This question already has answers here:
How to write an array to file in C
(3 answers)
Closed 3 years ago.
I would like to write an array of integers into a file using C. However, I get some gibberish in the file.
The code is about a function that converts a decimal number into binary then stores it into a file.
int * decToBinary(int n) //function to transform the decimal numbers to binary
{
static int binaryNum[16]; // array to store binary number
int i = 0; // counter for binary array
while (n > 0) {
binaryNum[i] = n % 2; // storing remainder in binary array
n = n / 2;
i++;
}
return binaryNum;
}
int main()
{
FILE *infile;
int i;
int *p;
int decimal= 2000;
int written = 0;
infile = fopen("myfile.txt","w");
p = decToBinary(decimal);
written = fwrite(p,sizeof(int),sizeof(p),infile) ;
if (written == 0) {
printf("Error during writing to file !");
}
fclose(infile);
return 0;
}
This is what I get in my file:
This is what I get when I write a text as a test, it does not have any problem with the text, but it has with the array.
char str[] = "test text --------- \n";
infile = fopen("myfile.txt","wb");
p=decToBinary(decimal);
fwrite(str , 1 , sizeof(str) , infile);
written = fwrite(p,sizeof(int),sizeof(p),infile) ;
And this is what I get when I make this change:
written = fwrite(&p,sizeof(int),sizeof(p),infile) ;
First, be aware that there are two interpretations for 'binary':
int n = 1012;
fwrite(&n, sizeof(n), 1, file);
This writes out the data just as is; as it is represented in form of bits, output is considered "binary" (a binary file).
Your question and the code you provided, though, rather imply that you actually want to have a file containing the numbers in binary text format, i. e. 7 being represented by string "111".
Then first, be aware that 0 and 1 do not represent the characters '0' and '1' in most, if not all, encodings. Assuming ASCII or compatible, '0' is represented by value 48, '1' by value 49. As C standard requires digits [0..9] being consecutive characters (this does not apply for any other characters!), you can safely do:
binaryNum[i] = '0' + n % 2;
Be aware that, as you want strings, you chose the bad data type, you need a character array:
static char binaryNum[X];
X??? We need to talk about required size!
If we create strings, we need to null-terminate them. So we need place for the terminating 0-character (really value 0, not 48 for character '0'), so we need at least one character more.
Currently, due to the comparison n > 0, you consider negative values as equal to 0. Do you really intend this? If so, you might consider unsigned int as data type, otherwise, leave some comment, then I'll cover handling negative values later on.
With restriction to positive values, 16 + 1 as size is fine, assuming int has 32 bit on your system! However, C standard allows int to be smaller or larger as well. If you want to be portable, use CHAR_BIT * sizeof(int) / 2 (CHAR_BIT is defined in <limits.h>; drop division by 2 if you switch to unsigned int).
There is one special case not covered: integer value 0 won't enter the loop at all, thus you'd end up with an empty string, so catch this case separately:
if(n == 0)
{
binaryNum[i++] = '0';
}
else
{
while (n > 0) { /.../ }
}
// now the important part:
// terminate the string!
binaryNum[i] = 0;
Now you can simply do (assuming you changed p to char*):
written = fprintf(file, "%s\n", p);
// ^^ only if you want to have each number on separate line
// you can replace with space or drop it entirely, if desired
Be aware that the algorithm, as is, prints out least significant bits first! You might want to have it inverse, then you'd either yet have to revert the string or (which I would prefer) start with writing the terminating 0 to the end and then fill up the digits one by one towards front - returning a pointer to the last digit (the most significant one) written instead of always the start of the buffer.
One word about your original version:
written = fwrite(p, sizeof(int), sizeof(p), infile);
sizeof(p) gives you the size of a pointer; this one is system dependent, but will always be the same on the same system, most likely 8 on yours (if modern 64-bit hardware), possibly 4 (on typical 32-bit CPU), other values on less common systems are possible as well. You'd need to return the number of characters printed separately (and no, sizeof(binaryNum) won't be suitable as it always returns 17, assuming 32-bit int and all changes shown above applied).
You probably want this:
...
int main()
{
int decimal = 2000;
int *p = decToBinary(decimal);
for (int i = 0; i< 16; i++)
{
printf("%d", p[i]);
}
return 0;
}
The output goes to the terminal instead into a file.
