Understanding pointers used for out-parameters in C/C++

Question

How do I return a char array from a function?

has the following code in the answers:

void testfunc(char* outStr){
  char str[10];
  for(int i=0; i < 10; ++i){
    outStr[i] = str[i];
  }
}

int main(){
  char myStr[10];
  testfunc(myStr);
  // myStr is now filled
}

Since I will be using an Arduino where memory is precious I dont want a "temporary variable" for storing some string and then copying it over. I also don't want the function to return anything. I want to use the idea above and do something like:

 void testfunc(char* outStr){
     outStr="Hi there.";
  }

int main(){
  char myStr[10];
  testfunc(myStr);

}

However, in this case myStr is empty!

Why doesn't this work? How do I fix it?

(I'm relatively new to C, and do have a basic understanding of pointers)

Thanks.

Answer 1

Why doesn't this work?

void testfunc(char* outStr){
     outStr="Hi there.";
 }

You have:

testfunc(myStr);

When testfunc is called, outStr is assigned the address of the first element of myStr . But in the testfunc function, you overwrite outStr and it now points to a string literal. You are only modifying ouStr , not myStr.

How do I fix it?

To copy a string, use strcpy function (or its secure sister strncpy ):

void testfunc(char* outStr){
     strcpy(ouStr, "Hi there."); 
 }

Answer 2

If you want memory reuse - and this is a dangerous place, has you must take really good care about allocation/deallocation responsibility, you should use pointer to string, ie:

#define STATIC_STRING "Hi there"

void testfunc(char**outStr){
    *outStr=STATIC_STRING;
}

int main(){
   char*myStr;
   testfunc(&myStr);
   //From now on, myStr is a string using preallocated memory.
}