I've got a programming question about the implementation of strcat() function.
I have been trying to solve that problem and I got some Access violation error.
My created function:
char str_cat(char str1, char str2)
{
return str1-'\0'+str2;
}
what is wrong in the above code?
One more question please,
is "iostream" a header file? where can I get it?
thanks
Unfortunately, everything is wrong with this function, even the return type and argument types. It should look like
char * strcat(const char *str1, const char *str2)
and it should work by allocating a new block of memory large enough to hold the concatenated strings using either malloc (for C) or new (for C++), then copy both strings into it. I think you've got your (home)work cut out for you, though, as I don't think you know much of what any of that means.
Nothing is right in the above code.
You need to take char * parameters
You need to return a char * if you have to return something (which isn't needed)
You'll need to loop over the string copying individual characters - no easy solution with + and -
You'll need to 0-terminate the result
E.g. like this:
void strcat(char * Dest, char const * Src) {
char * d = Dest;
while (*d++);
char const * s = Src;
while (*s) { *d++ = *s++; }
*d = 0;
}
Why do you need to do this? There's a perfectly good strcat in the standard library, and there's a perfectly good std::string which you can use + on.
Don't want to sound negative but there is not much right with this code.
Firstly, strings in C are char*, not char.
Second, there is no way to 'add' or 'subtract' them the way you would hope (which is sort of kind of possible in, say, python).
iostream is the standard I/O header for C++, it should be bundled with your distribution.
I would really suggest a tutorial on pointers to get you going - this I found just by googling "ansi c pointers" - I'm guessing the problem asks you for a C answer as opposed to C++, since in C++ you would use std::string and the overloaded operator+.
Related
There is a function which sends data to the server:
int send(
_In_ SOCKET s,
_In_ const char *buf,
_In_ int len,
_In_ int flags
);
Providing length seems to me a little bit weird. I need to write a function, sending a line to the server and wrapping this one such that we don't have to provide length explicitly. I'm a Java-developer and in Java we could just invoke String::length() method, but now we're not in Java. How can I do that, unless providing length as a template parameter? For instance:
void sendLine(SOCKET s, const char *buf)
{
}
Is it possible to implement such a function?
Use std string:
void sendLine(SOCKET s, const std::string& buf) {
send (s, buf.c_str(), buf.size()+1, 0); //+1 will also transmit terminating \0.
}
On a side note: your wrapper function ignores the return value and doesn't take any flags.
you can retrieve the length of C-string by using strlen(const char*) function.
make sure all the strings are null terminated and keep in mind that null-termination (the length grows by 1)
Edit: My answer originally only mentioned std::string. I've now also added std::vector<char> to account for situations where send is not used for strictly textual data.
First of all, you absolutely need a C++ book. You are looking for either the std::string class or for std::vector<char>, both of which are fundamental elements of the language.
Your question is a bit like asking, in Java, how to avoid char[] because you never heard of java.lang.String, or how to avoid arrays in general because you never heard of java.util.ArrayList.
For the first part of this answer, let's assume you are dealing with just text output here, i.e. with output where a char is really meant to be a text character. That's the std::string use case.
Providing lenght seems to me a little bit wierd.
That's the way strings work in C. A C string is really a pointer to a memory location where characters are stored. Normally, C strings are null-terminated. This means that the last character stored for the string is '\0'. It means "the string stops here, and if you move further, you enter illegal territory".
Here is a C-style example:
#include <string.h>
#include <stdio.h>
void f(char const* s)
{
int l = strlen(s); // l = 3
printf(s); // prints "foo"
}
int main()
{
char* test = new char[4]; // avoid new[] in real programs
test[0] = 'f';
test[1] = 'o';
test[2] = 'o';
test[3] = '\0';
f(test);
delete[] test;
}
strlen just counts all characters at the specified position in memory until it finds '\0'. printf just writes all characters at the specified position in memory until it finds '\0'.
So far, so good. Now what happens if someone forgets about the null terminator?
char* test = new char[3]; // don't do this at home, please
test[0] = 'f';
test[1] = 'o';
test[2] = 'o';
f(test); // uh-oh, there is no null terminator...