For writing into a file use fopen as in your code, and use fprintf instead of printf.
Concerning the decToBinary there is still room for improvement, especially you could transform the number directly into an array of char containing only chars 0 and 1 using the << and & operators.
Related
I wrote this function that performs a slightly modified variation of run-length encoding on text files in C.
I'm trying to generalize it to binary files but I have no experience working with them. I understand that, while I can compare bytes of binary data much the same way I can compare chars from a text file, I am not sure how to go about printing the number of occurrences of a byte to the compressed version like I do in the code below.
A note on the type of RLE I'm using: bytes that occur more than once in a row are duplicated to signal the next-to-come number is in fact the number of occurrences vs just a number following the character in the file. For occurrences longer than one digit, they are broken down into runs that are 9 occurrences long.
For example, aaaaaaaaaaabccccc becomes aa9aa2bcc5.
Here's my code:
char* encode(char* str)
{
char* ret = calloc(2 * strlen(str) + 1, 1);
size_t retIdx = 0, inIdx = 0;
while (str[inIdx]) {
size_t count = 1;
size_t contIdx = inIdx;
while (str[inIdx] == str[++contIdx]) {
count++;
}
size_t tmpCount = count;
// break down counts with 2 or more digits into counts ≤ 9
while (tmpCount > 9) {
tmpCount -= 9;
ret[retIdx++] = str[inIdx];
ret[retIdx++] = str[inIdx];
ret[retIdx++] = '9';
}
char tmp[2];
ret[retIdx++] = str[inIdx];
if (tmpCount > 1) {
// repeat character (this tells the decompressor that the next digit
// is in fact the # of consecutive occurrences of this char)
ret[retIdx++] = str[inIdx];
// convert single-digit count to string
snprintf(tmp, 2, "%ld", tmpCount);
ret[retIdx++] = tmp[0];
}
inIdx += count;
}
return ret;
}
What changes are in order to adapt this to a binary stream? The first problem I see is with the snprintf call since it's operating using a text format. Something that rings a bell is also the way I'm handling the multiple-digit occurrence runs. We're not working in base 10 anymore so that has to change, I'm just unsure how having almost never worked with binary data.
A few ideas that can be useful to you:
one simple method to generalize RLE to binary data is to use a bit-based compression. For example the bit sequence 00000000011111100111 can be translated to the sequence 0 9623. Since the binary alphabet is composed by only two symbols, you need to only store the first bit value (this can be as simple as storing it in the very first bit) and then the number of the contiguous equal values. Arbitrarily large integers can be stored in a binary format using Elias gamma coding. Extra padding can be added to fit the entire sequence nicely into an integer number of bytes. So using this method, the above sequence can be encoded like this:
00000000011111100111 -> 0 0001001 00110 010 011
^ ^ ^ ^ ^
first bit 9 6 2 3
If you want to keep it byte based, one idea is to consider all the even bytes frequencies (interpreted as an unsigned char) and all the odd bytes the values. If one byte occur more than 255 times, than you can just repeat it. This can be very inefficient, though, but it is definitively simple to implement, and it might be good enough if you can make some assumptions on the input.
Also, you can consider moving out from RLE and implement Huffman's coding or other sophisticated algorithms (e.g. LZW).
Implementation wise, i think tucuxi already gave you some hints.
You only have to address 2 problems:
you cannot use any str-related functions, because C strings do not deal well with '\0'. So for example, strlen will return the index of the 1st 0x0 byte in a string. The length of the input must be passed in as an additional parameter: char *encode(char *start, size_t length)
your output cannot have an implicit length of strlen(ret), because there may be extra 0-bytes sprinkled about in the output. You again need an extra parameter: size_t encode(char *start, size_t length, char *output) (this version would require the output buffer to be reserved externally, with a size of at least length*2, and return the length of the encoded string)
The rest of the code, assuming it was working before, should continue to work correctly now. If you want to go beyond base-10, and for instance use base-256 for greater compression, you would only need to change the constant in the break-things-up loop (from 9 to 255), and replace the snprintf as follows:
// before
snprintf(tmp, 2, "%ld", tmpCount);
ret[retIdx++] = tmp[0];
// after: much easier
ret[retIdx++] = tmpCount;
This question already has answers here:
What does void* mean and how to use it?
