Return is not working inside a char function properly.

Projects Programming
1

Hi.. if I use the Serial.println(vbuffer_Saida_geral).. It prints perfectly, but when I return the value it doesn't return correctly.. just some weird characters.

void setup()
{
	Serial.begin(9600);
}

char *completar_com_zeros(byte numero, int zeros)
{
	char vsaida[6] = "";

	char vbuffer_Saida_geral[20];
	char vinicio_dos_parametros[3] = "%0";

	char vbuffer_zeros_para_inserir[4];
	sprintf(vbuffer_zeros_para_inserir, "%01i", zeros); // converte o numero int para a string com nome de buffer_zero e inserindo  'zeros'
	strcat(vsaida, vinicio_dos_parametros);
	strcat(vsaida, vbuffer_zeros_para_inserir);
	strcat(vsaida, "d");//insere  o parametro 'd' no final da string vsaida

	sprintf(vbuffer_Saida_geral, vsaida, numero);
	return(vbuffer_Saida_geral);
}

void loop()
{
	int total_zeros = 4;
	int vvalor = 56;
	Serial.println(completar_com_zeros(vvalor, total_zeros));

	//Serial.println(vsaida);

	delay(1000);
}

Something related to convert vbuffer_Saida_geral which is 20 sized to the *completar_com_zeros from the function?

2

You have declared vbuffer_Saida_geral inside the function. It does not exist once the function returns.
One way to fix this is to make vbuffer_Saida_geral a global array.

Pete

3

Also, you are pointing to a buffer allocated on the stack which will disappear. Make it global or static.

4

el_supremo:
You have declared vbuffer_Saida_geral inside the function. It does not exist once the function returns.
One way to fix this is to make vbuffer_Saida_geral a global array.

Pete

Just change to static and worked fine!

Could you explain, why changing to static works?

Also, if I can print inside the function its because value is there(isn't it?). but why when I return it, the value goes away?

5

Could you explain, why changing to static works?

Can YOU google "static keyword"? I'm sure that you'll remember more of what you read if you put some effort into finding the data yourself.

6

PaulS:
Can YOU google "static keyword"? I'm sure that you'll remember more of what you read if you put some effort into finding the data yourself.

When the answer was from you, I regret myself to ask, and there was your "answer!"

But, ok!. tks everybory that helped!

7

If, anybody can help me to understand why inside the function, for printing the value was accessible and for return was not, I would appreciate.

Tks

8

If it is local to the function, the array is allocated on the stack when the function starts. When you return from the function, what was on the stack is no longer valid.

For more details, have a look for "scope of variables in C++"

Pete

9

I told you. Because it was on the stack.
http://c-faq.com/malloc/retaggr.html

10

el_supremo:
For more details, have a look for "scope of variables in C++"

Pete

Actually, scope does not matter here, but the lifetime. The scope is exactly the same between the static and auto variables declared/defined in the function.

11

An easy solution is to pass in a reference of a char array your defined outside of a function like

char mystring[10];
manipulate(mystring);

//


void manipulate( char *instring) {
    //do stuff
   instring[0] = 'a';

}
12

Thank you everybody!! Now its clear!

My misunderstood was in think that if I can print, I can return which is not real!

To me didn't make any sense but now its understood!

Tks @el_supremo, @KeithRB and @mistergreen!!