I have a few files named like so: file1, file2, file3, etc.
I have a function:
load(const char *file)
which I would call like so load(file1), load(file2), etc.
I am trying do this a bit more dynamically, based on the number of files imported.
So if I have more than 1 file do something like this:
if (NUM_OF_FILES > 1) {
for (int i = 2; i <= NUM_OF_FILES; i++) {
load("file" + i);
}
}
However, this is not working.
Is there a way of doing this?
The type of a string literal like "file" is char const[N] (with a suitable N) whic happily decays into a char const* upon the first chance it gets. Although there is no addition defeined between T[N] and int, there is an addition defined between char const* and int: it adds the int to the pointer. That isn't quite what you want.
You probably want to convert the int into a suitable std::string, combine this with the string literal you got, and get a char const* from that:
load(("file" + std::to_string(i)).c_str());
It depends on what version of C++ you are using. If it's C++11, the solution will involve std::to_string. If it's an older version of C++, you can convert an integer to a string like this:
#include <sstream>
// ...
std::ostringstream converter;
converter << i; // i is an int
std::string s(convert.str());
Now, the load function takes a const char *. Is it your own function? Then consider changing it so that it takes a std::string const& instead, and you'll be able to pass the string directly. Otherwise, this is how can pass the string's contents to it:
load(s.c_str());
Related
Today I was surprised when trying to concatenate an std::string with an int. Consider the following MWE:
#include <iostream>
#include <string>
void print(const std::string& text)
{
std::cout << "The string is: " << text << ".\n";
}
int main()
{
print("iteration_" + 1);
return 0;
}
Instead of printing
The string is: iteration_1.
which I would expect, it prints
The string is: teration_.
What exactly is going on in the background? Does the string for some reason get converted into char[] or something of the sort? The documentation of operator+ does not list any with an std::string and int.
And what is the proper way of concatenating an std::string with a number? Do I really have to throw them both into an std::stringstream or convert the number into std::string explicitely with std::to_string()?
Does the string for some reason get converted into char[]
Actually it is the other way around. "iteration_" is a char[11] which decays to a const char* when you add 1. Incrementing the pointer by one makes it point to the next character in the string. This is then used to construct a temporary std::string that contains all but the first character.
The documentation you link is for operator+ of std::string, but to use that you need a std::string first.
This line is the problem:
print("iteration_" + 1);
The string literal is decaying to a char*. You are adding 1 to this char*, moving it to the next character.
If you wanted to add the string "1" to the end of your literal, a fairly simple way is to pass the string literal to the std::string constructor and convert the 1 to a string manually. For example:
print(std::string("iteration_") + std::to_string(1));
"iteration_" is not std::string, but const char[]. Which decays to const char*, and "iteration_" + 1 just performs pointer arithmetic and move the pointer pointing to the next char (i.e. 't'), then you got the c-style string "teration_".
You can use std::to_string to convert int to std::string, then concatenate them. e.g.
print("iteration_" + std::to_string(1));
For this case std::operator+(std::basic_string) is called and the 1st argument "iteration_" is converted to std::string implicitly and then passed to operator+, then the concatenated std::string is passed to print.
LIVE
If you try to use the following:
std::string str = "iteration" + 1;
compiler will throw the warning:
warning: adding 'int' to a string does not append to the string
[-Wstring-plus-int]
It is because you are incrementing the pointer to "iteration" string by 1 which means that now "teration" string is being assigned to str variable.
The proper way of concatenating would be:
std::string str = "iteration" + std::to_string(1);
The expression "iteration_" + 1 is a const char[11] literal added to the int 1.
In that expression, "iteration_" decays to a const char* pointer to the first element of the array. + 1 then takes place in pointer arithmetic on that pointer. The entire expression evaluates to a const char* type (pointing to the first t) which is a valid NUL-terminated input to a std::string constructor! (The anonymous temporary std::string binds to the const std::string& function parameter.)
This is completely valid C++ and can occasionally be put to good use.
If you want to treat + as a concatenation, then
print("iteration_" + std::to_string(1));
is one way.
I'm writing a program that reads a content of a binary file (specificly Windows PE file. Wikipedia page and detailed PE structure).
Because of the binary data in the file, the characters often "fall out" of the ascii range (0-127) and that result in negative values.
