convert char *aa="test" to unsigned char[]={0x74 ,0x65 ,0x73 ,0x74} in C

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How to convert char *aa="test" to unsigned char[]={0x74 ,0x65 ,0x73 ,0x74} in C

Posted 19-Dec-13 20:28pm

kalaivanan from Bangalore, India

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CPallini · Answer 1 · 2013-12-19T21:57:00

Solution 2

As already noted, a simple cast is enough.
Please note aa should be declared const char *.

C

#include <stdio.h>

int main()
{
  const char * aa = "test";
  const unsigned char * bb = (const unsigned char *) aa;

  while ( *bb )
    printf("%02X ", *bb++);

  printf("\n");

  return 0;
}

Posted 19-Dec-13 21:57pm

CPallini

User 59241 · Answer 2 · 2013-12-19T21:40:00

You asked for C so the cast is C. cout is easier than printf to show contents of array.

C++

#include <iostream>
#include <conio.h>
#include <iomanip>


using namespace std;


int main()
{
char *aa="test";

unsigned char *bb = (unsigned char*) aa;

 int n = 0;

 while(bb[n])
 {
	 cout << hex << (int) bb[n] << endl;
	 n++;
 }


_getch();
return 0;
}