(10 answers)
Closed 5 years ago.
I'm trying to make a calculator that takes a void pointer called yourVal, look at the first byte and decide if it's a '*' or '/'. Based on the sign, i multiply bytes 3+4, 5+6, and 7+8. say i have *1234567. I multiply 23 * 45 * 67. With the division, I divide byte 5(45) by byte 3(23). I'm a novice with pointers in C, and I really have no idea how to even set a value to a void pointer. When I do the following in main
void *yourVal;
*yourVal = "*1234567";
printf("%s\n", yourVal);
I'm not able to dereference a void pointer. But I tried with a char pointer, and I have the same issue.
This is my code for the calculator function. Based on whether I use printf or not, I get different results.
int calculator(void *yourVal){
char *byteOne;
short int *byteThree, *byteFive, *byteSeven;
int value;
byteOne = (char *)yourVal;
byteThree = (short int *)yourVal+2;
byteFive = (short int *)yourVal+4;
byteSeven= (short int *)yourVal+6;
if(*byteOne == '*') {
value = *byteThree * *byteFive * *byteSeven;
printf("You multiplied\n");
}
else if(*byteOne == '/') {
if (*byteThree == 0) {
value = 0xBAD;
printf("Your input is invalid\n");
}
else {
value = *byteFive / *byteThree;
printf("You divided\n");
}
}
else {
value = 0xBAD;
printf("Your input is invalid\n");
}
}
The division isn't working at all, and the multiplication only grabs one digit. Any tips would be appreciated. I looked at various sources but I'm not seeing how to work with void pointers efficiently. Also, I can't use any library functions other than printf, and this is a school assignment, so try not to give too many spoilers or do it for me. We were given one hint, which is to cast yourVal to a structure. But I'm lost on that. Thanks
byteOne = (char *)payload;
byteThree = (short int *)yourVal+2;
byteFive = (short int *)yourVal+4;
byteSeven= (short int *)yourVal+6;
This doesn't do what you think it does. If you want to read the numbers at these positions, you need to do something like.
char* Value = yourValue;
unsigned byteOne, byteThree, byteFive, byteSeven;
byteOne = Value[0] - '0';
byteThree = Value[2] - '0';
byteFive = Value[4] - '0';
byteSeven = Value[6] - '0';
What I have done here is read the byte at that position and subtract the ASCII value of '0' to get the numerical value of that character. But again this will work only for a single character.
If you need to read more characters you will have to use library functions like sscanf or atoi.
The void pointer adds no functionality you need to solve this problem, it just complicates things. Use a char pointer instead.
"*1234567" is a string, not an array of integers. You cannot treat it as an array of integers. Each character would have to be converted to an integer before you do arithmetic. The easiest way to do that is to subtract by the ASCII character '0'.
"...i multiply bytes 3+4..." When counting bytes, you always start at 0. In the string "*1234567", the 2 is the byte with index 2 not 3.
"Based on the sign, i multiply bytes 3+4, 5+6, and 7+8. say i have *1234567. I multiply 23 * 45 * 67. With the division, I divide byte 5(45) by byte 3(23)"
I fail to see how this algorithm makes any sense. What is the 1 there for? Why aren't you using some conventional formatting such as prefix, postfix or just plainly typed-out equations?
Example:
int calculator (const char* yourVal)
...
int byte2 = yourVal[2] - '0';
I have to print 1,000,000 four digit numbers. I used printf for this purpose
for(i=0;i<1000000;i++)
{
printf("%d\n", students[i]);
}
and it turns out to be too slow.Is there a faster way so that I can print it.
You could create an array, fill it with output data and then print out that array at once. Or if there is memory problem, just break that array to smaller chunks and print them one by one.
Here is my attempt replacing printf and stdio stream buffering with straightforward special-case code:
int print_numbers(const char *filename, const unsigned int *input, size_t len) {
enum {
// Maximum digits per number. The input numbers must not be greater
// than this!
# if 1
DIGITS = 4,
# else
// Alternative safe upper bound on the digits per integer
// (log10(2) < 28/93)
DIGITS = sizeof *input * CHAR_BIT * 28UL + 92 / 93,
# endif
// Maximum lines to be held in the buffer. Tune this to your system,
// though something on the order of 32 kB should be reasonable
LINES = 5000
};
// Write the output in binary to avoid extra processing by the CRT. If necessary
// add the expected "\r\n" line endings or whatever else is required for the
// platform manually.