The result will be undefined behaviour. strlen will keep looking for '\0'. So will printf. The functions will try to read memory they are not supposed to. The program is allowed to do anything, including crashing. The evil thing is that most likely, nothing will happen for a while because a '\0' just happens to be stored there in memory, until one day you are not so lucky anymore.
That's why C functions are sometimes made safer by requiring you to explicitly specify the number of characters. Your send is such a function. It works fine even without null-terminated strings.
So much for C strings. And now please don't use them in your C++ code. Use std::string. It is designed to be compatible with C functions by providing the c_str() member function, which returns a null-terminated char const * pointing to the contents of the string, and it of course has a size() member function to tell you the number of characters without the null-terminated character (e.g. for a std::string representing the word "foo", size() would be 3, not 4, and 3 is also what a C function like yours would probably expect, but you have to look at the documentation of the function to find out whether it needs the number of visible characters or number of elements in memory).
In fact, with std::string you can just forget about the whole null-termination business. Everything is nicely automated. std::string is exactly as easy and safe to use as java.lang.String.
Your sendLine should thus become:
void sendLine(SOCKET s, std::string const& line)
{
send(s, line.c_str(), line.size());
}
(Passing a std::string by const& is the normal way of passing big objects in C++. It's just for performance, but it's such a widely-used convention that your code would look strange if you just passed std::string.)
How can I do that, unless providing lenght as a template parameter?
This is a misunderstanding of how templates work. With a template, the length would have to be known at compile time. That's certainly not what you intended.
Now, for the second part of the answer, perhaps you aren't really dealing with text here. It's unlikely, as the name "sendLine" in your example sounds very much like text, but perhaps you are dealing with raw data, and a char in your output does not represent a text character but just a value to be interpreted as something completely different, such as the contents of an image file.
In that case, std::string is a poor choice. Your output could contain '\0' characters that do not have the meaning of "data ends here", but which are part of the normal contents. In other words, you don't really have strings anymore, you have a range of char elements in which '\0' has no special meaning.
For this situation, C++ offers the std::vector template, which you can use as std::vector<char>. It is also designed to be usable with C functions by providing a member function that returns a char pointer. Here's an example:
void sendLine(SOCKET s, std::vector<char> const& data)
{
send(s, &data[0], data.size());
}
(The unusual &data[0] syntax means "pointer to the first element of the encapsulated data. C++11 has nicer-to-read ways of doing this, but &data[0] also works in older versions of C++.)
Things to keep in mind:
std::string is like String in Java.
std::vector is like ArrayList in Java.
std::string is for a range of char with the meaning of text, std::vector<char> is for a range of char with the meaning of raw data.
std::string and std::vector are designed to work together with C APIs.
Do not use new[] in C++.
Understand the null termination of C strings.
which is the best way to concat?
const char * s1= "\nInit() failed: ";
const char * s2 = "\n";
char buf[100];
strcpy(buf, s1);
strcat(buf, initError);
strcat(buf, s2);
wprintf(buf);
It gives error. What should be the correct way?
Thanks.
I think the correct way is:
std::string msg = std::string("Init() Failed ") + initError + "\n";
std::cout<<msg;
or
std::cout<<"Init() Failed "<<initError<<"\n";
Your big problem is that you're mixing data types. Use either char and associated functions or wchar and associated functions. If you need to mix them, use a conversion function. This makes no more sense than trying to pass a float to a function needing a string. (The compiler should be able to catch both problems, since the declaration of wprintf is something like int wprintf(const wchar_t *, ...).)
Another, more minor, issue is that printf and such are not the right functions to print out general strings, since if there are any percent signs in the strings you'll get undefined behavior. Use printf("%s",...) or puts(...) or related functions.
And, since this is C++, you'd be much better off using the std::string class. It isn't perfect, but it's far better than C-style strings.
Also, telling us what the error is would help. You haven't even told us whether it's a compiler error or a run-time error.
char * msg = new char[65546];
want to initialize to 0 for all of them. what is the best way to do this in C++?
char * msg = new char[65546]();
It's known as value-initialisation, and was introduced in C++03. If you happen to find yourself trapped in a previous decade, then you'll need to use std::fill() (or memset() if you want to pretend it's C).