To make sure I won't work with unwanted negative values, I can either pass const unsigned char * or convert the resulting char in the calculation to unsigned char.
On one hand, passing const unsigned char * makes sense because the data is non-ascii that has a numaric value and thus should be treated as positive.
In addition, it'll let me perform calculations without the need to cast the result to unsigned char.
On the other hand, I can't pass constant strings (const char *, such as pre-defined strings "MZ", "PE\0\0" etc.) to functions without first casting them to const unsigned char *.
What would be the better approach or best-practice in this scenario?
I think I'd use unsigned char, but avoid casting, and instead define a little class named ustring (or something similar). You have a couple of choices with that. One would be to instantiate std::basic_string over unsigned char. This can be useful (it gives you all of std::string's functionality, but with unsigned chars instead of chars. The obvious disadvantage is that it's probably overkill, and has essentially no compatibility with std::string, even though it's almost exactly the same thing.
The other obvious possibility would be to define your own class. Since you apparently care mostly about string literals, I'd probably go this way. The class would be initalized with a string literal, and it would just hold a pointer to the string, but as unsigned char * instead of just char *.
Then there's one more step to make life better: define a user defined literal operator named something like _us, so creating an object of your type from a string literal will look something like this: auto DOS_sig = "MZ"_us;
class ustring {
unsigned char const *data;
unsigned long long len;
public:
ustring(unsigned char const *s, unsigned long long len)
: data(s)
, len(len)
{}
operator char const *() const { return data; }
bool operator==(ustring const &other) const {
// note: memcmp treats what you pass it as unsigned chars.
return len == other.len && 0 == memcmp(data, other.data, len);
}
// you probably want to add more stuff here.
};
ustring operator"" _us(char const * const s, unsigned long long len) {
return ustring((unsigned char const *)s, len);
}
If I'm not mistaken, this should be pretty easy to work with. For example, let's assume you've memory mapped what you think is a PE file, with its base address at mapped_file. To see if it has a DOS signature, you might do something like this:
if (ustring(&mapped_file[0], 2) == "MZ"_us)
std::cerr << "File appears to be an executable.\n";
else
std::cerr << "file does not appear to be an executable.\n";
Caution: I haven't tested this, so fencepost errors and such are likely--for example, I don't remember whether the length passed to the user defined literal operator includes the NUL terminator or not. This isn't intended to represent finished code, just a sketch of a general direction that might be useful to explore.
I am trying to pull a specific char from a string and convert it to an int. I have tried the following code, but I am unclear why it doesn't work nor can I find a way to do the conversion.
int value = 0;
std::string s = "#/5";
value = std::atoi(s[2]); // want value == 5
You can create std::string from one char and use std::stoi to convert to integer.
#include <iostream>
#include <string.h>
using namespace std;
int main() {
int value = 0;
string s = "#/5";
value = stoi(string(1, s[2])); //conversion
cout << value;
}
You can write:
std::string s = "#/5";
std::string substring = s.substr(2, 1);
int value = std::stoi(substring);
Using the substr method of std::string to pull out the substring that you want to parse as an integer, and then using stoi (which takes a std::string) instead of atoi (which takes a const char *).
You should read the manual page for atoi() more carefully. The actual prototype is:
int atoi(const char *string)
You are attempting to pass a single character rather than a pointer to a character array. In other words, by using s[2] you are dereferencing the pointer. You could, instead, use:
value = std::atoi(s+2);
or alternatively:
value = std::atoi(&s[2]);
This code doesn't dereference the pointer.
The argument to std::atoi must be char*, but s[2] is char. You need to use its address. And to get a valid C string from a std::string, you need to use the c_str() method.
value = std::atoi(&(s.c_str()[2]));
You should have gotten an error saying that the argument wasn't of the correct type.
In Lua (apologise, I like working with it the best), the conversion between int and string is done automatically, so
"hi"..2
would result as
"hi2"
In C++ (cause I can't seem to get the default C++11 stoi() and to_string() methods to work) I defined these for myself:
int stoi(string str) {
char* ptr;
strtol(str.c_str(), &ptr, 10);
}
string to_string(int i) {
char* buf;
sprintf(buf, "%d", i);
return buf;
}
which are basically how the default ones are defined anyways.
