showCard(intx,inty,unsignedcharnuelseif(c>='0'

单片机51的C程序谁给解释下?
#include
#include
unsigned char code displaybit[]={0xfe,0xfd,0xfb,0xf7,
0xef,0xdf,0xbf,0x7f};
unsigned char code displaycode[]={0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71,0x00,0x40};
unsigned char code dotcode[32]={0,3,6,9,12,16,19,22,
25,28,31,34,38,41,44,48,
50,53,56,59,63,66,69,72,
75,78,81,84,88,91,94,97};
unsigned char displaycount;
unsigned char displaybuf[8]={16,16,16,16,16,16,16,16};
unsigned char timecount;
unsigned char readdata[8];
sbit DQ=P3^7;
bit sflag;
bit resetpulse(void)
{
unsigned char i;
DQ=0;
for(i=255;i>0;i--);
DQ=1;
for(i=60;i>0;i--);
return(DQ);
for(i=200;i>0;i--);
}
void writecommandtods18b20(unsigned char command)
{
unsigned char i;
unsigned char j;
for(i=0;i0;j--);
DQ=1;
}
else
{
DQ=0;
for(j=2;j>0;j--);
DQ=1;
for(j=33;j>0;j--);
}
command=_cror_(command,1);
}
}
unsigned char readdatafromds18b20(void)
{
unsigned char i;
unsigned char j;
unsigned char temp;
temp=0;
for(i=0;i0;j--);
if(DQ==1)
{
temp=temp | 0x80;
}
else
{
temp=temp | 0x00;
}
for(j=200;j>0;j--);
}
return(temp);
}
void main(void)
{
TMOD=0x01;
TH0=(65536-4000)/256;
TL0=(65536-4000)%256;
ET0=1;
EA=1;
while(resetpulse());
writecommandtods18b20(0xcc);
writecommandtods18b20(0x44);
TR0=1;
while(1)
{
;
}
}
void t0(void) interrupt 1 using 0
{
unsigned char x;
unsigned int result;
TH0=(65536-4000)/256;
TL0=(65536-4000)%256;
if(displaycount==2)
{
P0=displaycode[displaybuf[displaycount]] | 0x80;
}
else
{
P0=displaycode[displaybuf[displaycount]];
}
P2=displaybit[displaycount];
displaycount++;
if(displaycount==8)
{
displaycount=0;
}
timecount++;
if(timecount==150)
{
timecount=0;
while(resetpulse());
writecommandtods18b20(0xcc);
writecommandtods18b20(0xbe);
readdata[0]=readdatafromds18b20();
readdata[1]=readdatafromds18b20();
for(x=0;x255)
{
readdata[1]++;
}
}
readdata[1]=readdata[1]4;
x=x & 0x0f;
readdata[1]=readdata[1] | x;
x=2;
result=readdata[1];
while(result/10)
{
displaybuf[x]=result%10;
result=result/10;
x++;
}
displaybuf[x]=result;
if(sflag==1)
{
displaybuf[x+1]=17;
}
x=readdata[0] & 0x0f;
x=x
就是用DS18B20进行温度检测

无符号变量和有符号变量的区别
What`s the difference between the unsigned integer and the signed integer?(in c++)

★关于素数筛选法
编写一函数unsigned int *findPrime(unsigned int n),按照Eractosthenes筛法找出1-n之间的素数.然后在编写一函数void printPrime(unsigned int *p,int n)打印这些素数.(提示:在findPrime函数分配堆内存表示筛法中用到的数组,并返回该数组首地址),主函数可为:
void main()
{
unsigned int *sieve;
unsigned int n;
printf("enter a number:");
scanf("%u",&n);
sieve = findPrime(n);
printPrime(sieve,n);
free(sieve);
}

请编写一个unsigned fun(unsigned w）
请编写一个unsigned fun(unsigned w)，w是一个大于10的无符号整数，若w是n（n>=2）位的整数，函数求出w的后n-1位的数作为函数值返回。
例如：w值为5923，则函数返回923；w值为923则函数返回23。

