I need to read an integer one by one until i read a '$', and then to determine the largest, smallest and so on. I could use a character variable and do it, but it works for numbers from 0 to 9. But how do I read integers of two or more digits and at the same time, detect a '$' - I used a char *, but I guess it is equivalent to an array, which I should not use here. Also, char holds a single number / char, hence not suitable for larger numbers. What should I do?
No arrays, no pointers, no tricky char-by-char read & convert. Just plain scanf and getchar.
#include <stdio.h>
int main()
{
int newValue=0; /* value being acquired */
int max; /* current maximum value */
int min; /* current minimum value */
int firstAcquired=0; /* boolean flag set to 1 after first acquisition */
int ch; /* used as temporary storage for the getchar() */
for(;;)
{
/* scanf returns the number of successfully acquired fields; here if it
returns 0 means that the value couldn't be acquired */
if(scanf("%d",&newValue)==0)
{
/* scanf failed, but it's guaranteed it put the offending character
back into the stream, from where we can get it */
ch=getchar();
if(ch=='$' || ch==EOF)
break;
else
/* from here to the break it's just to handle invalid input and EOF
gracefully; if you are not interested you can replace this stuff
with a random curse to the user */
{
puts("Invalid input, retry.");
/* Empty the buffer */
while((ch=getchar())!='\n' && ch!=EOF)
;
}
/* if it's EOF we exit */
if(ch==EOF)
break;
}
else
{
/* Everything went better than expected */
if(!firstAcquired || newValue>max)
max=newValue;
if(!firstAcquired || newValue<min)
min=newValue;
firstAcquired=1;
}
}
if(firstAcquired)
{
printf("The maximum value was %d\n", max);
printf("The minimum value was %d\n", min);
}
return 0;
}
In the interest of spoiling all the fun, showing off, outright overkill and darn tooting fun:
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/karma.hpp>
#include <boost/spirit/include/support_istream_iterator.hpp>
namespace qi = boost::spirit::qi;
template <typename V>
void show_statistics(const V& data)
{
using namespace boost::spirit::karma;
std::cout << "data:\t"<< format('{' << auto_ % ", " << '}', data) << std::endl;
std::cout << "min:\t" << *std::min_element(data.begin(), data.end()) << std::endl;
std::cout << "max:\t" << *std::max_element(data.begin(), data.end()) << std::endl;
auto sum = std::accumulate(data.begin(), data.end(), 0);
std::cout << "sum:\t" << sum << std::endl;
std::cout << "avg:\t" << (1.0*sum) / data.size() << std::endl;
}
void dostats(const std::vector<int>& data) { show_statistics(data); }
int main()
{
std::cin.unsetf(std::ios::skipws);
auto f = boost::spirit::istream_iterator(std::cin);
decltype(f) l;
bool ok = qi::phrase_parse(f, l, +(+qi::int_ > "$") [ dostats ], qi::space);
if (f!=l)
std::cout << "Remaining input unparsed: " << std::string(f,l) << std::endl;
return ok? 0:255;
}
Demo:
Sample run:
sehe#natty:/tmp$ ./test2 <<< "1 2 3 4 5 $ 3 -9 0 0 0 $ 900 9000 $ unparsed trailing text"
data: {1, 2, 3, 4, 5}
min: 1
max: 5
sum: 15
avg: 3
data: {3, -9, 0, 0, 0}
min: -9
max: 3
sum: -6
avg: -1.2
data: {900, 9000}
min: 900
max: 9000
sum: 9900
avg: 4950
Remaining input unparsed: unparsed trailing text
You can use 'scanf("%s")' to read a group of characters. You can then check if the first character is a '%' and terminate if so. Otherwise, call atoi to convert to an integer. Store the largest and smallest in integer types, not character types.
Basically, the only time you have to deal with characters is when you read them in and check if it's a '$'. Otherwise, use integers all the way through.
If I'm getting what you want correctly it should be something like this:
int i = 0;
char c = getchar();
while (c != '$')
{
i = i * 10 + (c - '0');
c = getchar();
}
Hope it helped.
You can read char by char in a loop, check values and so on...
int i = 0;
char c = 0;
int size = 10;
int currentIndex = 0;
int* integers = malloc(sizeof(int) * size);
int counter = 0;
do
{
scanf("%c", &c);
if (c == ' ') // Match space, used a number separator
{
if (counter != 0 && i != 0)
{
if (currentIndex >= size)
{
size += 5;
integers = realloc(integers, size);
}
integers[currentIndex] = i;
currentIndex++;
}
counter = 0;
i = 0;
}
else if (c >= '0' && c <= '9')
{
i = (i * counter * 10) + (c - '0');
counter++;
}
}
while(c != '$');
Don't forget to free integers in the end!
