Read string from MEMPROG

Projects Programming
1

I need some help please.

Declared array:
const char Bet_ve[6][10] PROGMEM=
{
"text1","text2","text3 ","text4","text5 ",
};

char name[8];
name= char buffer [strlen_P(Bet_ve [text_number] ) ];
I got "Compilation error: expected primary-expression before 'char' "

Thanks.

2

As your topic does not relate directly to the installation or operation of the IDE it has been moved to the Programming category of the forum

Please post a small but complete sketch that illustrates your problem

3

While you omitted telling us exactly where said error occurred (snipperitis strikes again!), I'll wager it's in the above line. It confuses me, and I imagine the compiler is equally confused. Could you explain just what it is you thought it would do?

4

I dare say the line below will confuse it as well...

5

"name" needs an address or brackets as "buffer" has brackets.

This:

const char Bet_ve[6][10] PROGMEM = { "text1", "text2", "text3", "text4", "text5" };
char name[8];
char buffer[8];
int text_number = 3;

void setup() {
  name[text_number] = buffer[strlen_P(Bet_ve[text_number])];
}

void loop() {}

Yields this:

Sketch uses 444 bytes (1%) of program storage space. Maximum is 30720 bytes.
Global variables use 9 bytes (0%) of dynamic memory, leaving 2039 bytes for local variables. Maximum is 2048 bytes.

But this...

const char Bet_ve[6][10] PROGMEM = { "text1", "text2", "text3", "text4", "text5" };
char name[8];
char buffer[8];
int text_number = 3;

void setup() {
  name = char buffer[strlen_P(Bet_ve[text_number])];
}

void loop() {}

Yields this...

sketch_oct4a.ino:7:10: error: expected primary-expression before 'char'
   name = char buffer[strlen_P(Bet_ve[text_number])];
          ^~~~
exit status 1

Compilation error: expected primary-expression before 'char'
6

you can't write name = char buffer[...]. The compiler sees char where it expects an expression after =, hence "expected primary-expression before 'char'".

Note also that name is a fixed-size array of 8 elements and arrays can't be assigned to with = in C/C++ at all ➜ you need to copy the bytes into that array if you need a RAM based copy.

const char Bet_ve[6][10] PROGMEM = {
  "text1", "text2", "text3 ", "text4", "text5 ",
};

char name[10];  // big enough for the longest entry + null

void getText(uint8_t text_number) {
  strlcpy_P(name, (char*)Bet_ve[text_number], sizeof name);
}
7 
  strlcpy_P(name, (char*)Bet_ve[text_number], sizeof name);

There is no need for the cast to char*, Bet_ve[text_number] is already a const char*.

How is name being used? Many functions, such as print(), can take a char array directly from PROGMEM, making the ram copy unnecessary.

8

You are totally right

const char Bet_ve[6][10] PROGMEM = {
  "text1", "text2", "text3 ", "text4", "text5 ", ""
};

char name[10];

void getText(uint8_t text_number) {
  if (text_number >= sizeof Bet_ve / sizeof Bet_ve[0]) {
    name[0] = '\0';
    return;
  }
  strlcpy_P(name, Bet_ve[text_number], sizeof name);
}

And yes this is just in case