Related
My doubt is about the basic theory of "or logical operator". Especifically, logical OR returns true only if either one operand is true.
For instance, in this OR expression (x<O || x> 8) using x=5 when I evalute the 2 operand, I interpret it as both of them are false.
But I have an example that does not fit wiht it rule. On the contrary the expression works as range between 0 and 8, both included.
Following the code:
#include <stdio.h>
int main(void)
{
int x ; //This is the variable for being evaluated
do
{
printf("Imput a figure between 1 and 8 : ");
scanf("%i", &x);
}
while ( x < 1 || x > 8); // Why this expression write in this way determinate the range???
{
printf("Your imput was ::: %d ",x);
printf("\n");
}
printf("\n");
}
I have modified my first question. I really appreciate any helpo in order to clarify my doubt
In advance, thank you very much. Otto
It's not a while loop; it's a do ... while loop. The formatting makes it hard to see. Reformatted:
#include <stdio.h>
int main(void) {
int x;
// Execute the code in the `do { }` block once, no matter what.
// Keep executing it again and again, so long as the condition
// in `while ( )` is true.
do {
printf("Imput a figure between 1 and 8 : ");
scanf("%i", &x);
} while (x < 1 || x > 8);
// This creates a new scope. While perfectly valid C,
// this does absolutely nothing in this particular case here.
{
printf("Your imput was ::: %d ",x);
printf("\n");
}
printf("\n");
}
The block with the two printf calls is not part of the loop. The while (x < 1 || x > 8) makes it so that the code in the do { } block runs, so long as x < 1 or x > 8. In other words, it runs until x is between 1 and 8. This has the effect of asking the user to input a number again and again, until they finally input a number that's between 1 and 8.
Whenever I try to run the following code to count my characters I constantly get zero instead of the number of characters I have inserted.
#include <stdio.h>
void main() {
int c;
int count = 0;
while ( (c = getchar() != EOF) && c == 'a' ) {
count = count +1;
}
printf("Number of chara: %d", count);
}
I have altered the code to instead only count whenever 'a' comes up, but still only get zero when I enter my characters and hit the return key.
The condition of your while loop evaluates to false when the program reads a character that isn't an 'a', which ends the while loop. Since your program reads characters in the while loop, no more characters get read.
Try checking if the character is an 'a' inside the loop body before updating the counter instead of in the loop condition.
The loop will stop because you set the c=='a'. Try to remove that in your code.
Try this one:
int c;
int count = 0;
while (c = getchar() != EOF ) {
count = count +1;
}
printf("Number of chara: %d", count);
}`
I'm sorry if I duplicate, but I have tried EVERYTHING, and I can't figure out why this code keeps breaking. The highest-priority goal was to make this code handle input safely, or just anything that the user can type into the console, without it breaking. However, I also need it to be able to run more than once. fgets() won't let me do that since it keeps reading '\n' somewhere and preventing me from entering input more than once when it hits the end of the do/while loop. I have tried fflushing stdin, I have tried scanf("%d *[^\n]"); and just regular scanf("%d *[^\n]");, but none of those work, and, in fact, they break the code! I used this website to try to get the "Safely handling input" code to work, but I don't completely understand what they're doing. I tried to jerry-rig (spelling?) it as best I could, but I'm not sure if I did it right. Did I miss something? I didn't think a problem this seemingly simple could be so much of a headache! >_<
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
using namespace std;
#define BUF_LEN 100
#define SPACE 32
#define SPCL_CHAR1F 33
#define SPCL_CHAR1L 47
#define SPCL_CHAR2F 58
#define SPCL_CHAR2L 64
#define SPCL_CHAR3F 91
#define SPCL_CHAR3L 96
#define NUMF 48
#define NUML 57
#define UC_CHARF 65
#define UC_CHARL 90
#define LC_CHARF 97
#define LC_CHARL 122
void main ()