FILE *file = fopen(filename, "wb");
if(!file)
return EOF;
// Disable automatic file buffering in favor of our own
setbuf(file, NULL);
while(len) {
// Set up a write pointer for a buffer going back-to-front. This
// simplifies the reverse order of digit extraction
char buffer[(DIGITS + 1 /* for the newline */) * LINES];
char *tail = &buffer[sizeof buffer];
char *head = tail;
// Grab the largest set of lines still remaining to be printed which
// will safely fit in our buffer
size_t chunk = len > LINES ? LINES : len;
const unsigned int *input_chunk;
len -= chunk;
input += chunk;
input_chunk = input;
do {
// Convert the each number by extracting least-significant digits
// until all have been printed.
unsigned int number = *--input_chunk;
*--head = '\n';
do {
# if 1
char digit = '0' + number % 10;
number /= 10;
# else
// Alternative in case the compiler is unable to merge the
// division/modulo and perform reciprocal multiplication
char digit = '0' + number;
number = number * 0xCCCDUL >> 19;
digit -= number * 10;
# endif
*--head = digit;
} while(number);
} while(--chunk);
// Dump everything written to the present buffer
fwrite(head, tail - head, 1, file);
}
return fclose(file);
}
I fear this won't buy you much more than a fairly small constant factor over your original (by avoiding some printf format parsing, per-character buffering, locale handling, multithreading locks, etc.)
Beyond this you may want to consider processing the input and writing the output on-the-fly instead of reading /processing/writing as separate stages. Of course whether or not this is possible depends entirely on the operation to be performed.
Oh, and don't forget to enable compiler optimizations when building the application. A run through with a profiler couldn't hurt either.
So, I wrote a function converting a decimal number into a hexadecimal number by using recursion, but I can't seem to figure out how to add the prefix "0x" and leading zeros to my converted hexadecimal number. Let's say I pass the number 18 into the parameters of my function. The equivalent hexadecimal number should be 0x00000012. However, I only end up getting 12 as my hexidecimal number. The same applies when I pass in a hexidecimal number 0xFEEDDAD. I end up getting only FEEDDAD without the prefix as my answer. Can someone please help me figure this out? I've listed my code below. Also, I'm only allowed to use fputc to display my output.
const char digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
void hexout (unsigned long number, FILE * stream)
{
long quotient;
long remainder;
quotient = number / 16;
remainder = number % 16;
if(quotient != 0)
hexout(quotient,stream);
fputc(digits[remainder],stream);
}
void hexout (unsigned long number, FILE * stream)
{
fprintf(stream, "0x%08lX", number);
}
If you cannot use fprintf (neither sprintf), you can use this kind of code (no recursion, but a 8-chars array on the stack):
const char digits[] = "0123456789ABCDEF";
void hexout(unsigned long number, FILE * stream)
{
unsigned long int input = number;
unsigned long int quotient;
unsigned long int remainder;
unsigned short ndigit = 0;
char result[8] = {0};
// Compute digits
do
{
quotient = input / 16;
remainder = input % 16;
result[7-ndigit] = digits[remainder];
input = quotient;
ndigit++;
}
while (ndigit < 8);
// Display result
fputc('0', stream);
fputc('x', stream);
for (ndigit = 0; ndigit < 8; ndigit++)
{
fputc(result[ndigit], stream);
}
}
Of course, this can be improved a lot...
Add digits to a string, and print out string with zero-padding using fprintf. Or just use fprintf to begin with.
Your own hexout fails for obvious reasons. You cannot 'continue' to output a number of zeroes when the value reaches 0, because you don't know how much numbers you already emitted. Also, you don't know when to prepend "0x" -- it should be before you start to emit hex digits, but how can you know you are at the start?
The logical way¹ to do this is to not use recursion, but a simple loop instead. Then again -- unsaid, but a fair bet this is a homework assignment, and in that case any number of silly constraints are possible ("write a C program without using the character '{'" comes to mind). In your case it's "you must use recursion".
