I'm interested (for various reasons) format using sprintf with the result in a std::string object. The most syntactically direct way I came up with is:
char* buf;
sprintf(buf, ...);
std::string s(buf);
Any other ideas?
Don't use the printf line of functions for formatting in C++. Use the language feature of streams (stringstreams in this case) which are type safe and don't require the user to provide special format characters.
If you really really want to do that, you could pre-allocate enough space in your string and then resize it down although I'm not sure if that's more efficient than using a temporary char buffer array:
std::string foo;
foo.resize(max_length);
int num_bytes = snprintf(&foo[0], max_length, ...);
if(num_bytes < max_length)
{
foo.resize(num_bytes);
}
If you absolutely want to use sprintf, there is no more direct way than the one you wrote.
sprintf Writes into the array pointed by str a C string
If you want to transform this C string into a std::string, there is no better and safer way than the one you wrote.
Answered here. To get length of buffer needed, call:
snprintf( nullptr, 0, format, args... )
You can write short function which can be used as such:
std::string s = string_sprintf("%g, %g\n", 1.23, 0.001);
Example in linked stackoverflow answer.
Related
I'm getting warnings when I compile something like this...
std::string something = "bacon";
sprintf("I love %s a lot", something.c_str());
Where it says "warning: deprecated conversion from string constant to 'char *'. I tried converting the text to be...
const char *
instead but I get a different error. I'm not committed to sprintf if there is a better option.
For sprintf to work, you need to provide an array of char big enough to write the result to as the first argument.
However, you can (and should!) just use the far easier operator+ for C++ strings:
std::string res = "I love " + something + " a lot";
sprintf("I love %s a lot", something.c_str);
In that code, you should call something.c_str() with proper function call () syntax.
Note also that the above use of sprintf() is wrong, since you didn't provide a valid destination string buffer for the resulting formatted string.
Moreover, for security reasons, you should use the safer snprintf() instead of sprintf(). In fact, with snprintf() you can specify the size of the destination buffer, to avoid buffer overruns.
The following compilable code is an example of snprintf() usage:
#include <stdio.h>
#include <string>
int main()
{
std::string something = "bacon";
char buf[128];
snprintf(buf, sizeof(buf), "I love %s a lot", something.c_str());
printf("%s\n", buf);
}
P.S.
In general, in C++ you may consider string concatenation using std::string::operator+, e.g.:
std::string result = "I love " + something + " a lot";
It doesn't look like a correct use of sprintf.
First parameter is supposed to be a char * with already a backing memory.
For example:
char *str = malloc (BUFSIZ);
sprintf (str, "I love %s a lot", something.c_str);
I have been working with C++ strings and trying to load char * strings into std::string by using C functions such as strcpy(). Since strcpy() takes char * as a parameter, I have to cast it which goes something like this:
std::string destination;
unsigned char *source;
strcpy((char*)destination.c_str(), (char*)source);
The code works fine and when I run the program in a debugger, the value of *source is stored in destination, but for some odd reason it won't print out with the statement
std::cout << destination;
I noticed that if I use
std::cout << destination.c_str();
The value prints out correctly and all is well. Why does this happen? Is there a better method of copying an unsigned char* or char* into a std::string (stringstreams?) This seems to only happen when I specify the string as foo.c_str() in a copying operation.
Edit: To answer the question "why would you do this?", I am using strcpy() as a plain example. There are other times that it's more complex than assignment. For example, having to copy only X amount of string A into string B using strncpy() or passing a std::string to a function from a C library that takes a char * as a parameter for a buffer.
Here's what you want
std::string destination = source;
What you're doing is wrong on so many levels... you're writing over the inner representation of a std::string... I mean... not cool man... it's much more complex than that, arrays being resized, read-only memory... the works.
This is not a good idea at all for two reasons:
destination.c_str() is a const pointer and casting away it's const and writing to it is undefined behavior.
You haven't set the size of the string, meaning that it won't even necessealy have a large enough buffer to hold the string which is likely to cause an access violation.
std::string has a constructor which allows it to be constructed from a char* so simply write:
std::string destination = source
Well what you are doing is undefined behavior. Your c_str() returns a const char * and is not meant to be assigned to. Why not use the defined constructor or assignment operator.
std::string defines an implicit conversion from const char* to std::string... so use that.
