Concatenate char*

Using Arduino Programming Questions
#1

Hello!

I need a char* composed by another char* and a String.
char* myFunction(){
String my_str = "sent";
char* chary = "message";
int str_len = my_str.length() + 1;
char buf[str_len];
my_str.toCharArray(buf, str_len);
strcat(chary, buf);
return chary;
}

I tried that but is not working

Thanks!

#2

What is chary? You haven't declared it. Is it large enough to fit the combined strings?

Where do you use charie? You declare it as a constant c string, but you don't use it anywhere.

That code doesn't compile. Please post code that compiles.

#3

Just changed the first post. They were both chary.

#4

Forget Strings and use character arrays. Very straightforward and reliable.

void setup() {
  char buf[30];
  const char *first = "sent ";
  const char *second = "message";
  strcpy(buf,first);
  strcat(buf,second);
  Serial.begin(9600);
  Serial.println(buf);  //prints sent message
}
void loop(){
}
#5 
 char* chary = "message";

chary is an array that is EXACTLY large enough to hold 8 characters.

 strcat(chary, buf);

There is NOT enough room to add more characters.

  return chary;

When the function ends, chary goes out of scope. Returning a pointer to memory that can be overwritten at any time is a very bad idea.

#6

When the function ends, chary goes out of scope. Returning a pointer to memory that can be overwritten at any time is a very bad idea.

Paul's right and, even worse, the code may appear to work correctly...for a while.

#7

chary is a pointer to a string literal, which is constant. It's undefined behavior to write to a string literal.

#8

It's undefined behavior to write to a string literal.

It's not undefined at all. The effect may be unexpected, but the behavior (of the compiler) is not undefined.

#9

Chistop: They are not writing to a string literal, they are writing to ca char array that is initialised to a string literal.

Now this would be undefined:
strcat("message", "some more");

#10

PaulS:
It’s not undefined at all. The effect may be unexpected, but the behavior (of the compiler) is not undefined.

It is indeed undefined behavior. This is from the C++ 11 draft standard, section 2.14.5:

The effect of attempting to modify a string literal is undefined.

I don’t have a copy of the older C++ standards, but modifying a string literal has been undefined behavior since the days of C89 (the C++ standards borrow heavily from the C standards). This is from the C89 draft standard, section 3.1.4:

If the program attempts to modify a string literal of either form, the behavior is undefined.

(“either form” means either wide or non-wide.)

You can think of a string literal as being of type “const char *”, which obviously cannot be modified.

KeithRB:
Chistop: They are not writing to a string literal, they are writing to ca char array that is initialised to a string literal.

Nope. This declares chary to be a pointer to char and initializes it to point to a string literal:

char* chary = "message";

You cannot modify the contents of a string literal even through a non-const pointer to it. Besides, the compiler should give you a warning for that like “warning: deprecated conversion from string constant to ‘char*’”, which tells you that you’re probably doing something wrong (g++ version 4.7.4 does for me, even without -Wall).

On the other hand, this would initialize chary to be a string that you can modify:

char chary[] = "message";
#11