You must add a counter to your recursive function; when it reaches 0, you know you have output 0x, and if it's not 0 you need to output a hex digit, irrespective if your value is 0 or not. There are a couple of ways of adding a counter to a recursive function: a global variable (which would be the easiest and utterly ugliest way, so please don't stop reading here), a static variable -- only semantically better than a global --, or a pass-by-reference argument (of which some say is a myth, but then again the end result is the same).
Which method is best for you depends on how well you can defend why you used that method.
¹ So is printf("0x%08X") an "illogical" solution? Yes. It solves the problem but without any further insights. The purpose of this assignment is not to find out the existence of printf and its parameters, it's to learn how (and why) to use recursion.
n.b. I know that this question has been asked on StackOverflow before in a variety of different ways and circumstances, but the search for the answer I seek doesn't quite help my specific case. So while this initially looks like a duplicate of a question such as How can I convert an integer to a hexadecimal string in C? the answers given, are accurate, but not useful to me.
My question is how to convert a decimal integer, into a hexadecimal string, manually. I know there are some beat tricks with stdlib.h and printf, but this is a college task, and I need to do it manually (professor's orders). We are however, permitted to seek help.
Using the good old "divide by 16 and converting the remainder to hex and reverse the values" method of obtaining the hex string, but there must be a big bug in my code as it is not giving me back, for example "BC" for the decimal value "188".
It is assumed that the algorithm will NEVER need to find hex values for decimals larger than 256 (or FF). While the passing of parameters may not be optimal or desirable, it's what we've been told to use (although I am allowed to modify the getHexValue function, since I wrote that one myself).
This is what I have so far:
/* Function to get the hex character for a decimal (value) between
* 0 and 16. Invalid values are returned as -1.
*/
char getHexValue(int value)
{
if (value < 0) return -1;
if (value > 16) return -1;
if (value <= 9) return (char)value;
value -= 10;
return (char)('A' + value);
}
/* Function asciiToHexadecimal() converts a given character (inputChar) to
* its hexadecimal (base 16) equivalent, stored as a string of
* hexadecimal digits in hexString. This function will be used in menu
* option 1.
*/
void asciiToHexadecimal(char inputChar, char *hexString)
{
int i = 0;
int remainders[2];
int result = (int)inputChar;
while (result) {
remainders[i++] = result % 16;
result /= (int)16;
}
int j = 0;
for (i = 2; i >= 0; --i) {
char c = getHexValue(remainders[i]);
*(hexString + (j++)) = c;
}
}
The char *hexString is the pointer to the string of characters which I need to output to the screen (eventually). The char inputChar parameter that I need to convert to hex (which is why I never need to convert values over 256).
If there is a better way to do this, which still uses the void asciiToHexadecimal(char inputChar, char *hexString) function, I am all ears, other than that, my debugging seems to indicate the values are ok, but the output comes out like \377 instead of the expected hexadecimal alphanumeric representation.
Sorry if there are any terminology or other problems with the question itself (or with the code), I am still very new to the world of C.
Update:
It just occurred to me that it might be relevant to post the way I am displaying the value in case its the printing, and not the conversion which is faulty. Here it is:
char* binaryString = (char*) malloc(8);
char* hexString = (char*) malloc(2);
asciiToBinary(*(asciiString + i), binaryString);
asciiToHexadecimal(*(asciiString + i), hexString);
printf("%6c%13s%9s\n", *(asciiString + i), binaryString, hexString);
(Everything in this code snip-pit works except for hexString)
char getHexValue(int value)
{
if (value < 0) return -1;
if (value > 16) return -1;
if (value <= 9) return (char)value;
value -= 10;
return (char)('A' + value);
}
You might wish to print out the characters you get from calling this routine for every value you're interested in. :) (printf(3) format %c.)
When you call getHexValue() with a number between 0 and 9, you return a number between 0 and 9, in the ASCII control-character range. When you call getHexValue() with a number between 10 and 15, you return a number between 65 and 75, in the ASCII letter range.
The sermon? Unit testing can save you hours of time if you write the tests about the same time you write the code.
Some people love writing the tests first. While I've never had the discipline to stick to this approach for long, knowing that you have to write tests will force you to write code that is easier to test. And code that is easier to test is less coupled (or 'more decoupled'), which usually leads to fewer bugs!