{
char* buffer;
int SpcCounter=0, SpclCounter=0, NumCounter=0,LcCounter=0, UcCounter=0;
char line[BUF_LEN],response[4];
char*input="";
bool repeat=false;
do
{
for(int i=0;i<BUF_LEN;i++)
{
line[i]=NULL;
}
buffer=NULL;
printf("Enter your mess of characters.\n");
buffer=fgets(line,BUF_LEN,stdin);
//To handle going over the buffer limit: BROKEN
if(buffer!=NULL)
{
size_t last=strlen(line)-1;
if(line[last]=='\n')
line[last]='\0';
else
{
fscanf(stdin,"%c *[^\n]");
}
}
for(int i=0;i<BUF_LEN;i++)
{
char temp=buffer[i];
if(temp==SPACE||temp==255)
SpcCounter++;
else if((temp >= SPCL_CHAR1F && temp <= SPCL_CHAR1L)||/*Special characters*/
(temp >= SPCL_CHAR2F && temp <= SPCL_CHAR2L)||
(temp >= SPCL_CHAR3F && temp <= SPCL_CHAR3L))
SpclCounter++;
else if (temp >=NUMF && temp <= NUML)/*Numbers*/
NumCounter++;
else if (temp >= UC_CHARF && temp <= UC_CHARL)/*Uppercase letters*/
UcCounter++;
else if (temp >= LC_CHARF && temp <= LC_CHARL)/*Lowercase letters*/
LcCounter++;
}
printf("There were %i space%s, %i special character%s, %i number%s, and %i letter%s,\n"
"consisting of %i uppercase letter%s and %i lowercase.\n",
SpcCounter,(SpcCounter==1?"":"s"),SpclCounter,(SpclCounter==1?"":"s"), NumCounter,(NumCounter==1?"":"s"),UcCounter+LcCounter,
(UcCounter+LcCounter==1?"":"s"), UcCounter,(UcCounter==1?"":"s"), LcCounter);
printf("Would you like to do this again? (yes/no)");
input=fgets(response,4,stdin);
/*
ALL BROKEN
if(input!=NULL)
{
size_t last=strlen(response)-1;
if(response[last]=='\n')
response[last]='\0';
else
{
fscanf(stdin,"%*[^\n]");
fscanf(stdin,"%c");
}
}
*/
//To capitalize the letters
for(int i=0;i<4;i++)
{
char* temp=&response[i];
if (*temp >= LC_CHARF && *temp <= LC_CHARL)
*temp=toupper(*temp);//Capitalize it
}
//To set repeat: WORKS, BUT WEIRD
repeat=!strncmp(input,"YES",4);
}
while(repeat);
}
For safe, secure user input in C (and in C++ if I'm using C-style strings), I usually revert to an old favorite of mine, the getLine function:
// Use stdio.h and string.h for C.
#include <cstdio>
#include <cstring>
#define OK 0
#define NO_INPUT 1
#define TOO_LONG 2
static int getLine (char *prmpt, char *buff, size_t sz) {
int ch, extra;
// Output prompt then get line with buffer overrun protection.
if (prmpt != NULL) {
printf ("%s", prmpt);
fflush (stdout);
}
if (fgets (buff, sz, stdin) == NULL)
return NO_INPUT;
// If it was too long, there'll be no newline. In that case, we flush
// to end of line so that excess doesn't affect the next call.
if (buff[strlen(buff)-1] != '\n') {
extra = 0;
while (((ch = getchar()) != '\n') && (ch != EOF))
extra = 1;
return (extra == 1) ? TOO_LONG : OK;
}
// Otherwise remove newline and give string back to caller.
buff[strlen(buff)-1] = '\0';
return OK;
}
This function:
can output a prompt if desired.
uses fgets in a way that avoids buffer overflow.
detects end-of-file during the input.
detects if the line was too long, by detecting lack of newline at the end.
removes the newline if there.
"eats" characters until the next newline to ensure that they're not left in the input stream for the next call to this function.
It's a fairly solid piece of code that's been tested over many years and is a good solution to the problem of user input.
In terms of how you call it for the purposes in your question, I would add something very similar to what you have, but using the getLine function instead of directly calling fgets and fiddling with the results. First some headers and the same definitions:
#include <iostream>
#include <cstdlib>
#include <cctype>
#define BUF_LEN 100
#define SPACE 32
#define SPCL_CHAR1F 33
#define SPCL_CHAR1L 47
#define SPCL_CHAR2F 58
#define SPCL_CHAR2L 64
#define SPCL_CHAR3F 91
#define SPCL_CHAR3L 96
#define NUMF 48
#define NUML 57
#define UC_CHARF 65
#define UC_CHARL 90
#define LC_CHARF 97
#define LC_CHARL 122
Then the first part of main gathering a valid line (using the function) to be evaluated:
int main () {
int SpcCounter, SpclCounter, NumCounter, LcCounter, UcCounter;
char line[BUF_LEN], response[4];
bool repeat = false;
do {
SpcCounter = SpclCounter = NumCounter = LcCounter = UcCounter = 0;
// Get a line until valid.
