我目前正致力于使用微控制器和电路通过LED显示二进制十进制的分配。我必须使用三个按钮：一个增加，一个减少，最后一个重置。我完成了接线和配置。另一方面，我的代码有一些小问题，我无法修复，只有我的腰带上的C类介绍。我正在使用Code Composer。第一个问题：我的计数器在＆＃34; Tens＆＃34; section（MSB）不会在9处停止，而是开始显示二进制10-15。前四位是1（右）和后4位，数十（左）ex：1010 0001但最大值是1001 1001。问题二：如果我递减值，从零开始，计数器显示95而不是99。
例如：从0000 0000减量开始显示1001 0101。我尝试使用if语句为MSB制作一个条件，如果大于10，则代码不会运行。几乎我对代码所做的任何修改都会阻止它正常工作。
#include <msp430.h>
unsigned int dec2bcd(unsigned int num);
void delay(void);
int main(void) {
WDTCTL = WDTPW | WDTHOLD; // Stop watchdog timer
P1DIR = 0xFF;
P2DIR = 0xF0;
unsigned int c;
unsigned int count = 0x00;
while(1){
if((P2IN & BIT0) == BIT0){
count++;
c = dec2bcd(count);
}
else if((P2IN&BIT1) == BIT1){
count--;
c = dec2bcd(count);
}
else if((P2IN&BIT2) == BIT2){
count = 0x00;
c = dec2bcd(count);
}
delay();
P1OUT = c;
}
}
unsigned int dec2bcd(unsigned int num)
{
unsigned int ones = 0;
unsigned int tens = 0;
unsigned int temp = 0;
ones = num%10;
temp = num/10;
tens = temp<<4;
return (tens + ones);
}
void delay(void)
{
volatile unsigned int i, j;
for(i=10000; i>0; i--)
{
for(j=3; j>0; j--){
}
}
}
答案 0 :(得分：1)
I don't have a strip of LEDs handy right now, but this shows how to work within a decimal range, but only convert at the point of output. The actual value is kept in the range 0..99 (note how the modulus is done when I decrement). Then you split the value into BCD nibbles, and combine them for output.
#include <stdio.h>
void display(int num, char *msg)
{
int lsnib = num % 10; // decimal value of each nibble
int msnib = num / 10;
int ledpatt = (msnib << 4) + lsnib; // recombine as BCD
printf("%02X Expected %s\n", ledpatt, msg);
}
int main(void){
int value;
value = 42;
display (value, "forty two"); // display 42
value = 0;
display (value, "zero"); // display 0
value = 99;
display (value, "ninety nine"); // display 99
value = (value + 1) % 100; // increment from 99
display (value, "zero"); // should display 0
value = (value -1 + 100) % 100; // decrement from 0
display (value, "ninety nine"); // should display 99
return 0;
}
Program output
42 Expected forty two
00 Expected zero
99 Expected ninety nine
00 Expected zero
99 Expected ninety nine
答案 1 :(得分：1)
Why you see 95?
As mentioned by @Olaf, msp430 uses 16bit integer.
When you do from reportlab.pdfbase import pdfmetrics
from reportlab.pdfbase.ttfonts import TTFont
from reportlab.platypus import Paragraph
class MultiFontParagraph(Paragraph):
# Created by B8Vrede for http://stackoverflow.com/questions/35172207/
def __init__(self, text, style, fonts_locations):
font_list = []
for font_name, font_location in fonts_locations:
# Load the font
font = TTFont(font_name, font_location)
# Get the char width of all known symbols
font_widths = font.face.charWidths
# Register the font to able it use
pdfmetrics.registerFont(font)
# Store the font and info in a list for lookup
font_list.append((font_name, font_widths))
# Set up the string to hold the new text
new_text = u''
# Loop through the string
for char in text:
# Loop through the fonts
for font_name, font_widths in font_list:
# Check whether this font know the width of the character
# If so it has a Glyph for it so use it
if ord(char) in font_widths:
# Set the working font for the current character
new_text += u'<font name="{}">{}</font>'.format(font_name, char)
break
Paragraph.__init__(self, new_text, style)
, count acutally wraps to count = 0u -1;
65535Why tens can to beyond 10
Same problem, because your input is greater than 99.