补码运算的问题
unsigned short a;
short int b = -1;
a = b;
-1的补码形式为1111111111111111(即全部16个二进制位均为1)，将它传送给a，而a是无符号型变量，16个位全1是十进制的65535。如果b为正值，且在0~32767之间，则赋值后数值不变.
a第一位不是符号位么.?为什么16个位全是1,最后一位不算服符号位?

unsigned char discode[] ={0X03,0X9F,0X25,0X0D,0X99,0X41,0X1F,0X01,0X09};
x05x05x05unsigned char const positon[3]={ 0xdf,0xbf,0x7f};
x05x05x05unsigned char disbuff[4]x05 ={ 0,0,0,};
//
void Display(void)x05x05x05x05 //扫描数码管
x05{
x05 if(posit==0) //就是这句话
x05 {P0=(discode[disbuff[posit]])&0x7f;} //就是这句话
x05 elsex05x05x05x05x05x05x05x05 //就是这句话
x05 {P0=discode[disbuff[posit]];}x05x05 //就是这句话
P2=positon[posit];
x05 if(++posit>=3)
x05 posit=0;
x05}

感激 若能答好追加100分
#include
#include
/*函数Dto2(unsigned n)的功能是将无符号的10进制数n转换成二进制数并输出,要求使用栈结构,
请将函数补充完整,并调试运行*/
void Dto2(unsigned n)
{int stack[100];
int top=-1;
/*请在此处将本函数补充完整*/
while(n>0)
{stack[++top]=n%2;
n=n/2;
}
while(top>-1)
printf("%d",stack[top--]);
}
void main()
{unsigned n=0xA0B0;
printf("n=%xn",n);
printf("n=");
Dto2(n);
getch();
}
奇怪了n不是十六进制数吗?这个Dto2是将十进制转化成二进制
这个怎么能行呢
怎么不用将十六进制先转化成十进制在用Dto2函数呢?
放心虽然现在我只有56分,但会天天登录
一定会弄到100分来答谢的
不能理解,可否说详细些 unsigned是定义无符号的,难道无符号的就一定是十进制吗,unsigned n=0xA0B0;
printf("n=%xn",n); n还不是十六进制的吗
怎么没人理,那就先给50分,满意在加50

C语言问题 不正确的变量说明是
不正确的变量说明是（ ）.

A. unsigned int ui;

B. short int s;
C. double int A;
D. int i,j,k;

求助tmp1[0]=tmp1[1]=tmp1[2]=tmp1[3]=*(have i);Unsignedshortdestinati
beginunsignedcharmin;

关于一道程序填空题的疑惑
#include
#include
#include
unsigned long fun(char *s)
{
unsigned long t = 0;
int k;
int i = 0;
i = strlen(s);
for (--i; i>=0; i--)
{
k = s[i]-'0';
t = t*10+k;
}
return t;
}
main()
{
char str[8];
printf("Enter a string made up of '0' to '9' digital character :n");
gets(str);
printf("The string is :%sn",str);
if (strlen(str) > 8)
printf(" The string is too long ");
else
printf("The result :%lun",fun(str));
}
fun函数中的for循环里,为什么k=s[i]-'0'还有为什么t=k+10*t?
感激不尽

在Vhdl中,定义count是unsigned(3downto0),怎么理解count=count+1这个表达式,其中1怎么理解,
还有其中1为什么不能写成'1'?

请帮忙分析一下句子成分及主干,
A 32-bit unsigned fixed-point number (16.16) that indicates the rate at which the sound samples were obtained.
最好有翻译，

C语言位运算问题
下列程序的运行结果是_______.
　　main（）
　　{ unsigned c1=0xff，c2=0x00；
　　c1=c2|c1>>2；c2=c1^0236；
　　printf（“%x，%xn”，c1，c2）；
　　}
　A）0x3f，0XA1 B）3f，A1 C）ffff，61 D）3f，a1
求高人解释一下这个问题，我算的答案都不对，答案应该是选D

c++指针强制类型转换 unsigned int a = 0xFFFFFFF7; char*b = (char*)&a; printf("%08x",*b);输出的问题
unsigned int a = 0xFFFFFFF7;
char*b = (char*)&a;
printf("%08x",*b);
这段代码输出为FFFFFFF7,可以理解为*b输出了a存储区的数据,可是将char*换成unsigned char*之后,输出为000000F7,只输出了低字节,一个字节的内容,这是为什么?

