C++ – how to add a int to a char[] in c/c++

cchar

i have a code like this:

int i = 123;
char myString[100];
strcpy(myString, "my text");

and how i want to add the 123 after "my text". how to do this in c/c++?

at the end myString sould be my test123

Best Answer

In C++:

std::stringstream ss;
ss << "my text" << i;
std::string resultingString = ss.str();

There's no such thing as C/C++.

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

C++ – How to convert a std::string to const char* or char*

If you just want to pass a std::string to a function that needs const char* you can use

std::string str;
const char * c = str.c_str();

If you want to get a writable copy, like char *, you can do that with this:

std::string str;
char * writable = new char[str.size() + 1];
std::copy(str.begin(), str.end(), writable);
writable[str.size()] = '\0'; // don't forget the terminating 0

// don't forget to free the string after finished using it
delete[] writable;

Edit: Notice that the above is not exception safe. If anything between the new call and the delete call throws, you will leak memory, as nothing will call delete for you automatically. There are two immediate ways to solve this.

boost::scoped_array

boost::scoped_array will delete the memory for you upon going out of scope:

std::string str;
boost::scoped_array<char> writable(new char[str.size() + 1]);
std::copy(str.begin(), str.end(), writable.get());
writable[str.size()] = '\0'; // don't forget the terminating 0

// get the char* using writable.get()

// memory is automatically freed if the smart pointer goes 
// out of scope

std::vector

This is the standard way (does not require any external library). You use std::vector, which completely manages the memory for you.

std::string str;
std::vector<char> writable(str.begin(), str.end());
writable.push_back('\0');

// get the char* using &writable[0] or &*writable.begin()