我正在尝试将十六进制输出向左移动一个,这样我就可以在7段lcd上显示9以上的数字。
在C上编程,我使用的软件是NIOS II,因此我可以直接重新编程到DE0板上。
这个项目的目的是每次按下“button1”时将LCD的值增加1。我已经成功地做到了这一点,当然在9之后它需要向左移动并从1重新开始,用0替换它的位置。我做了一些研究,但没有任何运气所以任何帮助表示赞赏。感谢。
代码如下:
#include "sys/alt_stdio.h" //for the alt_putstr function below. Outputs to Eclipse console
#include "altera_avalon_pio_regs.h" //for the I/O functions in the while loop below
#include "sys/alt_timestamp.h" //see Nios II Software Developer’s Handbook, Timestamp Driver
#include "system.h"
#define setHeaderOuts HEADEROUTPUTS_BASE+0x10 //HEADEROUTPUTS_BASE is defined in system.h of the _bsp file. It refers to the base address in the Qsys design
//the hex offset (in this case 0x10, which is 16 in decimal) gives the number of bytes of offset
//each register is 32 bits, or 4 bytes
//so to shift to register 4, which is the outset register, we need 4 * (4 bytes) = 16 bytes
#define clearHeaderOuts HEADEROUTPUTS_BASE+0x14 //to shift to register 5 (the 'outclear' register) we need to shift by 5 * (4 bytes) = 20 bytes, (=0x14 bytes)
// offset of 5 corresponds to the 'outclear' register of the PIO.
int main(void)
{
alt_putstr("This is the ELEE1062 version of the NIOS processor");
int buttons = 0; //the buttons on the DE0
//int switches = 0; //the switches on the DE0
int count = 0; //general purpose counter
int hexd = 0;
while(1)
{
buttons=IORD_ALTERA_AVALON_PIO_DATA(PUSHBUTTONS1_2_BASE); //read the value of the pushbuttons
while((buttons & 0x01) == 1) // i.e. while pushbutton 1 is not pressed
{
buttons=IORD_ALTERA_AVALON_PIO_DATA(PUSHBUTTONS1_2_BASE); //read the value of the pushbuttons
}
count=count+1;
IOWR_ALTERA_AVALON_PIO_DATA(DE0_LEDS_BASE,count); //display the value of count in binary, using the green LEDs
while((buttons & 0x01) == 0) //i.e. while pushbutton 1 is pressed
{
buttons=IORD_ALTERA_AVALON_PIO_DATA(PUSHBUTTONS1_2_BASE); //read the value of the pushbuttons
}
if (count==0)
{
hexd=0x000000c0;
}
else if (count==1)
{
hexd=0xf9;
}
else if ( count==2)
{
hexd=0xa4;
}
else if ( count==3)
{
hexd=0xb0;
}
else if ( count==4)
{
hexd=0x99;
}
else if ( count==5)
{
hexd=0x92;
}
else if ( count==6)
{
hexd=0x82;
}
else if ( count==7)
{
hexd=0xd8;
}
else if ( count==8)
{
hexd=0x80;
}
else if ( count==9)
{
hexd=0x90;
}
else if ( count>9)
{
hexd= hexd & ~(1<<count);
}
//count=alt_timestamp_start(); //start the timer. Timer increments each clock cycle. Clock for ELEE1062_NIOS is 50MHz
//buttons=IORD_ALTERA_AVALON_PIO_DATA(PUSHBUTTONS1_2_BASE); //read the value of the pushbuttons
//switches=IORD_ALTERA_AVALON_PIO_DATA(DE0SWITCHES_BASE); //read the value of the switches
IOWR_ALTERA_AVALON_PIO_DATA(SSEG_BASE,hexd); //DE0 7 segment displays all off --notice that a logic '1' turns the segment off
IOWR_ALTERA_AVALON_PIO_DATA(SSEG_BASE,hexd); //DE0 7 segment displays all on
IOWR_ALTERA_AVALON_PIO_DATA(DE0_LEDS_BASE,0x000); //all off --for the green LEDs, a logic '0' turns the LED off
IOWR_ALTERA_AVALON_PIO_DATA(DE0_LEDS_BASE,0xfff); //all on
IOWR_ALTERA_AVALON_PIO_DATA(clearHeaderOuts,0x01); //turn off the first pin of the output port
IOWR_ALTERA_AVALON_PIO_DATA(setHeaderOuts,0x01); //turn on the first pin of the output port
//IOWR_ALTERA_AVALON_PIO_DATA(SSEG_BASE,switches); //light up the 7 segment display segments corresponding to how the DE0 switches are set
IOWR_ALTERA_AVALON_PIO_DATA(DE0_LEDS_BASE,buttons); //light up the green LEDs corresponding to which DE0 buttons are pushed
//count=alt_timestamp(); //record the value of the timer, and store in the 'count' variable
}
}
答案 0 :(得分:1)
仅仅转移是行不通的。 9 -> 10
(你可以称之为转变)但是19 -> 20
呢?由于它显然是家庭作业或其他形式的学习,我不会为你编写代码。您的最终目标是在7段LED显示屏上表示数字。想一想。因此,作为输入,您有二进制数(count
),输出应该是引脚信号。你的任务是将一个转换成另一个。 Led主要使用十进制基数运算,因此您首先需要将二进制数转换为十进制数字序列,然后将它们转换为引脚信号(此代码已经存在)。要使用所有四位数字,您需要将您的数字转换为格式0x11223344,其中数字表示指示位置。 0xF9A4B099是1234(如果我没记错的话)。
答案 1 :(得分:0)
以下函数将返回由四个字节组成的u32
值,其中LED值用于数字显示。请注意,我不知道您的显示器想要数字的顺序,因此您可能需要对返回值进行字节换算。此外,此功能将为LED显示屏提供前导零 - 您可能需要修改内容,以便在显示屏上显示为空白。
typedef unsigned int u32;
static char led_digits[] = { 0xc0, 0xf9, 0xa4, 0xb0, 0x99, 0x92, 0x82, 0xd8, 0x80, 0x90 };
u32 four_digits( int x)
{
unsigned char c[4];
int i;
for (i = 0; i < 4; ++i)
{
int digit = x % 10;
x = x / 10;
c[i] = led_digits[digit];
}
return (u32)(c[3] << 24) | (u32)(c[2] << 16) | (u32)(c[1] << 8) | (u32)c[0];
}