Re: How to revise a charactor in string?

"Giovanni Dicanio" <>
Thu, 31 Jan 2008 12:39:50 +0100
"Lorry Astra" <> ha scritto nel

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

using namespace std;

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

int count=(int)strlen(c);

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

The problem is that the "abcd" string built as you did is a "read-only"
You can modify 'c' pointer address, to e.g. point to a different string.
But you can't modify the data (characters) of the read-only string that you
used to initialize the 'c' pointer.

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.

The "algorithm" is good, the problem is the implementation, i.e. you need a
readable and *writable* character buffer to modify the string.

You may consider the following working code (compiled with VS2008):

#include <iostream>
#include <cstring>

using std::cout;
using std::endl;

int main()
    char * c = "abcd";
    int count = (int)strlen(c);

    // Alloc buffer on the stack for "mutable" string
    char s[ 100 ];

    // Copy source (read-only) string to buffer
    strcpy( s, c );
    // better using a safe string function, like StringCbCopy here...
    // this is just a quick "demo"

    // Modify the last character in auxiliary buffer (s)
    s[count-1] = 'e';

    cout << "Original string: " << c << endl;
    cout << "Modified string: " << s << endl;

    return 0;


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?

Yes, see previous point.
(I think that in VC6 it was possible to modify also that kind of strings...
but I'm not sure; however, with more modern VS2008 you can't anymore.)

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
a the length of string which is pointed by char*?

'c' is a *pointer*, so its size in bytes (the value returned by 'sizeof') is
the size of a pointer, and it is 4 bytes (4 bytes * 8 bits/byte = 32 bits on
32 bits platforms).
'z' is an array, so its size is <number of element in z> * sizeof< an
element of z > = 6 [=strlen(z)+1] * sizeof(char) = 6 * 1 = 6 bytes.

  (b) If I can only get a char* which points to a string, how can I
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?

You have several options.
For example, you can get the number of characters pointed by 'c', and
allocate memory on the heap using 'new []', something like this:


  ASSERT( c != NULL ); // check pointer

  // Alloc memory on the heap
  char * copy = new char[ strlen(c) + 1 ];

  // Copy to destination buffer
  strcpy( copy, c );

  // copy is a modifiable char array...

  // Release the buffer
  delete [] copy;
  copy = NULL; // avoid dangling references


However, I would use a robust C++ string class to manage strings, like
CString (from ATL), or std::string or std::wstring from standard C++ library
(but, IMHO, for Windows C++ programming, CString/A/W are better designed
than std:: string classes).

Moreover, if you need to manage dynamic array, using std::vector is better
than raw new[] calls (for several reasons, for example: std::vector does
bounds-checking on vector index, it releases its own memory when variable
goes out of scope, etc.).

And if you really need to access low-level C string functions, I think you
may consider safe string functions like StringCbCopy, etc.


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