LC-3 to Hexadecimal - c

Question

This program takes the input
1283
5105
and prints out the following, I was wondering how to do it the opposite way around so to take the LC-3 instruction as the input and display it in Hexadecimal as I've been stuck on that for quite awhile. Any help would be great !
`add r1,r2,r3
and r0,r4,r5`
 
#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main(int arga, char *argb[]){
int str; //string
int firstReg=0;
int secondReg=0;
int opCode=0;
int reg=0; //register
char hexa[7]; //hexadecimal value
FILE *fp;
fp = fopen(argb[1],"r");
// Statement 1
while((fscanf(fp,"%s",hexa))!=EOF){
int x =0;
str = 0;
// Statement 2
for (;x<=3;x++){
switch(hexa[x]){
case '0': str = str|0;
break;
case '1': str = str|1;
break;
case '2': str = str|2;
break;
case '3': str = str|3;
break;
case '4': str = str|4;
break;
case '5': str = str|5;
break;
case '6': str = str|6;
break;
case '7': str = str|7;
break;
case '8': str = str|8;
break;
case '9': str = str|9;
break;
case 'a': str = str|10;
break;
case 'b': str = str|11;
break;
case 'c': str = str|12;
break;
case 'd': str = str|13;
break;
case 'e': str = str|14;
break;
case 'f': str = str|15;
break;
default: printf("\nInvalid hexadecimal digit %c ",hexa[x]); return 0;
}
if(x != 3){
str = str <<4;
}
}
opCode = str&(15<<12);
opCode = opCode>>12;
// Statement 3
if(opCode == 5){
printf("and");
reg= str&(7<<9);
reg= reg>>9;
printf(" r%d",reg);
firstReg =str & (7<<6);
firstReg= firstReg>>6;
printf(", r%d",firstReg);
secondReg = str &(7);
printf(", r%d",secondReg);
printf("\n");
}
// Statement 4
if(opCode == 1) {
printf("add");
reg= str&(7<<9);
reg= reg>>9;
printf(" r%d",reg);
firstReg =str & (7<<6);
firstReg= firstReg>>6;
printf(", r%d",firstReg);
secondReg = str &(7);
printf(", r%d",secondReg);
printf("\n");
}
}
fclose(fp);
return 0;
}