错误：没有忽略 void 值，因为它应该在 Arduino 中

Question

This is a function to convert 1 digit number to 3 digit. For example convert '2' to '002'.

void loop() {
 int x = convertdigit(time);
}

void convertdigit(int num){
  char buffer[50];
  int n;
  n=sprintf (buffer, "%03d",num);
  return buffer;
}

Error: void value not ignored as it ought to be

/sketch/sketch.ino: In function 'void loop()':
/sketch/sketch.ino:33:30: error: void value not ignored as it ought to be
     int x = convertdigit(time);
                              ^
Error during build: exit status 1

May i know how to fix it?

Answer 1

When you write this you are telling the compiler that convertdigit does not return anything (void):

void convertdigit(int num)

When you write this, you are telling the compiler to use the return value of convertdigit and store it in x :

int x = convertdigit(num);

Those two things are in conflict: if convertdigit doesn't return anything, how can you store that nothing in x ? Your code is kind of confusing so I'm not sure what you actually intended, but now that I have explained that error message, I hope you are able to make progress.

Hint: If you want convertdigit to return an int , change void convertdigit(... to int convertdigit(... .

Answer 2

You should distinguish numbers ( 2 ) from texts "002"

void loop() {
 static byte n = 1; 
 char* txt = convertdigit(n); // Convert to a 2 character text with a leading zero, if necessary;
 Serial.println(txt);
 delay(100); 
 n++; 
 if (n > 99) n=0; 
}

char* convertdigit(byte num) {
   static char buffer[4];
   sprintf (buffer, "%02d", num);
   return buffer;
}

static is required to keep the variable available after function has ended (similar to global variables):

n keeps its incremented value in the next loop round.

buffer is available via return value after converdigit() returns.