Note that this won't work for any value other than zero. I think C++0x will offer a way to do that, but I'm a bit behind the times so I can't comment on that.
UPDATE: it seems my ruminations on the past and future of the language aren't entirely accurate; see the comments for corrections.
The "most C++" way to do this would be to use std::fill.
std::fill(msg, msg + 65546, 0);
Absent a really good reason to do otherwise, I'd probably use:
std::vector<char> msg(65546, '\0');
what is the best way to do this in
C++?
Because you asked it this way:
std::string msg(65546, 0); // all characters will be set to 0
Or:
std::vector<char> msg(65546); // all characters will be initialized to 0
If you are working with C functions which accept char* or const char*, then you can do:
some_c_function(&msg[0]);
You can also use the c_str() method on std::string if it accepts const char* or data().
The benefit of this approach is that you can do everything you want to do with a dynamically allocating char buffer but more safely, flexibly, and sometimes even more efficiently (avoiding the need to recompute string length linearly, e.g.). Best of all, you don't have to free the memory allocated manually, as the destructor will do this for you.
This method uses the 'C' memset function, and is very fast (avoids a char-by-char loop).
const uint size = 65546;
char* msg = new char[size];
memset(reinterpret_cast<void*>(msg), 0, size);
memset(msg, 0, 65546)
You can use a for loop. but don't forget the last char must be a null character !
char * msg = new char[65546];
for(int i=0;i<65545;i++)
{
msg[i]='0';
}
msg[65545]='\0';
The C-like method may not be as attractive as the other solutions to this question, but added here for completeness:
You can initialise with NULLs like this:
char msg[65536] = {0};
Or to use zeros consider the following:
char msg[65536] = {'0' another 65535 of these separated by comma};
But do not try it as not possible, so use memset!
In the second case, add the following after the memset if you want to use msg as a string.
msg[65536 - 1] = '\0'
Answers to this question also provide further insight.
If you panic and can not assign dynamic data to a const char* in a constructor you can insert each element of a dynamic buffer piece by piece. You can even snprintf() to the buffer before making the imprint.
client_id = new char[26] {
buf[0],buf[1],buf[2],buf[3],buf[4],buf[5],buf[6],buf[7],buf[8],buf[9],
buf[10],buf[11],buf[12],buf[13],buf[14],buf[15],buf[16],buf[17],buf[18],buf[19],
buf[20],buf[21],buf[22],buf[23],buf[24],'\0'
};
To cover up what you have been doing, maybe the editor has an option where you can set the forecolor same as the background?
Before being fired you can actually prime the const char in the header file declaration with enough space and then later assign real data in the constructor. Great!
const char* client_id = "\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0";
It is a const pointer and does not have to be initialized before the constructor deals with it.
const char* client_id;
NOTE:
You can write at the top of the page: using namespace std,
and thus avoid writing std:: at the beginning of each command.
char * msg = new char[65546]={0};
This command reset all the array to 0.
I made a function like this:
bool IsSameString(char* p1, char* p2)
{
return 0 == strcmp(p1, p2);
}
The problem is that sometimes, by mistake, arguments are passed which are not strings (meaning that p1 or p2 is not terminated with a null character).
Then, strcmp continues comparing until it reaches non-accessible memory and crashes.
Is there a safe version of strcmp? Or can I tell whether p1 (and p2) is a string or not in a safe manner?
No, there's no (standard) way to tell whether a char * actually points to valid memory.
In your situation, it is better to use std::string rather than char *s for all your strings, along with the overloaded == operator. If you do this, the compiler would enforce type safety.
EDIT: As per the comments below if you find yourself in a situation where you sometimes pass char *s that may or may not be valid strings to functions that expect null-terminated strings then something is fundamentally wrong with your approach, so basically
#janm's answer below.
In some cases std::strncmp can solve your problem:
int strncmp ( const char * str1, const char * str2, size_t num );
It compares up to num characters of the C string str1 to those of the C string str2.