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define BUFF_SIZE 16
#define DATA_MAX_SIZE 64
int main(){
char buff[BUFF_SIZE];
int data[DATA_MAX_SIZE];
int i,value,counter = 0;
char *buffp,*p;
while(NULL!=fgets(buff,BUFF_SIZE,stdin)){
buff[BUFF_SIZE - 1]='\0';
buffp = buff;
next: while(isspace(*buffp))
++buffp;
if(*buffp == '\0')
continue;
value = strtol(buffp, &p, 0);
if(counter == DATA_MAX_SIZE){
printf("over data max size!\n");
break;
} else if(p != buffp){
data[counter++]=value;
if(*p == '\0' || *p == '\r'|| *p == '\n')
continue;
buffp = p;
goto next;
} else {
if(*p == '$')
break;
printf("format error\n");
break;
}
}
//check code
for(i=0;i<counter;++i){
printf("data[%d]=%d\n",i, data[i]);
}
return 0;
}
OUTPUT:
1 2 3
123
456
99 $
data[0]=1
data[1]=2
data[2]=3
data[3]=123
data[4]=456
data[5]=99
12345
4
$
data[0]=12345
data[1]=4
Related
I have a issue with my number to string implemention. For some reason I keep getting 000 on my terminal even though. I couldn't find a solution, what is the potantial issue here?
Im now sure my code is broken but don't really see the problem.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
/* public domain code feel free to use as what you wish, no warranties given*/
char finddigits(unsigned n) {
char base = 6;
unsigned tester = 100000;
while(base % tester != 0) {
base--;
/* inefficient but works for now */
switch(tester) {
case 100000:
tester = 10000;
break;
case 10000:
tester = 1000;
break;
case 1000:
tester = 100;
break;
case 100:
tester = 10;
break;
case 10:
tester = 1;
break;
}
}
return base;
}
char* num2str(unsigned n) {
char size = finddigits(n);
char* tempbuf = malloc(size);
*tempbuf = 48 + (n / pow(10, size));
for(unsigned int i = 1; i < size; i++)
*(tempbuf + i) = 48 + (n % (10 * i));
return tempbuf;
}
int main(int argc, char* argv[]) {
int numbr = 210;
printf("%s \n", num2str(numbr));
/* expected 210 on string got 000 */
return 0;
}
You just want num2str to return the digit string for n.
A few issues:
finddigits is supposed to calculate the number of digits in n. But, [if it works at all], it uses an algorithm I've never seen.
finddigits isn't needed in num2str as num2str can be [much] simpler by filling the buffer in the reverse direction.
num2str is too complicated.
Calling num2str from printf leaks memory from the num2str call to malloc
Here's a refactored version:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
char
finddigits(unsigned n)
{
int count = 0;
if (n == 0)
count = 1;
for (; n != 0; n /= 10, ++count);
return count;
}
char *
num2str(unsigned n)
{
static char buf[100];
char *dst = &buf[sizeof(buf) - 1];
// add string terminator
*dst-- = 0;
// we must always output a 0
if (n == 0)
*dst-- = '0';
// work backwards in the array
for (; n != 0; n /= 10, --dst)
*dst = (n % 10) + '0';
// point to first digit in string
dst += 1;
return dst;
}
void
dotest(unsigned n)
{
printf("n=%u '%s'\n",n,num2str(n));
}
int
main(int argc, char *argv[])
{
dotest(210);
dotest(0);
dotest(12345);
return 0;
}
Here's the program output:
n=210 '210'
n=0 '0'
n=12345 '12345'
The computer does what you told it to do, which is to say, it does complete nonsense.
finddigits(210) returns 1, because 6 % 100000 isn't 0 (it's 6), 5%10000 isn't 0 (it's 5), 4 % 1000 isn't 0 (it's 4), 3 % 100 isn't 0 (it's 3), 2 % 10 isn't 0 (it's 2), but 1 % 1 is 0 so the loop stops and the function returns 1.
Then, num2str allocates 1 byte. In this 1 byte, it sets the first byte to 48 + (210 / 10) which is 69, ASCII code for the letter E. Since size is 1 the loop doesn't run at all and num2str returns this allocation. When you print it, it prints the letter E - possibly with more gibberish after it since the string is not terminated, although for me it just printed E.
I have no idea how you managed to get 000.
You need to write code that tells the computer to do what you want it to do. When you can't get it to do what you want it to, for one part of the code, don't just skip that part of the code and go onto the next one. It all has to be right or it won't work.
Mathematics is also "public domain". Here are two versions of one of your functions, shown with a main() that tests both versions with several values.