int stat = getLine ("\nEnter a line: ", line, BUF_LEN);
while (stat != OK) {
// End of file means no more data possible.
if (stat == NO_INPUT) {
cout << "\nEnd of file reached.\n";
return 1;
}
// Only other possibility is "Too much data on line", try again.
stat = getLine ("Input too long.\nEnter a line: ", line, BUF_LEN);
}
Note that I've changed where the counters are set to zero. Your method had them accumulating values every time through the loop rather than resetting them to zero for each input line. This is followed by your own code which assigns each character to a class:
for (int i = 0; i < strlen (line); i++) {
char temp=line[i];
if(temp==SPACE||temp==255)
SpcCounter++;
else if((temp >= SPCL_CHAR1F && temp <= SPCL_CHAR1L)||
(temp >= SPCL_CHAR2F && temp <= SPCL_CHAR2L)||
(temp >= SPCL_CHAR3F && temp <= SPCL_CHAR3L))
SpclCounter++;
else if (temp >=NUMF && temp <= NUML)
NumCounter++;
else if (temp >= UC_CHARF && temp <= UC_CHARL)
UcCounter++;
else if (temp >= LC_CHARF && temp <= LC_CHARL)
LcCounter++;
}
printf("There were %i space%s, %i special character%s, "
"%i number%s, and %i letter%s,\n"
"consisting of %i uppercase letter%s and "
"%i lowercase.\n",
SpcCounter, (SpcCounter==1?"":"s"),
SpclCounter, (SpclCounter==1?"":"s"),
NumCounter, (NumCounter==1?"":"s"),
UcCounter+LcCounter, (UcCounter+LcCounter==1?"":"s"),
UcCounter, (UcCounter==1?"":"s"),
LcCounter);
Then finally, a similar way as above for asking whether user wants to continue.
// Get a line until valid yes/no, force entry initially.
*line = 'x';
while ((*line != 'y') && (*line != 'n')) {
stat = getLine ("Try another line (yes/no): ", line, BUF_LEN);
// End of file means no more data possible.
if (stat == NO_INPUT) {
cout << "\nEnd of file reached, assuming no.\n";
strcpy (line, "no");
}
// "Too much data on line" means try again.
if (stat == TOO_LONG) {
cout << "Line too long.\n";
*line = 'x';
continue;
}
// Must be okay: first char not 'y' or 'n', try again.
*line = tolower (*line);
if ((*line != 'y') && (*line != 'n'))
cout << "Line doesn't start with y/n.\n";
}
} while (*line == 'y');
}
That way, you build up your program logic based on a solid input routine (which hopefully you'll understand as a separate unit).
You could further improve the code by removing the explicit range checks and using proper character classes with cctype(), like isalpha() or isspace(). That would make it more portable (to non-ASCII systems) but I'll leave that as an exercise for later.
A sample run of the program is:
Enter a line: Hello, my name is Pax and I am 927 years old!
There were 10 spaces, 2 special characters, 3 numbers, and 30 letters,
consisting of 3 uppercase letters and 27 lowercase.
Try another line (yes/no): yes
Enter a line: Bye for now
There were 2 spaces, 0 special characters, 0 numbers, and 9 letters,
consisting of 1 uppercase letter and 8 lowercase.
Try another line (yes/no): no
Imagine a std:vector, say, with 100 things on it (0 to 99) currently. You are treating it as a loop. So the 105th item is index 4; forward 7 from index 98 is 5.
You want to delete N items after index position P.
So, delete 5 items after index 50; easy.
Or 5 items after index 99: as you delete 0 five times, or 4 through 0, noting that position at 99 will be erased from existence.
Worst, 5 items after index 97 - you have to deal with both modes of deletion.
What's the elegant and solid approach?
Here's a boring routine I wrote
-(void)knotRemovalHelper:(NSMutableArray*)original
after:(NSInteger)nn howManyToDelete:(NSInteger)desired
{
#define ORCO ((NSInteger)[original count])
static NSInteger kount, howManyUntilLoop, howManyExtraAferLoop;
if ( ... our array is NOT a loop ... )
// trivial, if messy...
