How to convert string to char array in C

Question

I would like to convert string to char array but not char   I know how to convert string to char   by using malloc or the way I posted it in my code  - but that s not what I want  I simply want to convert string to char size  array  Is it possible    include  lt iostream gt   include  lt string gt   include  lt stdio h gt  using namespace std   int main            char to string     char tab 4       tab 0     c       tab 1     a       tab 2     t       tab 3      0       string tmp tab       cout  lt  lt  tmp  lt  lt    n           string to char  - but thats not what I want      char  c   const cast lt char  gt  tmp c str         cout  lt  lt  c  lt  lt    n          string to char     char tab2 1024                 return 0

User · Answer

If you don t know the size of the string beforehand  you can dynamically allocate an array   auto tab2   std  make unique lt char   gt  temp size     1   std  strcpy tab2 get    temp c str

User · Answer

Just copy the string into the array with strcpy

User · Answer

Try this way it should be work      string line  hello world   char   data   new char line size     1   copy line begin    line end    data   data line size        0

User · Answer

If you re using C  11 or above  I d suggest using std  snprintf over std  strcpy or std  strncpy because of its safety  i e   you determine how many characters can be written to your buffer  and because it null-terminates the string for you  so you don t have to worry about it   It would be like this    include  lt string gt   include  lt cstdio gt   std  string tmp    cat   char tab2 1024   std  snprintf tab2  sizeof tab2     s   tmp c str         In C  17  you have this alternative    include  lt string gt   include  lt cstdio gt   include  lt iterator gt   std  string tmp    cat   char tab2 1024   std  snprintf tab2  std  size tab2     s   tmp c str

User · Answer

Well I know this maybe rather dumb than and simple  but I think it should work   string n  cin gt  gt  n  char b 200   for  int i   0  i  lt  sizeof n   i          b i    n i       cout lt  lt  b i  lt  lt

User · Answer

Ok  i am shocked that no one really gave a good answer  now my turn  There are two cases    A constant char array is good enough for you so you go with    const char  array   tmp c str     Or you need to modify the char array so constant is not ok  then just go with this   char  array    amp tmp 0      Both of them are just assignment operations and most of the time that is just what you need  if you really need a new copy then follow other fellows answers

User · Answer

str copy cstr  str length   1      since C  11 cstr str copy cstr  str length         0       before C  11 cstr str copy cstr  sizeof cstr -1       0       before C  11  safe    It s a better practice to avoid C in C    so std  string  copy should be the choice instead of strcpy

User · Answer

You could use strcpy    like so   strcpy tab2  tmp c str       Watch out for buffer overflow

User · Answer

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

User · Answer

Easiest way to do it would be this  std  string myWord    myWord   char myArray myWord size   1    as 1 char space for null is also required strcpy myArray  myWord c str

User · Answer

Try strcpy    but as Fred said  this is C    not C

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