Don't convert a signed integer value to unsigned for no particular reason. If you need/want the absolute value of a (possibly) negative number, C provides a function to achieve that.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int finddigits1( int n ) {
int count = n == 0; // ensure 0 is counted as 1 digit
while( n ) count++, n /= 10;
return count;
}
int finddigits2( int n ) {
// log of zero is undefined
return ( n == 0 ) ? 1 : (int)log10(abs(n))+1;
}
char *num2str( int n, char buf[] ) {
printf( "finddigits1( %d ) counted %d\n", n, finddigits1( n ) );
printf( "finddigits2( %d ) returns %d\n", n, finddigits2( n ) );
strcpy( buf, "Hello world!" ); // Left as an exercise to fix your own code
return buf; // returning allows direct usage by caller
}
int main() {
int tests[] = { 123, -123, 0, 100, 54321 };
char buf[ 30 ]; // better to pass large buffer to function than use malloc()
for( int i = 0; i < sizeof tests/sizeof tests[0]; i++ ) {
int n = tests[i];
printf( "n = %d '%s'\n", n, num2str( n, buf ) );
}
return 0;
}
finddigits1( 123 ) counted 3
finddigits2( 123 ) returns 3
n = 123 'Hello world!'
finddigits1( -123 ) counted 3 // NB: Account for negative sign!
finddigits2( -123 ) returns 3
n = -123 'Hello world!'
finddigits1( 0 ) counted 1
finddigits2( 0 ) returns 1
n = 0 'Hello world!'
finddigits1( 100 ) counted 3
finddigits2( 100 ) returns 3
n = 100 'Hello world!'
finddigits1( 54321 ) counted 5
finddigits2( 54321 ) returns 5
n = 54321 'Hello world!'
If you're trying to convert a number to a string, why not just use sprintf()?
See here:
How to convert an int to string in C?
I wrote this code to find the prime factorization of a number. I just cannot figure out the last part. If x is entered as a double or float, the program should print an error message and terminate. How do I achieve this?
#include <stdio.h>
int main()
{
int x, i;
printf("Enter an integer: ");
scanf("%d", &x);
if (x <= 1)
{
return 1;
}
printf("The prime factorization of %d is ", x);
if (x > 1)
{
while (x % 2 == 0)
{
printf("2 ");
x = x / 2;
}
for (i = 3; i < 1009; i = i + 2)
{
while (x % i == 0)
{
printf("%d ", i);
x = x / i;
}
}
}
return 0;
}
Your starting point should cover all desired and undesired cases so you should take float number from a user, not int. Then, you should check whether whole decimal part of the number is 0. That is, if all of them equals 0, the user want to provide an int number instead of float.
First step is to declare a float number:
float y;
After, take its value from the user:
scanf("%f", &y);
Second step is to check whether it is int or float. There are many ways for this step. For example, I find roundf() function useful. It takes a float number and computes the nearest integer to this number. So if the nearest integer is the number itself then the number has to be int. Right?
if(roundf(y)!=y)
If you are sure it is an int, you can move onto the third step: convert float type to int type. It is called type-casting. This step is required for the remaining part of your program because in your algorithm you manipulate the number with int type so just convert it to int:
x = (int)y;
After adding the line above, you can use the rest of code which you typed. I give the whole program:
#include <stdio.h>
#include <math.h>
int main()
{
int x,i;
float y;
printf("Enter an integer: ");
scanf("%f", &y);
if(roundf(y)!=y){
printf("%f is not an integer!",y);
return 1;
}
else{
x = (int)y;
}
if (x <= 1)
{
printf("%d <= 1",x);
return 1;
}
else
{
printf("The prime factorization of %d is ", x);
while (x%2 == 0)
{
printf("2 ");
x = x / 2;
}
for ( i = 3; i < 1009; i = i + 2)
{
while (x%i == 0)
{
printf("%d ",i);
x = x / i;
}
}
}
return 0;
}
The use of scanf() is a bit tricky, I would avoid it to scan user generated input at almost all cost. But nevertheless here is a short overview for how to get the errors of scanf()
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <string.h>
int main(void)
{
int x, i, scanf_return;
printf("Enter an integer: ");
/* Reset "errno". Not necessary here, just in case. */
errno = 0;
/* scanf() returns a value in case of an error */
scanf_return = scanf("%d", &x);
/*
* scanf() returns "EOF" if it didn't find all what you wanted or
* and error happened.
* It sets "errno" to the value of the actual error. See manpage
* for all of the details.
*/
if (scanf_return == EOF) {
/*
* The error is connected to the stream, so we can differ between
* an error within scanf() and and error with the input stream
* (here: stdin)
*/
if (ferror(stdin)) {
fprintf(stderr, "Something went wrong while reading stdin: %s\n", strerror(errno));
exit(EXIT_FAILURE);
} else {
/* e.g. a conversion error, a float instead of an integer, letters
instead of a decimal number */
fprintf(stderr, "Something went wrong within scanf()\n");
exit(EXIT_FAILURE);
}
}
/*
* If no error occurred, the return holds the number of objects
* scanf() was able to read. We only need one, but it would throw an
* error if cannot find any objects, so the check is here for
* pedagogical reasons only.
