Re: How to revise a charactor in string?

From:

David Wilkinson <no-reply@effisols.com>

Newsgroups:

microsoft.public.vc.language

Date:

Thu, 31 Jan 2008 07:00:19 -0500

Message-ID:

<u8#pdDAZIHA.5784@TK2MSFTNGP03.phx.gbl>

Lorry Astra wrote:

Hi,please help me to resolve these questions. thanks.

#include<iostream>
#include<string>
using namespace std;

int main()
{
    char* c="abcd";

    int count=(int)strlen(c);

    \\ *(c+count-1)='e'; !! This line is wrong

    cout << c << endl;
}

questions:
1. The comment line is what I want to do, I want to revise the last
charactor in this string,please tell me what I should do.

2.If a char* point to a string (just like char* c="abcd"), I think the
"abcd" is a const value for the char*, is that right?

3.I always wonder about char* and char array.
   For example:
                          char* c="abcde";
                          char z[]="abcde";

   (a) if I want to get length of these two string, I type sizeof(c) and
sizeof(z), but these two values is totally different. why? and how can I get
a the length of string which is pointed by char*?

   (b) If I can only get a char* which points to a string, how can I convert
char* to char array. I mean:
                    void chartrim(char* c)
                   {
                        \\ here,I want to define a char array whose content
is same with char* c. please tell me how to do?
                   }

Lorry:

You should always write

const char* c = "abcd";

Then the compiler will stop you from trying to modify a read-only string. To
make a modifiable string you should do

int main()
{
    char c[] = "abcd";
    size_t count = strlen(c);
    *(c+count-1) = 'e';
    cout << c << endl;
}

[Note that if the use the correct type (size_t) you do not need a cast.]

The usage

char* c = "abcd";

is only allowed for compatibility with legacy code (a mistake, IMHO).

--
David Wilkinson
Visual C++ MVP