CareerCup

Suppose that int is 32 bit (if not choose an integer-type that is)

1.

#include <stdio.h>

char *to_hex(int x) {
    static char buffer[9];
    sprintf(buffer, "%08X", x);
    return buffer;
}

int main() {
    int x = 0x1234ABCD;
    printf("ToHEX: %s\n", to_hex(x));
    return 0;
}

2.

include <stdio.h>

char *to_hex(int x) {
    static char digits[] = "0123456789ABCDEF";
    static char buffer[9];
    int i;
    for (i=0; i<8; i++) {
        buffer[7-i] = digits[x & 0xF];
        x >>= 4;
    }
    return buffer;
}

int main() {
    int x = 0xF1234ABC;
    printf("ToHEX: %s\n", to_hex(x));
    return 0;
}

static String toHex(int n) {
		int mask = 15;
		StringBuffer sb = new StringBuffer();
		while (n != 0) {
			int index = (n & mask);
			char c = (char) ((index < 10) ? (index + '0') : ('A' + index - 10));
			sb.append(c);
			n = n >>> 4;
		}
		sb.reverse();
		return sb.toString();
	}

I will ask the interviewer, if the integer is a big integer (4 bytes int can't express it)

is there anything wrong with this code(assuming that its 32 bit integer that has to be converted to hexadecimal)
logic: each set of four bits corresponds to a hexadecimal digit, thus the four rightmost bit are collected through bit manipulation and converted to hexadecimal using lookup table.

char[] convert(int n){
mask= ~((~0)<<4);//makes rightmost 4 bits 1
char hex[8];//to store 8 digit hexadecimal
int digit=7;
while(n!=0){
int hex_bit=0;
hex_bit=mask&&no;//stores 4 righmost bits in hex_bit
hex[digit--]=convert_to_hex(hex_bit)//this function implements just a look up from
// 0 to 16 to convert decimal to hexa
n=n>>4;
}
while(digit>-1) hex[digit--]='0';//rest left most hexa digits are set to zero
}

is there anything wrong with this code(assuming that its 32 bit integer that has to be converted to hexadecimal)
logic: each set of four bits corresponds to a hexadecimal digit, thus the four rightmost bit are collected through bit manipulation and converted to hexadecimal using lookup table.

char[] convert(int n){
mask= ~((~0)<<4);//makes rightmost 4 bits 1
char hex[8];//to store 8 digit hexadecimal
int digit=7;
while(n!=0){
int hex_bit=0;
hex_bit=mask&&no;//stores 4 righmost bits in hex_bit
hex[digit--]=convert_to_hex(hex_bit)//this function implements just a look up from
// 0 to 16 to convert decimal to hexa
n=n>>4;
}
while(digit>-1) hex[digit--]='0';//rest left most hexa digits are set to zero
return hex;
}

is there anything wrong with this code(assuming that its 32 bit integer that has to be converted to hexadecimal)
logic: each set of four bits corresponds to a hexadecimal digit, thus the four rightmost bit are collected through bit manipulation and converted to hexadecimal using lookup table.

char[] convert(int n){
mask= ~((~0)<<4);//makes rightmost 4 bits 1
char hex[8];//to store 8 digit hexadecimal
int digit=7;
while(n!=0){
int hex_bit=0;
hex_bit=mask&&no;//stores 4 righmost bits in hex_bit
hex[digit--]=convert_to_hex(hex_bit)//this function implements just a look up from
// 0 to 16 to convert decimal to hexa
n=n>>4;
}
while(digit>-1) hex[digit--]='0';//rest left most hexa digits are set to zero
return hex;
}