如何将char *复制到char [] []

Question

Before start, let me show you my code:

int     main(int argc, char *argv[])
{                  
  t_arbre       *node;
  char          str[3][1];

  &(*(str[0])) = strdup("1"); //lvalue required as left operand of assignment
  str[1] = strdup("2"); //incompatible types when assigning to type ‘char[1]’ from type ‘char *’
  str[2] = strdup("3");
  if ((node = malloc(sizeof(t_arbre))) == NULL)
    {
      printf("Error: malloc in the main function failed\n");
      return (-1);
    }
  create_node(node, str);
  return (0);
}

So, as you can see, I'm currently working in C. And I'm trying to assign a (char*) to a (char[3][1]). Otherwise the compiler throw me for error:

incompatible types when assigning to type 'char[1]' from type 'char *'

when I'm trying to assign by the following way:

str[2] = strdup("2"); 

I'm know that if I declare a string like this

char str[5];

and then try to assign it a string using for example strdup() function, it could trow me the same error. However in this case I know that I can "dereference" it like:

str[0] 

Which will give me a (char) variable. And given I want it to store a string. I have to "transform" it into a (char*) by using the '&' charactere. Which will give me

&str[0]

In this case, it will work. because I have a pointer to the first character. By doing so, I'll be able to copy a (char*) into a (char[]).

Otherwise, in the upper code I provide you, I want to copy a (char*) into a (char[3][1]) (Don't ask me why, it's a bit long to explain haha) So as you can see, I tried my way to solve the error like

str[2] = strdup("2");

which throw me this error message

incompatible types when assigning to type 'char[1]' from type 'char *'

And by this way

&(*(str[0])) = strdup("1"); //or &(*(str[0]))

which throw me this error message

lvalue required as left operand of assignment

Does anyone can tell me how can I copy a (char*) into a (char[][]) ? Or is it just imposible to do what I want to do ? I don't speak english very well, hopeing I used the right term.

Thank you :)

Answer 1

• You can't assign to things that aren't intended for assignment, that's what "lvalue required as left operand of assignment" is telling you, because you're taking the address of something with & , which is essentially just a value, not somewhere you can store something (eg a variable), and then trying to assign to it.
• A char* is not a char[] , that's what the incompatible types error is about.
• C arrays do not support assignment.
• Don't forget that in C, arrays start at index 0, so the last valid index in an array of length 3 is 2 (eg arr[2] ), not 3.

Use strcpy() instead to copy a string into an array:

char str[3][2]; // must allow space for the null terminator
strcpy(str[0], "1");
strcpy(str[1], "2");
strcpy(str[2], "3");

Or use an array of pointers and strdup() :

char *str[3];
str[0] = strdup("1");
...

If you do that, don't forget to free them when you're done, because strdup() allocates memory, eg

// free one pointer from the array:
free(str[0]);
// or if you have allocated to every value in the array:
size_t arr_len = sizeof(str) / sizeof(*str);
for (size_t n = 0; n < arr_len; ++n)
    free(str[n]);

Answer 2

strdup returns a pointer-to-a-char. If you want to store that pointer in an array, then you need an array of pointers, like this

char *str[3];           // an array of three pointer-to-char
str[0] = strdup("1");
str[1] = strdup("2");
str[2] = strdup("3");

for ( int i = 0; i < 3; i++ )
    printf( "%s\n", str[i] );

Answer 3

If you want to copy the entire string content you can use strcpy , however note that it copies null terminated strings. It is needed to have place for the last null symbol. So, the length of string arrays must be at least 2 ( char str[3][2] ) to copy strings like "1".