You decided to cast away an error as c_str() returns a const char*, i.e., it does not allow for writing to its underlying buffer. You did everything you could to get around that and it didn't work (you shouldn't be surprised at this).
c_str() returns a const char* for good reason. You have no idea if this pointer points to the string's underlying buffer. You have no idea if this pointer points to a memory block large enough to hold your new string. The library is using its interface to tell you exactly how the return value of c_str() should be used and you're ignoring that completely.
Do not do what you are doing!!!
I repeat!
DO NOT DO WHAT YOU ARE DOING!!!
That it seems to sort of work when you do some weird things is a consequence of how the string class was implemented. You are almost certainly writing in memory you shouldn't be and a bunch of other bogus stuff.
When you need to interact with a C function that writes to a buffer there's two basic methods:
std::string read_from_sock(int sock) {
char buffer[1024] = "";
int recv = read(sock, buffer, 1024);
if (recv > 0) {
return std::string(buffer, buffer + recv);
}
return std::string();
}
Or you might try the peek method:
std::string read_from_sock(int sock) {
int recv = read(sock, 0, 0, MSG_PEEK);
if (recv > 0) {
std::vector<char> buf(recv);
recv = read(sock, &buf[0], recv, 0);
return std::string(buf.begin(), buf.end());
}
return std::string();
}
Of course, these are not very robust versions...but they illustrate the point.
First you should note that the value returned by c_str is a const char* and must not be modified. Actually it even does not have to point to the internal buffer of string.
In response to your edit:
having to copy only X amount of string A into string B using strncpy()
If string A is a char array, and string B is std::string, and strlen(A) >= X, then you can do this:
B.assign(A, A + X);
passing a std::string to a function from a C library that takes a char
* as a parameter for a buffer
If the parameter is actually const char *, you can use c_str() for that. But if it is just plain char *, and you are using a C++11 compliant compiler, then you can do the following:
c_function(&B[0]);
However, you need to ensure that there is room in the string for the data(same as if you were using a plain c-string), which you can do with a call to the resize() function. If the function writes an unspecified amount of characters to the string as a null-terminated c-string, then you will probably want to truncate the string afterward, like this:
B.resize(B.find('\0'));
The reason you can safely do this in a C++11 compiler and not a C++03 compiler is that in C++03, strings were not guaranteed by the standard to be contiguous, but in C++11, they are. If you want the guarantee in C++03, then you can use std::vector<char> instead.
I've been wondering about the following issue: assume I have a C style function that reads raw data into a buffer
int recv_n(int handle, void* buf, size_t len);
Can I read the data directly into an std:string or stringstream without allocating any temporal buffers? For example,
std::string s(100, '\0');
recv_n(handle, s.data(), 100);
I guess this solution has an undefined outcome, because, afaik, string::c_str and string::data might return a temporal location and not necessarily return the pointer to the real place in the memory, used by the object to store the data.
Any ideas?
Why not use a vector<char> instead of a string? That way you can do:
vector<char> v(100, '\0');
recv_n(handle, &v[0], 100);
This seems more idiomatic to me, especially since you aren't using it as a string (you say it's raw data).
Yes, after C++11.
But you cant use s.data() as it returns a char const*
Try:
std::string s(100, '\0');
recv_n(handle, &s[0], 100);
Depending on situation, I may have chosen a std::vector<char> especially for raw data (though it would all depend on usage of the data in your application).
I want to copy a string into a char array, and not overrun the buffer.
So if I have a char array of size 5, then I want to copy a maximum of 5 bytes from a string into it.
what's the code to do that?
This is exactly what std::string's copy function does.
#include <string>
#include <iostream>
int main()
{
char test[5];
std::string str( "Hello, world" );
str.copy(test, 5);
std::cout.write(test, 5);
std::cout.put('\n');
return 0;
}
If you need null termination you should do something like this:
str.copy(test, 4);
test[4] = '\0';
First of all, strncpy is almost certainly not what you want. strncpy was designed for a fairly specific purpose. It's in the standard library almost exclusively because it already exists, not because it's generally useful.