Write tests early and often. :)
Update: After you included your output code, I had to comment on this too :)
char* binaryString = (char*) malloc(8);
char* hexString = (char*) malloc(2);
asciiToBinary(*(asciiString + i), binaryString);
asciiToHexadecimal(*(asciiString + i), hexString);
printf("%6c%13s%9s\n", *(asciiString + i), binaryString, hexString);
hexString has been allocated one byte too small to be a C-string -- you forgot to leave room for the ASCII NUL '\0' character. If you were printing hexString by the %c format specifier, or building a larger string by using memcpy(3), it might be fine, but your printf() call is treating hexString as a string.
In general, when you see a
char *foo = malloc(N);
call, be afraid -- the C idiom is
char *foo = malloc(N+1);
That +1 is your signal to others (and yourself, in two months) that you've left space for the NUL. If you hide that +1 in another calculation, you're missing an opportunity to memorize a pattern that can catch these bugs every time you read code. (Honestly, I found one of these through this exact pattern on SO just two days ago. :)
Is the target purely hexadecimal, or shall the function be parametizable. If it's constrained to hex, why not exploit the fact, that a single hex digit encodes exactly four bits?
This is how I'd do it:
#include <stdlib.h>
#include <limits.h> /* implementation's CHAR_BIT */
#define INT_HEXSTRING_LENGTH (sizeof(int)*CHAR_BIT/4)
/* We define this helper array in case we run on an architecture
with some crude, discontinous charset -- THEY EXIST! */
static char const HEXDIGITS[0x10] =
{'0', '1', '2', '3', '4', '5', '6', '7',
'8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
void int_to_hexstring(int value, char result[INT_HEXSTRING_LENGTH+1])
{
int i;
result[INT_HEXSTRING_LENGTH] = '\0';
for(i=INT_HEXSTRING_LENGTH-1; value; i--, value >>= 4) {
int d = value & 0xf;
result[i] = HEXDIGITS[d];
}
for(;i>=0;i--){ result[i] = '0'; }
}
int main(int argc, char *argv[])
{
char buf[INT_HEXSTRING_LENGTH+1];
if(argc < 2)
return -1;
int_to_hexstring(atoi(argv[1]), buf);
puts(buf);
putchar('\n');
return 0;
}
I made a librairy to make Hexadecimal / Decimal conversion without the use of stdio.h. Very simple to use :
char* dechex (int dec);
This will use calloc() to to return a pointer to an hexadecimal string, this way the quantity of memory used is optimized, so don't forget to use free()
Here the link on github : https://github.com/kevmuret/libhex/
You're very close - make the following two small changes and it will be working well enough for you to finish it off:
(1) change:
if (value <= 9) return (char)value;
to:
if (value <= 9) return '0' + value;
(you need to convert the 0..9 value to a char, not just cast it).
(2) change:
void asciiToHexadecimal(char inputChar, char *hexString)
to:
void asciiToHexadecimal(unsigned char inputChar, char *hexString)
(inputChar was being treated as signed, which gave undesirable results with %).
A couple of tips:
have getHexValue return '?' rather than -1 for invalid input (make debugging easier)
write a test harness for debugging, e.g.
int main(void)
{
char hexString[256];
asciiToHexadecimal(166, hexString);
printf("hexString = %s = %#x %#x %#x ...\n", hexString, hexString[0], hexString[1], hexString[2]);
return 0;
}
#include<stdio.h>
char* inttohex(int);
main()
{
int i;
char *c;
printf("Enter the no.\n");
scanf("%d",&i);
c=inttohex(i);
printf("c=%s",c);
}
char* inttohex(int i)
{
int l1,l2,j=0,n;
static char a[100],t;
while(i!=0)
{
l1=i%16;
if(l1>10)
{
a[j]=l1-10+'A';
}
else
sprintf(a+j,"%d",l1);
i=i/16;
j++;
}
n=strlen(a);
for(i=0;i<n/2;i++)
{
t=a[i];
a[i]=a[n-i-1];
a[n-i-1]=t;
}
//printf("string:%s",a);
return a;
//
}
In complement of the other good answers....
If the numbers represented by these hexadecimal or decimal character strings are huge (e.g. hundreds of digits), they won't fit in a long long (or whatever largest integral type your C implementation is providing). Then you'll need bignums. I would suggest not coding your own implementation (it is tricky to make an efficient one), but use an existing one like GMPlib