Question

I have an int array that represents a very large number such as:

// ...
unsigned int n1[200];
// ...

n1 = {1,3,4,6,1,...} ==> means my number is 13461...

How can I convert that large number to its hex value?

Answer 1

So here is my take on the problem:

1. You have an array of digits.
2. You want to build an unsigned int from this array of digits.
3. The array of digits could be either HEX digits, or DECIMAL digits.

To build this unsigned long long, assuming an array of DECIMAL digits:

unsigned long long myNum = 0;
unsigned int n1[200];

for (int i=0; i < n1.length ; i++ ){
    myNum += pow(10,i) * n1[n1.length - i];
}

To build this unsigned long long, assuming an array of HEX digits:

for (int i=0; i < n1.length ; i++ ){
    myNum += pow(16,i)* n1[n1.length - i];
}

(Notice the base 16)

Disclaimer: limited to exactly 16 digits MAX stored in your array. After that you will overrun the buffer

If it is just a matter of DISLAYING the number in the correct format…

Well, an int is an int is an int… (in memory).

There are 10 fingers on my hands whether or not I call that number 10, or A.

If you want to format the number for DISPLAY in hex, then try something like:

unsigned int i = 10;
//OR
unsigned int i = 0xA;

printf("My number in hex: %x", i);
printf("My number in decimal: %d", i);