Also, take a look, what the US DHS National Cyber Security Division recommends on this matter:
Ensure that strings are null terminated before passing into strcmp. This can be enforced by always placing a \0 in the last allocated byte of the buffer.
char str1[] ="something";
char str2[] = "another thing";
/* In this case we know strings are null terminated. Pretend we don't. */
str1[sizeof(str1)-1] = '\0';
str2[sizeof(str2)-1] = '\0';
/* Now the following is safe. */
if (strcmp(str1, str2)) { /* do something */ } else { /* do something else */ }
If you are passing strings to strcmp() that are not null terminated you have already lost. The fact that you have a string that is not null terminated (but should be) indicates that you have deeper issues in your code. You cannot change strcmp() to safely deal with this problem.
You should be writing your code so that can never happen. Start by using the string class. At the boundaries where you take data into your code you need to make sure you deal with the exceptional cases; if you get too much data you need to Do The Right Thing. That does not involve running off the end of your buffer. If you must perform I/O into a C style buffer, use functions where you specify the length of the buffer and detect and deal with cases where the buffer is not large enough at that point.
There's no cure for this that is portable. The convention states that there's an extra character holding a null character that belongs to the same correctly allocated block of memory as the string itself. Either this convention is followed and everything's fine or undefined behaviour occurs.
If you know the length of the string you compare against you can use strncmp() but his will not help if the string passed to your code is actually shorter than the string you compare against.
you can use strncmp, But if possible use std::string to avoid many problems :)
You can put an upper limit on the number of characters to be compared using the strncmp function.
There is no best answer to this as you can't verify the char* is a string. The only solution is to create a type and use it for string for example str::string or create your own if you want something lighter. ie
struct MyString
{
MyString() : str(0), len(0) {}
MyString( char* x ) { len = strlen(x); str = strdup(x); }
âMyString() { if(str) free(str); }
char* str;
size_t len;
};
bool IsSameString(MyString& p1, MyString& p2)
{
return 0 == strcmp(p1.str, p2.str);
}
MyString str1("test");
MyString str2("test");
if( IsSameString( str1, str2 ) {}
You dont write, what platform you are using. Windows has the following functions:
IsBadStringPtr
IsBadReadPtr
IsBadWritePtr
IsBadStringPtr might be what you are looking for, if you are using windows.
Const-correctness in C++ is still giving me headaches. In working with some old C code, I find myself needing to assign turn a C++ string object into a C string and assign it to a variable. However, the variable is a char * and c_str() returns a const char []. Is there a good way to get around this without having to roll my own function to do it?
edit: I am also trying to avoid calling new. I will gladly trade slightly more complicated code for less memory leaks.
C++17 and newer:
foo(s.data(), s.size());
C++11, C++14:
foo(&s[0], s.size());
However this needs a note of caution: The result of &s[0]/s.data()/s.c_str() is only guaranteed to be valid until any member function is invoked that might change the string. So you should not store the result of these operations anywhere. The safest is to be done with them at the end of the full expression, as my examples do.
Pre C++-11 answer:
Since for to me inexplicable reasons nobody answered this the way I do now, and since other questions are now being closed pointing to this one, I'll add this here, even though coming a year too late will mean that it hangs at the very bottom of the pile...
With C++03, std::string isn't guaranteed to store its characters in a contiguous piece of memory, and the result of c_str() doesn't need to point to the string's internal buffer, so the only way guaranteed to work is this:
std::vector<char> buffer(s.begin(), s.end());
foo(&buffer[0], buffer.size());
s.assign(buffer.begin(), buffer.end());
This is no longer true in C++11.
There is an important distinction you need to make here: is the char* to which you wish to assign this "morally constant"? That is, is casting away const-ness just a technicality, and you really will still treat the string as a const? In that case, you can use a cast - either C-style or a C++-style const_cast. As long as you (and anyone else who ever maintains this code) have the discipline to treat that char* as a const char*, you'll be fine, but the compiler will no longer be watching your back, so if you ever treat it as a non-const you may be modifying a buffer that something else in your code relies upon.
If your char* is going to be treated as non-const, and you intend to modify what it points to, you must copy the returned string, not cast away its const-ness.
I guess there is always strcpy.
Or use char* strings in the parts of your C++ code that must interface with the old stuff.
Or refactor the existing code to compile with the C++ compiler and then to use std:string.
