In my project I need to convert a string to a char array. From various examples I concluded that toCharArray() will convert a string to a char array. But I failed to do so. The error is:
'a' does not name a type
The code I am using is:
String a = "45317";
char b[6];
a.toCharArray(b,6);
Resources are https://www.arduino.cc/en/Reference/StringToCharArray and http://forum.arduino.cc/index.php?topic=199362.0
If you're trying to use a method of a in the global scope, that's doomed to failure. You can only call methods within functions.
If all you want is a char array with "45317" in it then just use:
char *b = "45317";
If you want to convert a string that is built at runtime into a char array, then your current method is correct - you just have to do it in the right place.
There's a built in conversion which will return the underlying string-contents as a NULL terminated character array:
String foo = "Steve was here"
char *text = foo.c_str();
That is probably all you need, unless you do want to copy into a buffer. In that case you can use the standard C library to do that:
// Declare a buffer
char buf[100];
// Copy this string into it
String foo = "This is my string"
snprintf( buf, sizeof(buf)-1, "%s", foo.c_str() );
// Ensure we're terminated
buf[sizeof(buf)] = '\0';
(You might prefer strcpy, memcpy, etc to snprintf.)
c_str() is no replacement for toCharArray(): your first example fails to compile with “error: invalid conversion from ‘const char*’ to ‘char*’”. Of course, if you only need a const char *, then c_str() is to be preferred. 2) In terms of code size, snprintf() is a very inefficient way of copying a string. 3) No need to subtract one from sizeof(buf) in the second argument to snprintf().
– Edgar Bonet
Mar 20 '17 at 18:46
