How to Convert String to Char Array in C++

How to convert a string to character array in c (or) how to extract a single char form string?

In C, a string is actually stored as an array of characters, so the 'string pointer' is pointing to the first character. For instance,

char myString[] = "This is some text";

You can access any character as a simple char by using myString as an array, thus:

char myChar = myString[6];
printf("%c\n", myChar); // Prints s

Hope this helps!
David

How to convert string to char array in C++?

Simplest way I can think of doing it is:

string temp = "cat";
char tab2[1024];
strcpy(tab2, temp.c_str());

For safety, you might prefer:

string temp = "cat";
char tab2[1024];
strncpy(tab2, temp.c_str(), sizeof(tab2));
tab2[sizeof(tab2) - 1] = 0;

or could be in this fashion:

string temp = "cat";
char * tab2 = new char [temp.length()+1];
strcpy (tab2, temp.c_str());

How to Convert a String Array to a Char * Array c++

You'll need to get the address of the data inside your std::strings. Note, that these are not required to be null-terminated, i.e., you'll need to make sure that all strings are null terminated. Also, the array passed as argv also needs to have the last element to be a null pointer. You could use code along the lines of this:

std::string array[] = { "s1", "s2" };
std::vector<char*> vec;
std::transform(std::begin(array), std::end(array),
               std::back_inserter(vec),
               [](std::string& s){ s.push_back(0); return &s[0]; });
vec.push_back(nullptr);
char** carray = vec.data();

When compiling with C++03, there are some changes necessary:

1. Instead of using the lambda expression, you need to create a suitable function or function object doing the same transformation.
2. Instead of using nullptr you need to use 0.
3. In C++03 std::string is not guaranteed to be contiguous, i.e., you need an additional, auxiliary std::vector<char> to hold a contiguous sequence of characters.

4. There are no functions begin() and end() deducing the size of an array but they can easily be implemented in C++03:

   template <typename T, int Size> T* begin(T (&array)[Size]) { return array; }
   template <typename T, int Size> T* end(T (&array)[Size]) { return array + Size; }
   

5. The C++03 std::vector<T> doesn't have a data() member, i.e., you also need to take the address of the first element.

c++ How to convert string array to char array

It's simple using string class .Run a loop to execute
productCode[i] = str[i].substr(0, 3) + str[i].substr(str[i].length() - 3);
and your work is done.

How does conversion from string to char array work in C++?

The thing to note is that operator[] returns a reference.

cppreference:

reference operator[]( size_type pos );

Returns a reference to the character at specified location pos.

This is important because it means that b[0] is essentially an alias to the first character in the string. Taking its address will then give you a pointer to the underlying buffer the string's data is stored in. It doesn't matter where it is (stack, heap, etc). This is also what lets you modify the string's data by using e.g. b[0] = 'a';.

Note that you can also use c_str() and data() to get a pointer to the first character in the string. Since C++17 data() can return a non-const pointer.

How to convert array of char into array of int in C Programming?

You are confusing single characters with character strings. The latter are sequences of characters (arrays) terminated with a nul (zero value) character, and that is what the atoi function expects as its input.

To convert a single character digit to its numerical value, you just need to subtract the value of the zero digit ('0') from that character's value (the values of the numerical digits are guaranteed by the Standard to be contiguous).

So, rather than:

array[i] = atoi(&parray[i]);

use:

array[i] = parray[i] - '0'; // Will work if (and only if) parray[i] is a digit.

---

What is happening in your code is that (by chance) there is a zero byte immediately after the end of your 5-character array (but you can not rely on this), so each atoi(&parray[i]) call is passing a character string starting with, respectively, the '7', '1', '5', '4' and '2' characters, and ending only after the '2'. Thus, you are getting values that represent the numbers formed by the concatenation of your individual array digits. But I repeat: you cannot rely on there being a zero-value character after your array!

---

Related Topics