Then I did this:
string operator+ (string& stuff, int why) {
stuff.append(to_string(why));
}
I tried it on the following code:
void print(string str) {
cout << str << endl;
}
int main() {
cout << stoi("1") + 2 << endl;
print("die" + 1);
return 0;
}
And it outputs
3
ie
Why is this so, and how can I fix it?
EDIT:
Here's what the code looks like now:
using namespace std;
string to_string(int i) {
char* buf;
sprintf(buf, "%d", i);
return buf;
}
string operator+ (string stuff, int why) {
stuff.append(to_string(why));
return stuff;
}
int main() {
cout << string("die") + 2 << endl;
return 0;
}
And it just keeps giving me stackdumps.
Replace print("die" + 1); with cout << std::string("die") + 1;
print() doesn't know what to do with strings. Use std::cout. "die" is a char*, +1 will increment the pointer.
std::string to_string(int i) {
char buf[(sizeof(int)*CHAR_BIT+2)/3+3];
sprintf(buf, "%d", i);
return buf;
}
You need to make an actual buffer to print to. The math is a quick over-estimate of big the largest decimal int is in characters; 3 bits can fit in 1 decimal character, plus null, plus negation, plus rounding, plus 1 for good measure. Hopefully I did not err: do some testing.
Also use snprintf instead of sprintf while you are at it: buffer overflows are not to be toyed with.
The next problem is that "hello" is not a std::string, It is a char const[6] -- an array of 6 char. It can be converted tomstd::string, but +1 will instead convert it to a pointer to the first character, then +1 it to the 2nd character.
Cast it to std::string before.
Finally, it is ambiguous in the standard (really) of pverloading an operator on std::string + int is legal. It is definitely poor practice, as you cannot do it in std legally, and you should overload operators in the type's namespace (so ADL works): these two conflict. On top of that, if std in the future adds such a + your code starts behaving strangely. On top of that, operators are part of a class's interface, and modifying the interface of a class you do not 'own' is rude and a bad habit.
Write your own string class that owns a std::string rather. Or a string view.
Finally, consider telling your compiler to use c++11, you probably just need to pass a flag to it like -std=c++11.
std::string s1("h1");
std::string s2("2");
s1 += s2;
If you are using C++11 compatible compiler you can convert int to string like this:
int i = 2;
std::string s = std::to_string(i);
If you are using Boost library:
#include <boost/lexical_cast.hpp>
int i = 2;
std::string s = boost::lexical_cast<std::string>(i);
Please do not use raw char pointers in C++ for strings.
overloading the operator+ on other than your own types it at best dangerous.
Just use std::to_string in conjunction with operator+ or +=, e.g.
std::string x = "hi";
x += std::to_string(2);
C++14 introduces a user-defined literal that takes a string literal (conversions are applied to make this a pointer) and returns a std::string. In C++11, you can just write your own (this is taken from libstdc++):
inline std::string
operator""_s(const char* str, size_t len)
{
return std::string{str, len};
}
(Note: UDLs without a preceding underscore are reserved names)
And you can use it like this:
// Assumes operator+ is overloaded
print("die"_s + 1);
Demo
"die" is not a std::string. It's a string literal.
Thus when you add 1 to the string literal, it decays to a const char* and the + 1 simply increments that pointer — to next char, 'i'.
Then you call print with the incremented pointer, which causes a std::string to be constructed using that pointer. Since it pointed to the 'i' character, to constructed string is initialized to "ie".
You must first make a std::string out of your string literal to make it call your operator+:
std::cout << std::string("die") + 1;
And then make a few fixes to your operator+:
string operator+ (string stuff, int why) {
return stuff.append(to_string(why));
}
Now it works.
I have a string with non-ASCII characters, for example std::string word ("żółć"); or std::string word ("łyżwy");
I need to convert it properly to const char * in order to call system(my_String_As_A_Const_Char_Pointer);
I'm working on Linux.
How can I do it?
You can use the std::string::c_str member function. It will return a const char * that can be used in functions that accept that type of argument. Here's an example:
int main(int, char*[]) {
std::string word("żółć");
const char* x = word.c_str();
std::cout << x;
}
And here's a live example.
With these conversions the only thing to care about is mixing wide chars with normal chars (which fails horribly). You are using a std:string, so c_str() is fine for pulling out a const char* to pass to some other library call.