There's always const_cast...
std::string s("hello world");
char *p = const_cast<char *>(s.c_str());
Of course, that's basically subverting the type system, but sometimes it's necessary when integrating with older code.
If you can afford extra allocation, instead of a recommended strcpy I would consider using std::vector<char> like this:
// suppose you have your string:
std::string some_string("hello world");
// you can make a vector from it like this:
std::vector<char> some_buffer(some_string.begin(), some_string.end());
// suppose your C function is declared like this:
// some_c_function(char *buffer);
// you can just pass this vector to it like this:
some_c_function(&some_buffer[0]);
// if that function wants a buffer size as well,
// just give it some_buffer.size()
To me this is a bit more of a C++ way than strcpy. Take a look at Meyers' Effective STL Item 16 for a much nicer explanation than I could ever provide.
You can use the copy method:
len = myStr.copy(cStr, myStr.length());
cStr[len] = '\0';
Where myStr is your C++ string and cStr a char * with at least myStr.length()+1 size. Also, len is of type size_t and is needed, because copy doesn't null-terminate cStr.
Just use const_cast<char*>(str.data())
Do not feel bad or weird about it, it's perfectly good style to do this.
It's guaranteed to work in C++11. The fact that it's const qualified at all is arguably a mistake by the original standard before it; in C++03 it was possible to implement string as a discontinuous list of memory, but no one ever did it. There is not a compiler on earth that implements string as anything other than a contiguous block of memory, so feel free to treat it as such with complete confidence.
If you know that the std::string is not going to change, a C-style cast will work.
std::string s("hello");
char *p = (char *)s.c_str();
Of course, p is pointing to some buffer managed by the std::string. If the std::string goes out of scope or the buffer is changed (i.e., written to), p will probably be invalid.
The safest thing to do would be to copy the string if refactoring the code is out of the question.
std::string vString;
vString.resize(256); // allocate some space, up to you
char* vStringPtr(&vString.front());
// assign the value to the string (by using a function that copies the value).
// don't exceed vString.size() here!
// now make sure you erase the extra capacity after the first encountered \0.
vString.erase(std::find(vString.begin(), vString.end(), 0), vString.end());
// and here you have the C++ string with the proper value and bounds.
This is how you turn a C++ string to a C string. But make sure you know what you're doing, as it's really easy to step out of bounds using raw string functions. There are moments when this is necessary.
If c_str() is returning to you a copy of the string object internal buffer, you can just use const_cast<>.
However, if c_str() is giving you direct access tot he string object internal buffer, make an explicit copy, instead of removing the const.
Since c_str() gives you direct const access to the data structure, you probably shouldn't cast it. The simplest way to do it without having to preallocate a buffer is to just use strdup.
char* tmpptr;
tmpptr = strdup(myStringVar.c_str();
oldfunction(tmpptr);
free tmpptr;
It's quick, easy, and correct.
In CPP, if you want a char * from a string.c_str()
(to give it for example to a function that only takes a char *),
you can cast it to char * directly to lose the const from .c_str()
Example:
launchGame((char *) string.c_str());
C++17 adds a char* string::data() noexcept overload. So if your string object isn't const, the pointer returned by data() isn't either and you can use that.
Is it really that difficult to do yourself?
#include <string>
#include <cstring>
char *convert(std::string str)
{
size_t len = str.length();
char *buf = new char[len + 1];
memcpy(buf, str.data(), len);
buf[len] = '\0';
return buf;
}
char *convert(std::string str, char *buf, size_t len)
{
memcpy(buf, str.data(), len - 1);
buf[len - 1] = '\0';
return buf;
}
// A crazy template solution to avoid passing in the array length
// but loses the ability to pass in a dynamically allocated buffer
template <size_t len>
char *convert(std::string str, char (&buf)[len])
{
memcpy(buf, str.data(), len - 1);
buf[len - 1] = '\0';
return buf;
}
Usage:
std::string str = "Hello";
// Use buffer we've allocated
char buf[10];
convert(str, buf);
// Use buffer allocated for us
char *buf = convert(str);
delete [] buf;
// Use dynamic buffer of known length
buf = new char[10];
convert(str, buf, 10);
delete [] buf;