Chap10[D]e Displaying data in structure with condition

There is a structure called employee that holds information like employee code, name, date of joining. Write a program to create an array of the structure and enter some data into it. Then ask the user to enter current date. Display the names of those employees whose tenure is 3 or more than 3 years according to the given current date.

This program doesn't check whether your date is correct so even if you put 99990001 it is still valid.

void main()
{
int i,k;
char date[8];
char j[1];

struct em
{
int code;
char name[20];
char datej[8];
};

struct em e[5];

for(i=0;i<5;i++)
{
printf("\nKey in code, name and date(ddmmyyy) of employment for %d) employee\n",i+1);
scanf("%d%s%s",&e[i].code,e[i].name,e[i].datej);
if(strlen(e[i].datej)!=8)
{
printf("\nYou keyed the date wrongly, key again");
i--;
continue;
}
}

while(1)
{
clrscr();
printf("\nEnter current date\n");
scanf("%s",date);


for(i=0;i<5;i++)
{
k=0;
k=funccheck(date,e[i].datej);

if(k==1)
printf("\nEmployee code= %d Employee name= %s has tenure more than or equal 3 years",e[i].code,e[i].name);
}

printf("\nKey q to quit or any other key to continue\n");
scanf("%s",j);

if(j[0]==113)
break;
}
}

funccheck(char *date,char *datej)
{
int a,a1,i,t;

for(i=7,t=1,a=0;i>3;i--,t*=10)
{
a=a+((*(date+i)-48)*t);
}

for(i=7,t=1,a1=0;i>3;i--,t*=10)
{
a1=a1+((*(datej+i)-48)*t);
}

if(a-a1>3)
return(1);
else if(a-a1<0)
return(0);

for(i=3,a=0,t=1;i>1;i--,t*=10)
{
a=a+((*(date+i)-48)*t);
}

for(i=3,a1=0,t=1;i>1;i--,t*=10)
{
a1=a1+((*(datej+i)-48)*t);
}

if(a-a1>0)
return(1);
else if(a-a1<0)
return(0);

for(i=1,a=0,t=1;i>=0;i--,t*=10)
{
a=a+((*(date+i)-48)*t);
}

for(i=1,a1=0,t=1;i>=0;i--,t*=10)
{
a1=a1+((*(datej+i)-48)*t);
}

if(a-a1>=0)
return(1);
else if(a-a1<0)
return(0);
}