为什么1602第二行只显示4个0？是不是需要定义一个0-9的16进制数组，把temp转化到数组里？

Question

为什么1602第二行只显示4个0？是不是需要定义一个0-9的16进制数组，把temp转化到数组里？

#include "reg52.h"
#include "xpt2046.h"
#include "lcd.h"

         
typedef unsigned int u16;
typedef unsigned char u8;

unsigned char code table1[] = "Voltage:";
uint temp;

sbit k1 = P3^2; 
sbit relay = P3^3;

/*void delay(uint i)
{
    while (i--);
}  */


void datapros() 
{
    uint temp;
    static u8 i;
    if (i == 50)
    {
        i = 0;
        temp = Read_AD_Data(0x94);
    }
    i++;

}

void  main()
{
    char i;
    init1602();
    datapros();
    writeinstruction(0x80);                      
    delay(5);
    
    
    EA = 1;
    EX0 = 1;
    EX1 = 1;
    
    for (i = 0; i < 8; ++i)                     
    {
    writedata(table1[i]);
    delay(20);

    }
    writeinstruction(0x02);                     
    writeinstruction(0x80 + 0x40);                  
    delay(10);

    
    writechar(temp);
     
    
    delay(10);
    /*砸瓦鲁多 */
    while (1)
     {
     }                                
    }
       
      void Display()interrupt 0 using 1
{
    
            if (k1 == 0)
            delay(100);
            if (k1 == 0)
            {    
                relay=0;
                delay(500);
                relay = 1;
                
                delay(100);
                
                
                while (!k1);                    
        }
          }

这是1602转化ASCII码程序

void  writechar(unsigned char z)
{
    unsigned char A;
    unsigned char B;
    unsigned char C;
    unsigned char D;
    RS = 1;
    EN = 0;
    RW = 0;
    A = z / 1 % 10;                                   
    B = z / 10 % 10;                               
    C = z / 100 % 10;
    D = z / 1000 % 10;                                
    writedata(D + 48);
    delay(5);
    writedata(C + 48);
    delay(5);
    writedata(B + 48);                            
    delay(5);
    writedata(A + 48);                            
    delay(5);

}

Answer 1

先要确认一下参数z的值是什么，用printf函数打印出来看看。