"C language" any decimal digit to hex

Approximate idea:
Enter any decimal digit first to determine the number of digits;

uses an array of shapes to store hexadecimal digits; a[0] represents the number of digits in hexadecimal digits;

#include <stdio.h>#include<stdlib.h>#defineMAXSIZE 10int* Ahex (intDec) {    intn=1; intm= -; int*A; A= (int*)malloc(sizeof(int) *maxsize+1);  while(Dec >=m) {m= m* -; N++; }//printf ("number of positions:%d\n", N); *a =N;  while(n>0)    {         * (a+n) = dec/(m/ -); Dec= dec-* (a+n) * (m/ -); N--; M= m/ -; }    returnA;}voidPrint_hex (int*p) {    intn = *p; printf ("hex:0x");  while(n>0)    {        Switch(* (p+N)) {             Case Ten:p rintf ("A"); Break;  Case  One:p rintf ("B"); Break;  Case  A:p rintf ("C"); Break;  Case  -:p rintf ("D"); Break;  Case  -:p rintf ("E"); Break;  Case  the:p rintf ("F"); Break; default: printf ("%d", * (P+n)); Break; } N--; } printf ("\ n");}intMain () {intRec=0; int*p;  while(1) {printf ("Please enter a decimal number:"); scanf ("%d",&rec); P=Ahex (REC);    Print_hex (P); }    return 0;}

Final effect:

"C language" any decimal digit to hex