Simple C to C++ [modified]

[Software2007]

Hi, I would like to convert this function from C to C++.I attempted the conversion below, but not sure if I got it correctly. - note: debugger would crash at msg[i] = nul in the c-like code. -I also have the feeling this function can be written in 2 lines with strings! Thanks

#define NUL 0
char z_buf[4095]

void replace_html_delimiters(char *msg)
{
for(i=0; ; i++)
{
if(msg[i]== NUL)
break;
if(msg[i]=='<')
{
msg[i] = NUL;
strcpy(z_buf,msg);
strcat(z_buf,"<");
strcat(z_buf,msg+i+1);//confusing me
strcpy(msg,z_buf);
}
}
}

//C++

<pre lang="c++">
#define NUL 0
char z_buf[4095]

void replace_html_delimiters(string msg)
{
for(i=0; ; i++) //Why no upper limit here?
{
if(msg[i]== NUL)
break;
if(msg[i]=='<')
{
msg[i] = NUL;
strcpy(z_buf,msg.c_str());
z_buf += "<";
strcat(z_buf,msg.rightOf[i]);
strcpy(msg,z_buf);
}
}
}</pre>

modified on Wednesday, September 14, 2011 2:30 PM

[Lost User]

You could do it like this:

string replace_html_delimiters(string msg)
{
for(int i=0;msg[i];i++)
if(msg[i] == '<')
msg.replace(i,1,"<",0,4); // "& l t ;", CodeProject won't display this properly but I'm guessing this is what you want.

return msg;

}

The msg[i] == NUL has been moved to the for-loop and the replace[^] function is used for replacing '<'-characters with & l t ;. Note that unlike with a char array, a string will not be modified when passed to a function like this. I solved this by returning the output string, but you could also convert your function to accept a string pointer as argument.

[Software2007]

Great, I think this is what I needed. I ma not sure about the for(int i=0;msg[i];i++) as a terminating condition. Did you mean to put something else instead of msg[i]?

[Chris Losinger]

when msg[i] = 0, the loop will stop. ('false' = 0, in C/C++)

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[Software2007]

for(int i=0;msg[i]=0;i++) crashes. for(int i=0;msg[i] !=' ';i++) seems to be okay.

[Chris Losinger]

Software2007 wrote:

for(int i=0;msg[i]=0;i++) crashes.

msg[i]=0 is an assignment. (sorry, i wasn't speaking C, when i typed it, in my comment above) 'msg[i]' in a conditional is equivalent to 'msg[i]!=0'

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[Software2007]

Sorry, but I am a bit confused. I am on C++ compiler now, and both of the following crash. for(int i=0;msg[i]=0;i++) //I understand this is no good, it's an assignment for(int i=0;msg[i]!=0;i++)//but shouldn't this work? "Expression string out of range"

[Chris Losinger]

Software2007 wrote:

but shouldn't this work?

yes, it should. what's the value of i when it crashes ? and how long is the string?

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[Software2007]

I send in a string "Test"; 19 characters long. The loop crashes at index 25 for(int i=0;msg[i];i++) { if(msg[i] == '<') msg.replace(i,1,"<",0,4); } The string actually becomes 25 characters after replacing '<' with //"<", so the very last iteration, it increments i to 25, it tries to check the condition msg[25], I believe it crashes since the string is only 24 chars long ?

modified on Wednesday, September 14, 2011 4:09 PM

[David Crow]

Software2007 wrote:

The loop crashes at index 25

How are you calling replace_html_delimiters()?

Software2007 wrote:

msg.replace(i,1,"<",0,4);

Crashing aside, why are you replacing < with the string equivalent? Wouldn't a simple msg[i] = '<'; suffice?

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[Software2007]

I am trying to replace the C code in my original post. Unfortuantely, CodeProject keeps replacing my "& l t ;" with "<". I had to put spaces just now to display it, but in the code, there is no space between the 4 letters.

