How can you convert hexadecimal to binary without using arrays in C program?

/* C Program to convert decimal to binary without using array */ #include int main(){  int n, i, j, bin_rev=0,bin=0;  printf("Enter a number to convert : ");  scanf("%d",&n); /*****************************/  i=1;  while(n){  bin_rev=bin_rev+(n%2)*i;  i=i*10;  n=n/2;  }    i=1;  while(bin_rev){  bin=bin+(bin_rev%10)*i;  i=i*10;  bin_rev=bin_rev/10;  }  /*****************************/ printf("nBinary is %d",bin);    getch();  return 0; } read less

The moment a (C) program is loaded into a computer’s memory by the OS’s loader (in laymen terms - started) two concepts apply to the entity that we call data: storage representation On a separate note - as far as a computer is concerned everything it acts is a number: even what we call code eventually becomes a carefully sequenced set of instructions each of which is still a number. As such, when a number comes to life it is stored in a format that can be acted on or manipulated by a given processor. When a human being wants to look at a number then the number’s representation becomes relevant: there is normally a one-to-many relationship between the storage and the representation since a number in an internal format can be rendered in various numerical basis, for example. A number in an internal, traditionally - binary, format is usually rendered in base sixteen in a textual form - as a sequence of characters each of which belongs to the base sixteen alphabet of 0 - 9 and A - F. If you have access to the variable of an integral binary type, such as int or long, for example, you can render that variable in a binary format directly. Assume that bits is a variable of the unsigned long type whose contents are to be rendered in a binary format on a 64 -bit machine: int i; char bd; unsigned long mask = 1UL << 63; for ( i = 0; i < 64; i++ ) { bd=bits & mask ? '1' : '0'; printf( "%c", bd ); mask>>= 1; } If you want to convert a base sixteen representation of a number to a binary representation then it can be done a multitude of ways. For example, convert the contents of the base sixteen string into an integral binary format via strtol() family of functions and then use the code above to render it visually in base two. If you are absolutely sure that the number rendered in base sixteen as string will fit into an int sized variable integer then: sscanf( string, "%x", &integer ); Adjust the sizes accordingly. If you want to convert a base sixteen character into a base two representation as an academic exercise then one way to do that is to: construct a base sixteen alphabet yourself in b16alph variable as follows: char b16alph[] = { '0', '1', '2', ..., 'F' }; scan the base sixteen string an a left-to-right fashion for each input base sixteen character hexch: find the location of hexch inside b16alph, say it's hexndx the contents of hexndx will be equal to the base ten internal representation of hexch render the contents of hexndx into base two representation as shown above (showing only a nibble, four rightmost bits, at a time: by shifting the mask leftward only 3 times (mask = 1 << 3)) read less