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?
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Do you mean the hex value as a string?geofftnz– geofftnz2011-03-29 23:18:25 +00:00Commented Mar 29, 2011 at 23:18
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Where is the hexadecimal involved...dreamlax– dreamlax2011-03-29 23:19:26 +00:00Commented Mar 29, 2011 at 23:19
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for example, my number is 55. that means it will be located in my int array like this : n1[0]=5 , n2[1]=5. now i want to find the hex value of 55 which is 37 (that s what i need). my number may be very large. if i made some mistakes when i was explaning, sorry about my english or explanation :)Buzzinga– Buzzinga2011-03-30 00:23:24 +00:00Commented Mar 30, 2011 at 0:23
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3 Answers
So here is my take on the problem:
- You have an array of digits.
- You want to build an unsigned int from this array of digits.
- 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);
7 Comments
dreamlax
The
^ operator is actually the bitwise exclusive-or operator. You need to use something like pow() to perform that type of operation.
J T
@dreamlax: Thanks for the catch. It was sort-of pseudo coded, but good call nonetheless.
Thomas Matthews
Your solution shows how to represent a large hex number using a similar scheme to the OP's. However, conversion between the two schemes is different. Remember that
9 * pow(16,3) is not equal to 9 * pow(10, 3).J T
@Thomas: I believe that is the point. The two schemes are supposed to be different. He wants to rebuild a unsigned int from an array of digits. But he wants to imagine that the array of digits represent hex digits, not decimal.
Ben Voigt
The question says "very large number". I don't think passing a single
unsigned int to printf will handle that. |
I'm unsure if you want the hexadecimal represented as a string. If that's the case, here's some code:
#include <iostream>
#include <stack>
using namespace std;
string toHexa(int num){
string digit = "0123456789ABCDEF", numStr = "";
stack<char> s;
do {
s.push(digit[num%16]);
num /= 16;
} while (num != 0);
while (!s.empty()){
numStr += s.top();
s.pop();
}
return numStr;
}
int main(){
int num = 235; // EB in hexa
cout << num << " to hexadecimal: " << toHexa(num) << endl;
return 0;
}
2 Comments
Eddie Flores
Nice tutorial: Converting decimal to hexadecimal.
Ben Voigt
I doubt that will work for a "very large number" as described in the question.
You could use the GMP library to make this relatively straightforward.
- Use
basic_stringstream<unsigned int>to wrap your array. - Use
operator <<to read it into ampz_tvariable. - Create another
basic_stringstream<unsigned int>for your result. Usestd::hexandoperator >>to write the variable back out in hexadecimal.
That would work on ASCII digits, but yours aren't. You can still use GMP, but you'll want to use the mpn_get_str and mpn_set_str functions instead. You'll need to copy your digits into an unsigned char[] and then you can specify the base for conversion to mp_limb_t and back to a string of digits.