unsigned int dec2bcd(unsigned int num) // num is now 65535
{
unsigned int ones = 0;
unsigned int tens = 0;
unsigned int temp = 0;
ones = num%10; // 65535%10 = 5
temp = num/10; // 65535/10 = 6553
tens = temp<<4; // what's displayed is by tens is actually the lower
// 4 bits of tens, so tens is 6553%16=9
return (tens + ones);// so the result is 95
}
What you should do?
In your code, limit the range to 0-99, you have the choice of either wrap-around (99+1=0, and 0-1=99)or saturation (99+1=99, 0-1=0). But whatevenr you need, you need to write it yourself: C language does not provide it.
答案 2 :(得分：0)
使用它有任何限制：
if((P2IN & BIT0) == BIT0){
count++;
if (count == 100)
count = 0;
c = dec2bcd(count);
}
else if((P2IN&BIT1) == BIT1){
if (count == 0)
count = 100;
count--;
c = dec2bcd(count);
}
答案 3 :(得分：0)
Let us keep count in countBDC.
unsigned countBCD = 0;
unsigned count = 0;
while(1){
if((P2IN & BIT0) == BIT0){
count++;
countBCD++;
if ((countBCD & 0xF) >= 0xA) {
// of course can combine these 2 lines
countBCD -= 0xA;
countBCD += 0x10;
if ((countBCD & 0xF0) >= 0xA0) {
countBCD -= 0xA0;
countBCD += 0x100;
...
// similar code for decrement
or use a BCD increment function
unsigned BCD_inc(unsigned x) {
unsigned carry = 1;
unsigned mask10 = 10;
unsigned maskNibble = 15;
while (carry) {
x += carry;
if ((x & maskNibble) >= mask10) {
x -= mask10;
mask10 <<= 4;
maskNibble <<= 4;
carry <<= 4;
} else {
carry = 0;
}
}
return x;
}
答案 4 :(得分：0)
以下是解决了我之前遇到的问题的新代码。它可能不是最优的，但我设法让它工作。我添加了两个创建环绕的if语句，并通过实现模数修改了dec2bcd转换器。 “十”部分的模数有助于摆脱显示问题。 谢谢大家帮助我！现在我理解了语法错误，并感谢我向我展示了我需要注意的更广泛的范围。祝我好运，希望得到额外的荣誉。直到下一次。
#include <msp430g2553.h>
unsigned int dec2bcd(unsigned int num);
void delay(void);
int main(void) {
WDTCTL = WDTPW | WDTHOLD; // Stop watchdog timer
P1DIR = 0xFF;
P2DIR = 0x00;
unsigned int c;
unsigned int count = 0x00;
//Here are some modified portions to create wrap around
while(1){
if((P2IN & BIT0) == BIT0){
count++;
if(count == (99+1)){
count = 0x00;
}
c = dec2bcd(count);
}
else if((P2IN&BIT1) == BIT1){
count--;
if(count == (0-1)){
count = 0x63;
}
c = dec2bcd(count);
}
else if((P2IN&BIT2) == BIT2){
count = 0x00;
c = dec2bcd(count);
}
delay();
P1OUT = c;
}
}
unsigned int dec2bcd(unsigned int num)
{
unsigned int ones = 0;
unsigned int tens = 0;
unsigned int temp = 0;
ones = num%10;
temp = num/10;
//I used modulus here to get rid of the improper display issue
tens = (temp%10)<<4;
return (tens + ones);
}
void delay(void)
{
volatile unsigned int i, j;
for(i=10000; i>0; i--)
{
for(j=3; j>0; j--){
}
}
}