[Chris Losinger]

this works for me:

#include <string>

void repl(std::string &msg)
{
for(int i=0;msg[i];i++)
{
if(msg[i] == '<')
msg.replace(i,1,"<",0,4);
}
}

int _tmain(int argc, _TCHAR* argv[])
{
std::string f = "<title>blah</title>";

repl(f);

return 0;

}

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[Software2007]

The same code you showed me crashes at index 25! I am running just simple console app in VS2010. It works on VS2008! Oh well! Thanks for your help

modified on Wednesday, September 14, 2011 4:31 PM

[Chris Losinger]

how about a generic std::string find/replace fn:

void repl2(std::string &str, const char *find, const char *repl)
{
// should probably quit if either of these are 0!
size_t findLen = strlen(find);
size_t replLen = strlen(repl);

size_t index = 0;
while (true) {
/* Locate the substring to replace. */
index = str.find(find, index);
if (index == std::string::npos) break;

  /\* Make the replacement. \*/
  str.replace(index, findLen, repl);

  /\* Advance index forward so the next iteration doesn't pick it up as well. \*/
  index+=replLen;

}
}

that's based on something from this thread[^].

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[Chuck OToole]

Well, I'm not even sure the straight C code works. 'msg' is a 'char *', meaning it points to a string of characters owned by the caller of the function. Presumably that string is in some buffer allocated by the caller. When you complete the replace using the temporary 'z_buf' buffer (presumably large enough, but assume it is), you them move the new (possibly longer) string back into the buffer pointed to by 'msg'. How do you know you aren't overwriting that buffer? How do you know it's large enough to receive the replaced string? I'm just saying, you could be clobbering something important.

[Chuck OToole]

for (int i=0; msg[i]; i++) is equivalent to: for (int i=0; msg[i]!=0; i++) The end condition is the character at msg[i] being the null at the end of the string. The compiler didn't like msg[i]=0 because that's an assignment, not an equality test. Need another '=' in there.

[Chuck OToole]

In the C++ example, just what does

strcpy(msg,z_buf);

do to the string object? Assuming it compiles, are you clobbering the object? I don't use the std:: classes much, but if this were MFC/ATL CString, the strcpy argument would use a typecasting to get a pointer to the internal buffer of CString (char *) which you are *NEVER ALLOWED TO OVERWRITE*. Yet the strcpy() call does exactly that. I would imagine that std::string has a similar internal structure and would be very upset if you started overwriting its content.

[Chuck OToole]

It is my understanding that std::string does *NOT* include a NULL ('\0') character at the end of the string. One cannot assume a null termination. So, basically you are using C style assumptions on C++ string objects. The way to deal with std::string is through the member functions string.length(), string.replace(), etc. The examples others have shown you work because they stay within the object's definition of operative functions. There is a string.c_str() member function that returns a pointer to a C style null terminated char * (http://www.cplusplus.com/reference/string/string/c_str/[^]) but that too cannot be modified by the receiving program. If you're going to convert from C to C++, you should go all the way and avoid those old char * uses and move to some string class, either std::string or MFC/ATL CString, depending on your project's needs.

[enhzflep]

The reason there's no upper limit on "i" in either of the for loops is that this would take more code - it would require a strlen be performed once before the loop in addition to checking to see if i is equal to this length. It's less clear to read and more prone to induce error during maintenance, but I believe it to be for this reason that the way the loop is exited with a break. Not sure why you'd go the trouble of #defining NUL as 0x0.. It would be clearer if the already provided NULL was used (less code too, since there's only 4 references to 'NUL' - 4 cases of simply adding another 'L'. In any case, the executable code is identical - it is just the source-code that suffers from reduced readability, unlike the loop-terminating-condition check, which produces a smaller executable when done this way than the more readable alternative of checking the strlen first then using a terminarting condition of ichar *htmlStr = ""; replace_html_delimiters(htmlStr); while I can see this succeeding

char *htmlStr = "";
char htmlStrCopy = strdup(htmlStr);
replace_html_delimiters(htmlStrCopy);
..
.. other actions on htmlStrCopy
..
free(htmlStrCopy);

Add to that the fact that the "char z_buf[4095]" statement isn't terminated with a ';' in either case and you have rather a problem, considering that the C function called with a string containing "" returns the same string. Studying the function, it appears to scan through a string, quitting upon end of string (0x0), while perplexingly, when it intercepts a '<' character it copies all of the text except for this character, then it appends the '<' explicitly. It's 5am here, and I can't think of a circumstance that the output sring would be different to the input string.

[Chuck OToole]

You're right in that the code looks odd but I think it was because the Code Project Editor messed it up. The OP was trying to replace the < character with the sequence ampersand-l-t, a sequence which if typed into this editor will yield a <, making the code look wrong. He's really making the string bigger with the replacements.