英语翻译
void
GPIO_PORT_D_ISR(void)
{
unsigned char ucKey;
unsigned long ulStatus;
ulStatus = GPIOPinIntStatus(GPIO_PORTD_BASE,true);
GPIOPinIntClear(GPIO_PORTD_BASE,ulStatus);
if(ulStatus & GPIO_PIN_7)
{
ucKey = GPIOPinRead(GPIO_PORTC_BASE,GPIO_PIN_4);
ucKey = ucKey>>1;
ucValue = ucValue + ucKey;
//ucValue = ucValue

USE IEEE.STD_LOGIC_UNSIGNED.ALL; 报错
LIBRARY IEEE;
USE IEEE.STD_LOGIC_1164.ALL
USE IEEE.STD_LOGIC_UNSIGNED.ALL;
ENTITY CNT10 IS
PORT (CLK,RST,EN,LOAD:IN STD_LOGIC;
DATA:IN STD_LOGIC_VECTOR(3 DOWNTO 0);
DOUT:OUT STD_LOGIC_VECTOR(3 DOWNTO 0);
COUT:OUT STD_LOGIC);
END CNT10;
ARCHITECTURE behav OF CNT10 IS
BEGIN
PROCESS(CLK,RST,EN,LOAD)
VARIABLE Q:STD_LOGIC_VECTOR(3 DOWNTO 0);
BEGIN
IF RST='0' THEN Q:=(OTHERS=>'0');
ELSIF CLK' EVENT AND CLK='1' THEN
IF EN='1' THEN
IF (LOAD='0') THEN Q:=DATA;ELSE
IF Q'0');
ENDIF;
ENDIF;
ENDIF;
ENDIF;
IF Q="1001" THEN COUT

一个简单的C++问题,
64-bit unsigned integer是什么意思,在C++中用什么表示?
原C++题如下：
Description
This is a classical problem.Your job is to calculate the combination formula ,
C(n,m) = / (m!(n - m)!)
since n,m and C(n,m) fit a 64-bit unsigned integer.
Input
The first line of input there is one integer T (T

这个程序不懂,
#include
#define uchar unsigned char
#define uint unsigned int
sbit BEEP = P3^7;
uchar code DSY_CODE[]=
{
0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90,0x88,0x83,0xc6,0xa1,0x86,0x8e,0x00
};
uchar Pre_KeyNO = 16,KeyNO = 16;（这个不懂）
void DelayMS(uint ms)
{
uchar t;
while(ms--)
{
for(t=0;t> 4 ^ 0x0f;（这个不懂）
switch(Tmp)
{
case 1:KeyNO += 0; break;（为什么有+)
case 2:KeyNO += 4; break;
case 4:KeyNO += 8; break;
case 8:KeyNO += 12;
}
}
void Beep()
{
uchar i;
for(i=0;i

C遗传算法
struct individual
{
unsigned char *chrom;
double fitness;
double varible;
int xsite;
int parent[2];
int *utility;
};
struct individual *oldpop;
struct individual *newpop;
void initmalloc()
{
unsigned char nbytes;
int j;
nbytes=popsize*sizeof(struct individual);
if((oldpop=(struct individual*)malloc(nbytes))==NULL)
nomemory("oldpop");
if((newpop=(struct individual*)malloc(nbytes))==NULL)
nomemory("newpop");
nbytes=chromsize*sizeof(unsigned char);
for(j=0;j

编写具有如下原型的函数：int f(unsigned long x,int n,int& Lxn);
它负责将整数x的第n位（从左边数第n位,n>0）的数值放到引用Lxn之中（将作为结果返回到主调函数的对应实参变量中）,并将倒数第n位（从右边数第n位,n>0）的数值作为函数结果返回去.并编制主函数对它进行调用以验证其正确性.
例如,当x=123456789,n=7时,执行语句“Rxn=f(x,n,Lxn);”将使返回的Lxn为7,并使Rxn变为3；而执行语句“Rxn=f(12345,6,Lxn);”将使Lxn与Rxn都变为为0（超出数的“长度”即总位数时返回0）