{
for ( kount = 1; kount<=desired; ++kount )
{
if ( (nn+1) >= ORCO )
return;
[original removeObjectAtIndex:( nn+1 )];
}
return;
}
else // our array is a loop
// messy, confusing and inelegant. how to improve?
// here we go...
{
howManyUntilLoop = (ORCO-1) - nn;
if ( howManyUntilLoop > desired )
{
for ( kount = 1; kount<=desired; ++kount )
[original removeObjectAtIndex:( nn+1 )];
return;
}
howManyExtraAferLoop = desired - howManyUntilLoop;
for ( kount = 1; kount<=howManyUntilLoop; ++kount )
[original removeObjectAtIndex:( nn+1 )];
for ( kount = 1; kount<=howManyExtraAferLoop; ++kount )
[original removeObjectAtIndex:0];
return;
}
#undef ORCO
}
Update!
InVariant's second answer leads to the following excellent solution. "starting with" is much better than "starting after". So the routine now uses "start with". Invariant's second answer leads to this very simple solution...
N times do if P < currentsize remove P else remove 0
-(void)removeLoopilyFrom:(NSMutableArray*)ra
startingWithThisOne:(NSInteger)removeThisOneFirst
howManyToDelete:(NSInteger)countToDelete
{
// exception if removeThisOneFirst > ra highestIndex
// exception if countToDelete is > ra size
// so easy thanks to Invariant:
for ( do this countToDelete times )
{
if ( removeThisOneFirst < [ra count] )
[ra removeObjectAtIndex:removeThisOneFirst];
else
[ra removeObjectAtIndex:0];
}
}
Update!
Toolbox has pointed out the excellent idea of working to a new array - super KISS.
Here's an idea off the top of my head.
First, generate an array of integers representing the indices to remove. So "remove 5 from index 97" would generate [97,98,99,0,1]. This can be done with the application of a simple modulus operator.
Then, sort this array descending giving [99,98,97,1,0] and then remove the entries in that order.
Should work in all cases.
This solution seems to work, and it copies all remaining elements in the vector only once (to their final destination).
Assume kNumElements, kStartIndex, and kNumToRemove are defined as const size_t values.
vector<int> my_vec(kNumElements);
for (size_t i = 0; i < my_vec.size(); ++i) {
my_vec[i] = i;
}
for (size_t i = 0, cur = 0; i < my_vec.size(); ++i) {
// What is the "distance" from the current index to the start, taking
// into account the wrapping behavior?
size_t distance = (i + kNumElements - kStartIndex) % kNumElements;
// If it's not one of the ones to remove, then we keep it by copying it
// into its proper place.
if (distance >= kNumToRemove) {
my_vec[cur++] = my_vec[i];
}
}
my_vec.resize(kNumElements - kNumToRemove);
There's nothing wrong with two loop solutions as long as they're readable and don't do anything redundant. I don't know Objective-C syntax, but here's the pseudocode approach I'd take:
endIdx = after + howManyToDelete
if (Len <= after + howManyToDelete) //will have a second loop
firstloop = Len - after; //handle end in the first loop, beginning in second
else
firstpass = howManyToDelete; //the first loop will get them all
for (kount = 0; kount < firstpass; kount++)
remove after+1
for ( ; kount < howManyToDelete; kount++) //if firstpass < howManyToDelete, clean up leftovers
remove 0
This solution doesn't use mod, does the limit calculation outside the loop, and touches the relevant samples once each. The second for loop won't execute if all the samples were handled in the first loop.
