LED 4x7 segment display from China (3641bs)

I bought from china chip display and had an problem with this display,
it took a lot of time, to found code this is why i give my code with exaplmle and small correction:

// Updated (and added example by Marcin P. Chuć 2016.04.16 marcin [delete it and space] chuc (at) gmail.com
// primary version downloaded from Box

unsigned char LED_0F =
{// 0 1 2 3 4 5 6 7 8 9 A(10) b(11) C(12) d(13) E(14) F(15) - (16) empty(17) dot(18) L1(19) l2(20) l3 (21) l4(22) a1(23), a2 (24)
0xc0, 0xF9,0xA4,0xB0,0x99,0x92,0x82,0xF8,0x80,0x90,0x88,0x83,0xC6,0xA1,0x86,0x8e, 0xFF,0xbf,0x7f, 0xde, 0xe7,0xf3, 0xfc, 0xd2, 0xe4 };
// How it works:
// 0 = segment is on
// 1 = segment if off
// segments ar described as: hgfedcba
// example : “0” = on are: a,b,c,d,e,f = 11000000 binary = 0xc0 Hex
// Wyświetlacz siedmiosegmentowy – Wikipedia, wolna encyklopedia

unsigned char LED[4]; //用于LED的4位显示缓存
int cykl[4];
int cyklDelay=150;

int SCLK = 5; /////////////////////////// Connect to SLCK
int RCLK = 6; /////////////////////////// Connect to RCLK
int DIO = 4; //////////////////////////// Connect to DATA
void setup ()
{
pinMode(SCLK,OUTPUT);
pinMode(RCLK,OUTPUT);
pinMode(DIO,OUTPUT); //让三个脚都是输出状态
}

void CheckLed(unsigned char &l, int &i)
{
i++;
if(i>cyklDelay )
{
i=0;
l=l+1;
}

if(l>24)
{
l=0;
}
}

void loop()
{
LED[0]=3;
LED[1]=2;
LED[2]=1;
LED[3]=0;

cykl[0]=0;
cykl[1]=0;
cykl[2]=0;
cykl[3]=0;

while(1)
{
CheckLed(LED[0],cykl[0]);
CheckLed(LED[1],cykl[1]);
CheckLed(LED[2],cykl[2]);
CheckLed(LED[3],cykl[3]);

LED4_Display ();
}

}

void LED4_Display (void)
{
unsigned char *led_table; // 查表指针
unsigned char i;
//显示第1位
led_table = LED_0F + LED[0];
i = *led_table;
LED_OUT(i);
LED_OUT(0x01);
digitalWrite(RCLK,LOW);
digitalWrite(RCLK,HIGH);
//显示第2位
led_table = LED_0F + LED[1];
i = *led_table;
LED_OUT(i);
LED_OUT(0x02);
digitalWrite(RCLK,LOW);
digitalWrite(RCLK,HIGH);
//显示第3位
led_table = LED_0F + LED[2];
i = *led_table;
LED_OUT(i);
LED_OUT(0x04);
digitalWrite(RCLK,LOW);
digitalWrite(RCLK,HIGH);
//显示第4位
led_table = LED_0F + LED[3];
i = *led_table;
LED_OUT(i);
LED_OUT(0x08);
digitalWrite(RCLK,LOW);
digitalWrite(RCLK,HIGH);
}

void LED_OUT(unsigned char X)
{
unsigned char i;
for(i=8;i>=1;i–)
{
if (X&0x80)
{
digitalWrite(DIO,HIGH);
}
else
{
digitalWrite(DIO,LOW);
}
X<<=1;
digitalWrite(SCLK,LOW);
digitalWrite(SCLK,HIGH);
}

}

And, your question is ?