Probably the simplest way to do what you want is with something like:
sprintf(buffer, "%.4s", your_string.c_str());
Unlike strncpy, this guarantees that the result will be NUL terminated, but does not fill in extra data in the target if the source is shorter than specified (though the latter isn't a major issue when the target length is 5).
Use function strlcpybroken link, and material not found on destination site if your implementation provides one (the function is not in the standard C library), yet it is rather widely accepted as a de-facto standard name for a "safe" limited-length copying function for zero-terminated strings.
If your implementation does not provide strlcpy function, implement one yourself. For example, something like this might work for you
char *my_strlcpy(char *dst, const char *src, size_t n)
{
assert(dst != NULL && src != NULL);
if (n > 0)
{
char *pd;
const char *ps;
for (--n, pd = dst, ps = src; n > 0 && *ps != '\0'; --n, ++pd, ++ps)
*pd = *ps;
*pd = '\0';
}
return dst;
}
(Actually, the de-facto accepted strlcpy returns size_t, so you might prefer to implement the accepted specification instead of what I did above).
Beware of the answers that recommend using strncpy for that purpose. strncpy is not a safe limited-length string copying function and is not supposed to be used for that purpose. While you can force strncpy to "work" for that purpose, it is still akin to driving woodscrews with a hammer.
Update: Thought I would try to tie together some of the answers, answers which have convinced me that my own original knee-jerk strncpy response was poor.
First, as AndreyT noted in the comments to this question, truncation methods (snprintf, strlcpy, and strncpy) are often not a good solution. Its often better to check the size of the string string.size() against the buffer length and return/throw an error or resize the buffer.
If truncation is OK in your situation, IMHO, strlcpy is the best solution, being the fastest/least overhead method that ensures null termination. Unfortunately, its not in many/all standard distributions and so is not portable. If you are doing a lot of these, it maybe worth providing your own implementation, AndreyT gave an example. It runs in O(result length). Also the reference specification returns the number of bytes copied, which can assist in detecting if the source was truncated.
Other good solutions are sprintf and snprintf. They are standard, and so are portable and provide a safe null terminated result. They have more overhead than strlcpy (parsing the format string specifier and variable argument list), but unless you are doing a lot of these you probably won't notice the difference. It also runs in O(result length). snprintf is always safe and that sprintf may overflow if you get the format specifier wrong (as other have noted, format string should be "%.<N>s" not "%<N>s"). These methods also return the number of bytes copied.
A special case solution is strncpy. It runs in O(buffer length), because if it reaches the end of the src it zeros out the remainder of the buffer. Only useful if you need to zero the tail of the buffer or are confident that destination and source string lengths are the same. Also note that it is not safe in that it doesn't necessarily null terminate the string. If the source is truncated, then null will not be appended, so call in sequence with a null assignment to ensure null termination: strncpy(buffer, str.c_str(), BUFFER_LAST); buffer[BUFFER_LAST] = '\0';
Some nice libc versions provide non-standard but great replacement for strcpy(3)/strncpy(3) - strlcpy(3).
If yours doesn't, the source code is freely available here from the OpenBSD repository.
void stringChange(string var){
char strArray[100];
strcpy(strArray, var.c_str());
}
I guess this should work. it'll copy form string to an char array.
i think snprintf() is much safe and simlest
snprintf ( buffer, 100, "The half of %d is %d", 60, 60/2 );
null character is append it end automatically :)
The most popular answer is fine but the null-termination is not generic. The generic way to null-terminate the char-buffer is:
std::string aString = "foo";
const size_t BUF_LEN = 5;
char buf[BUF_LEN];
size_t len = aString.copy(buf, BUF_LEN-1); // leave one char for the null-termination
buf[len] = '\0';
len is the number of chars copied so it's between 0 and BUF_LEN-1.
std::string my_string("something");
char* my_char_array = new char[5];
strncpy(my_char_array, my_string.c_str(), 4);
my_char_array[4] = '\0'; // my_char_array contains "some"
With strncpy, you can copy at most n characters from the source to the destination. However, note that if the source string is at most n chars long, the destination will not be null terminated; you must put the terminating null character into it yourself.
A char array with a length of 5 can contain at most a string of 4 characters, since the 5th must be the terminating null character. Hence in the above code, n = 4.
std::string str = "Your string";
char buffer[5];
strncpy(buffer, str.c_str(), sizeof(buffer));
buffer[sizeof(buffer)-1] = '\0';
The last line is required because strncpy isn't guaranteed to NUL terminate the string (there has been a discussion about the motivation yesterday).