The common way to do this in DSP is with a circular buffer. This is just a fixed length buffer with two associated counters:
//make sure BUFSIZE is a power of 2 for quick mod trick
#define BUFSIZE 1024
int CircBuf[BUFSIZE];
int InCtr, OutCtr;
void PutData(int *Buf, int count) {
int srcCtr;
int destCtr = InCtr & (BUFSIZE - 1); // if BUFSIZE is a power of 2, equivalent to and faster than destCtr = InCtr % BUFSIZE
for (srcCtr = 0; (srcCtr < count) && (destCtr < BUFSIZE); srcCtr++, destCtr++)
CircBuf[destCtr] = Buf[srcCtr];
for (destCtr = 0; srcCtr < count; srcCtr++, destCtr++)
CircBuf[destCtr] = Buf[srcCtr];
InCtr += count;
}
void GetData(int *Buf, int count) {
int srcCtr = OutCtr & (BUFSIZE - 1);
int destCtr = 0;
for (destCtr = 0; (srcCtr < BUFSIZE) && (destCtr < count); srcCtr++, destCtr++)
Buf[destCtr] = CircBuf[srcCtr];
for (srcCtr = 0; srcCtr < count; srcCtr++, destCtr++)
Buf[destCtr] = CircBuf[srcCtr];
OutCtr += count;
}
int BufferOverflow() {
return ((InCtr - OutCtr) > BUFSIZE);
}
This is pretty lightweight, but effective. And aside from the ctr = BigCtr & (SIZE-1) stuff, I'd argue it's highly readable. The only reason for the & trick is in old DSP environments, mod was an expensive operation so for something that ran often, like every time a buffer was ready for processing, you'd find ways to remove stuff like that. And if you were doing FFT's, your buffers were probably a power of 2 anyway.
These days, of course, you have 1 GHz processors and magically resizing arrays. You kids get off my lawn.
Another method:
N times do {remove entry at index P mod max(ArraySize, P)}
Example:
N=5, P=97, ArraySize=100
1: max(100, 97)=100 so remove at 97%100 = 97
2: max(99, 97)=99 so remove at 97%99 = 97 // array size is now 99
3: max(98, 97)=98 so remove at 97%98 = 97
4: max(97, 97)=97 so remove at 97%97 = 0
5: max(96, 97)=97 so remove at 97%97 = 0
I don't program iphone for know, so I image std::vector, it's quite easy, simple and elegant enough:
#include <iostream>
using std::cout;
#include <vector>
using std::vector;
#include <cassert> //no need for using, assert is macro
template<typename T>
void eraseCircularVector(vector<T> & vec, size_t position, size_t count)
{
assert(count <= vec.size());
if (count > 0)
{
position %= vec.size(); //normalize position
size_t positionEnd = (position + count) % vec.size();
if (positionEnd < position)
{
vec.erase(vec.begin() + position, vec.end());
vec.erase(vec.begin(), vec.begin() + positionEnd);
}
else
vec.erase(vec.begin() + position, vec.begin() + positionEnd);
}
}
int main()
{
vector<int> values;
for (int i = 0; i < 10; ++i)
values.push_back(i);
cout << "Values: ";
for (vector<int>::const_iterator cit = values.begin(); cit != values.end(); cit++)
cout << *cit << ' ';
cout << '\n';
eraseCircularVector(values, 5, 1); //remains 9: 0,1,2,3,4,6,7,8,9
eraseCircularVector(values, 16, 5); //remains 4: 3,4,6,7
cout << "Values: ";
for (vector<int>::const_iterator cit = values.begin(); cit != values.end(); cit++)
cout << *cit << ' ';
cout << '\n';
return 0;
}
However, you might consider:
creating new loop_vector class, if you use this kind of functionality enough
using list if you perform many deletions (or few deletions (not from end, that's simple pop_back) but large array)
If your container (NSMutableArray or whatever) is not list, but vector (i.e. resizable array), you most definitely don't want to delete items one by one, but whole range (e.g. std::vector's erase(begin, end)!
Edit: reacting to comment, to fully realize what must be done by vector, if you erase element other than the last one: it must copy all values after that element (e.g. 1000 items in array, you erase first, 999x copying (moving) of item, that is very costly).
Example:
#include <iostream>
#include <vector>
#include <ctime>
using namespace std;
int main()
{
clock_t start, end;
vector<int> vec;
const int items = 64 * 1024;
cout << "using " << items << " items in vector\n";
for (size_t i = 0; i < items; ++i) vec.push_back(i);
start = clock();
while (!vec.empty()) vec.erase(vec.begin());
end = clock();
cout << "Inefficient method took: "
<< (end - start) * 1.0 / CLOCKS_PER_SEC << " ms\n";
for (size_t i = 0; i < items; ++i) vec.push_back(i);
start = clock();
vec.erase(vec.begin(), vec.end());
end = clock();
cout << "Efficient method took: "
<< (end - start) * 1.0 / CLOCKS_PER_SEC << " ms\n";
return 0;
}
Produces output:
using 65536 items in vector
Inefficient method took: 1.705 ms
Efficient method took: 0 ms
Note it's very easy to get inefficient, look e.g. have at http://www.cplusplus.com/reference/stl/vector/erase/
Say I have a large number (integer or float) like 12345 and I want it to look like 12,345.
