Pointer memory error

Question

The problem

I can't get the data from the Flash memory when using the function that return the address of the pattern desired in the Flash (simplified in the example below with only 1 constant : PATTERN_P).

Some code before explication

The type patternp is defined as

typedef prog_uchar patternp[NUM_ROWS];

The global PATTERN_P variable is an array of type patternp, defined as

const patternp PATTERN_P PROGMEM = {
   0b11110000 ,
   0b10010000 ,
   0b10010000 ,
   0b10010000 ,
   0b11110000 ,
   0b10000000 ,
   0b10000000 ,
   0b10000000 
};

getpattern():

const patternp * getPattern()
{
    //... 
      return &PATTERN_P;
}

main():

const patternp *bufferPattern = getPattern();

uint8_t rowPatternData[NUMBER_ROW_PER_MATRIX];
const patternp *bufferPattern = getPattern(s[iLetter]);  
for(int iRow = 0; iRow<NUMBER_ROW_PER_MATRIX; iRow++)
{  
 rowPatternData[iRow]=pgm_read_byte( &PATTERN_P[iRow] );   // <--- WORK!
 rowPatternData[iRow]=pgm_read_byte( bufferPattern[iRow] ); // Not Working! 

}

Explications

As you can see, the code get the pattern (in this example, it will return PATTERN_P every time... than I use pgm_read_byte to get the data from the Flash memory. This use the AVR pgmspace (link below). It takes an address and return the data. The code above work when I use the direct access of a template : &PATTERN_P[iRow], but won't work when I use bufferPattern[iRow] or &bufferPattern[iRow]. Any idea?

Reference : pgm_read_byte is defined in pgmspace

Answer 1

bufferPattern is a pointer to an array. When you write bufferPattern[iRow], this does NOT evaluate to a pointer to entry iRow of patternp; the [] operation is acting on the pointer, not the array it points to. What you appear to want to write is &((*bufferPattern)[iRow]).

That will fix the immediate problem. However, the code is a bit confusing. It may be that your code would be simplified by passing the array directly (C does not pass arrays by value; so it won't copy the array - you don't need to make a pointer to the array to avoid this).

Answer 2

this

&PATTERN_P[iRow]

is

&(PATTERN_P[iRow])

when your working line gives this:

(&PATTERN_P)[iRow]