stm32f4 write/read 16bits into two 8-bit gpio port - stm32

Question

I try to write 16 bit data into two selected 8-bits gpio ports . I must split data for
LSB and MSB :
void LCD_write_command(uint16_t cmd) {
GPIOD->ODR = cmd & 0x00ff; //lsb
GPIOA->ODR = (GPIOA->ODR & 0x00ff) | (cmd >> 8); //msb
}
and read data :
uint16_t LCD_read_data(void) {
(here is instruct gpio as input)
volatile uint16_t data = 0;
data = (uint16_t)GPIOD->IDR & 0x00ff; //lsb
data |= (uint16_t)GPIOA->IDR << 8 ; // msb
(here is instruct gpio as output)
return data;
}
When i use one 16bit gpio to write and read everything is fine:
void LCD_write_command(uint16_t cmd) {
GPIOD->ODR = cmd & 0xffff;
}
uint16_t LCD_read_data(void) {
volatile uint16_t data = 0;
data = (uint16_t)GPIOD->IDR & 0xffff;
return data;
}
I relay dont know what im missing.

Answer 1

wtite_bits(uint16_t cmd)
{
uint32_t data = GPIOA -> ODR;
data &= ~(0x1fff);
data |= cmd & 0x1fff;
GPIOA -> ODR = data;
data = GPIOB -> ODR;
data &= ~(0x0007);
data |= (cmd & 0x8fff) >> 13;
GPIOB -> ODR = data;
}
preserve other bits in the register

Answer 2

You need to learn a bit more about bitwise operations.
Writing
void LCD_write_command(uint16_t cmd) {
uint32_t tmp = GPIOD->ODR;
tmp &= ~(0xff);
tmp |= (cmd & 0x00ff);
GPIOD->ODR = tmp; //lsb
tmp = GPIOA->ODR;
tmp &= ~(0xff);
tmp |= (cmd >> 8);
GPIOA->ODR = tmp; //msb
}
or
void LCD_write_command(uint16_t cmd) {
*(volatile uint8_t *)&GPIOD->ODR = cmd & 0xff;
*(volatile uint8_t *)&GPIOA->ODR = cmd >> 8; //msb
}
forcing the compiler to use 8 bit store instructions.
Before using non word access to the registers check in the RM if your micro allows it:

Answer 3

Non of yours sugested code works for me , below is original source code of handling with LCD:
void LCD_write_command(uint16_t cmd) {
GPIOB->BRR = LCD_CS; // LCD_CS low (chip select pull)
GPIOB->BRR = LCD_RS; // LCD_RS low (register select = instruction)
//GPIOA->ODR = cmd; // put cmd to PortA (full length)
// put cmd [0..12] bits to PortA (actual LCD_DB00..LCD_DB12)
// put cmd [13..15] bits to PortB (actual LCD_DB13..LCD_DB15)
GPIOA->ODR = cmd & 0x1fff;
GPIOB->ODR = (GPIOB->ODR & 0xfff8) | (cmd >> 13);
GPIOB->BRR = LCD_WR; // pull LCD_WR to low (write strobe start)
// Write strobe 66ns long by datasheet. GPIO speed on STM32F103 at 72MHz slower -> delay is unnecessary
// asm volatile ("nop");
GPIOB->BSRR = LCD_WR; // pull LCD_WR to high (write strobe end)
GPIOB->BSRR = LCD_CS; // LCD_CS high (chip select release)
}
I've checked RM and i can operate on 8-bit ,half word and word.

Answer 4

I had similar problem: need to write some bits to port at same time.
My output bits are on port B: 0,1,2,4,5,6,7,8,9 (see missing bit 3).
It is possible to read whole port (ODR), and/or bits and write it back.
Faster version is to set all bits:
GPIOB ->BSRR =0b00000000000000000000001111111011;
and then clear only ones needed:
GPIOB ->BRR =some_bits_to_clear;
It was like some custom 74LS154 chip.
Or even try to clear and set bits at same time using:
GPIOB ->BSRR =bits_to clear<<16 | bits_to_set;
Depending on situation, there are many ways to optimize code.