How would I accomplish that?
I'm trying to do this for an iPhone app, so something in Objective-C or C would be nice.
Here is the answer.
NSNumber* number = [NSNumber numberWithDouble:10000000];
NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setNumberStyle:kCFNumberFormatterDecimalStyle];
[numberFormatter setGroupingSeparator:#","];
NSString* commaString = [numberFormatter stringForObjectValue:number];
[numberFormatter release];
NSLog(#"%# -> %#", number, commaString);
Try using an NSNumberFormatter.
This should allow you to handle this correctly on an iPhone. Make sure you use the 10.4+ style, though. From that page:
"iPhone OS: The v10.0 compatibility mode is not available on iPhone OS—only the 10.4 mode is available."
At least on Mac OS X, you can just use the "'" string formatter with printf(3).
$ man 3 printf
`'' Decimal conversions (d, u, or i) or the integral portion
of a floating point conversion (f or F) should be
grouped and separated by thousands using the non-mone-
tary separator returned by localeconv(3).
as in printf("%'6d",1000000);
Cleaner C code
// write integer value in ASCII into buf of size bufSize, inserting commas at tousands
// character string in buf is terminated by 0.
// return length of character string or bufSize+1 if buf is too small.
size_t int2str( char *buf, size_t bufSize, int val )
{
char *p;
size_t len, neg;
// handle easy case of value 0 first
if( val == 0 )
{
a[0] = '0';
a[1] = '\0';
return 1;
}
// extract sign of value and set val to absolute value
if( val < 0 )
{
val = -val;
neg = 1;
}
else
neg = 0;
// initialize encoding
p = buf + bufSize;
*--p = '\0';
len = 1;
// while the buffer is not yet full
while( len < bufSize )
{
// put front next digit
*--p = '0' + val % 10;
val /= 10;
++len;
// if the value has become 0 we are done
if( val == 0 )
break;
// increment length and if it's a multiple of 3 put front a comma
if( (len % 3) == 0 )
*--p = ',';
}
// if buffer is too small return bufSize +1
if( len == bufSize && (val > 0 || neg == 1) )
return bufSize + 1;
// add negative sign if required
if( neg == 1 )
{
*--p = '-';
++len;
}
// move string to front of buffer if required
if( p != buf )
while( *buf++ = *p++ );
// return encoded string length not including \0
return len-1;
}
I did this for an iPhone game recently. I was using the built-in LCD font, which is a monospaced font. I formatted the numbers, ignoring the commas, then stuck the commas in afterward. (The way calculators do it, where the comma is not considered a character.)
Check out the screenshots at RetroJuJu. Sorry--they aren't full-sized screenshots so you'll have to squint!
Hope that helps you (it's in C) :
char* intToFormat(int a)
{
int nb = 0;
int i = 1;
char* res;
res = (char*)malloc(12*sizeof(char));
// Should be enough to get you in the billions. Get it higher if you need
// to use bigger numbers.
while(a > 0)
{
if( nb > 3 && nb%3 == 0)
res[nb++] = ',';
// Get the code for the '0' char and add it the position of the
// number to add (ex: '0' + 5 = '5')
res[nb] = '0' + a%10;
nb++;
a /= 10;
}
reverse(&res);
return res;
}
There might be a few errors I didn't see (I'm blind when it comes to this...)
It's like an enhanced iToA so maybe it's not the best solution.
Use recursion, Luke:
#include <stdio.h>
#include <stdlib.h>
static int sprint64u( char* buffer, unsigned __int64 x) {
unsigned __int64 quot = x / 1000;
int chars_written;
if ( quot != 0) {
chars_written = sprint64u( buffer, quot);
chars_written += sprintf( buffer + chars_written, ".%03u", ( unsigned int)( x % 1000));
}
else {
chars_written = sprintf( buffer, "%u", ( unsigned int)( x % 1000));
}
return chars_written;
}
int main( void) {
char buffer[ 32];
sprint64u( buffer, 0x100000000ULL);
puts( buffer);
return EXIT_SUCCESS;
}