str)
{
    *str = "S";
}

Which does indeed change the original variable passed, but remember, this is p not the array. What you did is change where p points. It now points to a constant "S" .

Answer 2

In the first version of the changeStr function, the argument str is considered a local variable, and as all local variables changes to them are only changed inside the actual function.

The second version passes the string as a reference. Basically this is what happens when you specify that a variable should be passed as reference with the & specification:

void changeStr(char *&str)

Internally the compiler passes references as pointers.

Answer 3

For the value of str to change you have to pass it by reference.

It must be like

void changeStr(char* &str)

this will ensure that variable is passed by reference, thus if you make any changes to the variable in the function it will be a global change.

Answer 4

you shouldn't use assignment operator to store string in a variable even if its a pointer. use strcpy after including string.h

void changeStr(char *str)
{
    strcpy(str, "D");
}

this should change the data to which str pointer is pointing...even in the scope where the function call was made...