Thread: Can someone explain a line of code for me please?

You are printing the first string contained in what appears to be an array. In that string, there is a formatting character (maybe %d or %i or %u) that is expecting the defined size of your array, less the null terminating character.

That doesn't make sense. First of all, (char*)&mytext[0] can be rewritten as just mytext. Secondly, since you are passing a second parameter to printf(), the first parameter should contain a format specifier (like %d). The text " dsPIC33F Demo " doesn't have any format specifiers, so why are you passing a second parameter to printf()?

Given that string the result will be the same as:

Code:

printf(" dsPIC33F Demo ",15);

This will just print the mytext string. The second parameter will be ignored. The conversion to char * is pointless here.

This is how it originally was and the I added the printf.

Code:

void Update_LCD( void );
const char mytext[] =  " dsPIC33F Demo ";
const char mytext1[] = "Press S3 to cont";
const char time_msg[] = "Time 00: 00: 00 ";
const char adc_msg1[] =" RP5 = 0.00 Vdc  ";


int main ( void )
{

	/* Welcome message */
	home_clr();
	puts_lcd( (char*) &mytext[0], sizeof(mytext) -1 );
	line_2();
	puts_lcd( (char*) &mytext1[0], sizeof(mytext1) -1 );

Code:

void puts_lcd( unsigned char *data, unsigned char count ) 
{
  	while ( count )
	{
		lcd_data( *data++ );
		count --;
	}	
}

Well "puts_lcd" is not the same thing as printf. It looks more like a fwrite style...


Quzah.

It's probably not an accident that puts_lcd starts with "puts", which is a standard function for printing a line of text.