If you used wide strings, instead of sizeof(buffer) you'd use sizeof(buffer)/sizeof(*buffer), or, even better, a macro like
#define ARRSIZE(arr) (sizeof(arr)/sizeof(*(arr)))
/* ... */
buffer[ARRSIZE(buffer)-1]='\0';
char mystring[101]; // a 100 character string plus terminator
char *any_input;
any_input = "Example";
iterate = 0;
while ( any_input[iterate] != '\0' && iterate < 100) {
mystring[iterate] = any_input[iterate];
iterate++;
}
mystring[iterate] = '\0';
This is the basic efficient design.
If you always have a buffer of size 5, then you could do:
std::string s = "Your string";
char buffer[5]={s[0],s[1],s[2],s[3],'\0'};
Edit:
Of course, assuming that your std::string is large enough.
I'm working in a C++ unmanaged project.
I need to know how can I take a string like this "some data to encrypt" and get a byte[] array which I'm gonna use as the source for Encrypt.
In C# I do
for (int i = 0; i < text.Length; i++)
buffer[i] = (byte)text[i];
What I need to know is how to do the same but using unmanaged C++.
Thanks!
If you just need read-only access, then c_str() will do it:
char const *c = myString.c_str();
If you need read/write access, then you can copy the string into a vector. vectors manage dynamic memory for you. You don't have to mess with allocation/deallocation then:
std::vector<char> bytes(myString.begin(), myString.end());
bytes.push_back('\0');
char *c = &bytes[0];
std::string::data would seem to be sufficient and most efficient. If you want to have non-const memory to manipulate (strange for encryption) you can copy the data to a buffer using memcpy:
unsigned char buffer[mystring.length()];
memcpy(buffer, mystring.data(), mystring.length());
STL fanboys would encourage you to use std::copy instead:
std::copy(mystring.begin(), mystring.end(), buffer);
but there really isn't much of an upside to this. If you need null termination use std::string::c_str() and the various string duplication techniques others have provided, but I'd generally avoid that and just query for the length. Particularly with cryptography you just know somebody is going to try to break it by shoving nulls in to it, and using std::string::data() discourages you from lazily making assumptions about the underlying bits in the string.
Normally, encryption functions take
encrypt(const void *ptr, size_t bufferSize);
as arguments. You can pass c_str and length directly:
encrypt(strng.c_str(), strng.length());
This way, extra space is allocated or wasted.
In C++17 and later you can use std::byte to represent actual byte data. I would recommend something like this:
std::vector<std::byte> to_bytes(std::string const& s)
{
std::vector<std::byte> bytes;
bytes.reserve(std::size(s));
std::transform(std::begin(s), std::end(s), std::back_inserter(bytes), [](char c){
return std::byte(c);
});
return bytes;
}
From a std::string you can use the c_ptr() method if you want to get at the char_t buffer pointer.
It looks like you just want copy the characters of the string into a new buffer. I would simply use the std::string::copy function:
length = str.copy( buffer, str.size() );
If you just need to read the data.
encrypt(str.data(),str.size());
If you need a read/write copy of the data put it into a vector. (Don;t dynamically allocate space that's the job of vector).
std::vector<byte> source(str.begin(),str.end());
encrypt(&source[0],source.size());
Of course we are all assuming that byte is a char!!!
If this is just plain vanilla C, then:
strcpy(buffer, text.c_str());
Assuming that buffer is allocated and large enough to hold the contents of 'text', which is the assumption in your original code.
If encrypt() takes a 'const char *' then you can use
encrypt(text.c_str())
and you do not need to copy the string.
You might go with range-based for loop, which would look like this:
std::vector<std::byte> getByteArray(const string& str)
{
std::vector<std::byte> buffer;
for (char str_char : str)
buffer.push_back(std::byte(str_char));
return buffer;
}
I dont think you want to use the c# code you have there. They provide System.Text.Encoding.ASCII(also UTF-*)
string str = "some text;
byte[] bytes = System.Text.Encoding.ASCII.GetBytes(str);
your problems stem from ignoring the encoding in c# not your c++ code