Automatically Concatenate Strings and Int C++ - c++

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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);

Answer 5

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

Answer 6

"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.