*/
if (scanf_return != 1) {
fprintf(stderr, "Something went wrong within scanf(): wrong number of objects read.\n");
exit(EXIT_FAILURE);
}
if (x <= 1) {
fprintf(stderr, "Input must be larger than 1!\n");
exit(EXIT_FAILURE);
}
printf("The prime factorization of %d is ", x);
/* No need for that test, x is already larger than one at this point. */
/* if (x > 1) { */
while (x%2 == 0) {
printf("2 ");
x = x / 2;
}
for (i = 3; i < 1009; i = i + 2) {
while (x%i == 0) {
printf("%d ",i);
x = x / i;
}
}
/* } */
/* Make it pretty. */
putchar('\n');
exit(EXIT_SUCCESS);
}
Does it work?
$ ./stackoverflow_003
Enter an integer: 1234
The prime factorization of 1234 is 2 617
$ factor 1234
1234: 2 617
$ ./stackoverflow_003
Enter an integer: asd
Something went wrong within scanf(): wrong number of objects read.
$ ./stackoverflow_003
Enter an integer: 123.123
The prime factorization of 123 is 3 41
No, it does not work. Why not? If you ask scanf() to scan an integer it grabs all consecutive decimal digits (0-9) until there is no one left. The little qualifier "consecutive" is most likely the source of your problem: a floating point number with a fractional part has a decimal point and that is the point where scanf() assumes that the integer you wanted ended. Check:
$ ./stackoverflow_003
Enter an integer: .123
Something went wrong within scanf(): wrong number of objects read
How do you find out? #weather-vane gave one of many ways to do so: check if the next character after the integer is a period (or another decimal separator of your choice):
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <string.h>
int main(void)
{
int x, i, scanf_return;
char c = -1;
printf("Enter an integer: ");
/* Reset "errno". Not necessary here, just in case. */
errno = 0;
/* scanf() returns a value in case of an error */
scanf_return = scanf("%d%c", &x, &c);
/*
* scanf() returns "EOF" if it didn't find all what you wanted or
* and error happened.
* It sets "errno" to the value of the actual error. See manpage
* for all of the details.
*/
if (scanf_return == EOF) {
/*
* The error is connected to the stream, so we can differ between
* an error within scanf() and and error with the input stream
* (here: stdin)
*/
if (ferror(stdin)) {
fprintf(stderr, "Something went wrong while reading stdin: %s\n", strerror(errno));
exit(EXIT_FAILURE);
} else {
/* e.g. a conversion error, a float instead of an integer, letters
instead of a decimal number */
fprintf(stderr, "Something went wrong within scanf()\n");
exit(EXIT_FAILURE);
}
}
/*
* If no error occurred, the return holds the number of objects
* scanf() was able to read. We can use this information now.
* If there is a period (actually any character) after the integer
* it returns 2 (assuming no error happened, of course)
*/
/* If no integer given, the following character ("%c") gets ignored. */
if (scanf_return == 0) {
fprintf(stderr, "Something went wrong within scanf(): no objects read.\n");
exit(EXIT_FAILURE);
}
/* Found two objects, check second one which is the character. */
if (scanf_return == 2) {
if (c == '.') {
fprintf(stderr, "Floating point numbers are not allowed.\n");
exit(EXIT_FAILURE);
}
}
if (x <= 1) {
fprintf(stderr, "Input must be larger than 1!\n");
exit(EXIT_FAILURE);
}
printf("The prime factorization of %d is ", x);
/* No need for that test, x is already larger than one at this point. */
/* if (x > 1) { */
while (x%2 == 0) {
printf("2 ");
x = x / 2;
}
for (i = 3; i < 1009; i = i + 2) {
while (x%i == 0) {
printf("%d ",i);
x = x / i;
}
}
/* } */
/* Make it pretty. */
putchar('\n');
exit(EXIT_SUCCESS);
}
Check:
$ ./stackoverflow_003
Enter an integer: 123
The prime factorization of 123 is 3 41
$ ./stackoverflow_003
Enter an integer: 123.123
Floating point numbers are not allowed.
$ ./stackoverflow_003
Enter an integer: .123
Something went wrong within scanf(): no objects read.
Looks good enough for me. With one little bug:
$ ./stackoverflow_003
Enter an integer: 123.
Floating point numbers are not allowed
But I think I can leave that as an exercise for the dear reader.
You can try this simple C99 implementation of Pollard Rho algorithm :
// Integer factorization in C language.
// Decompose a composite number into a product of smaller integers.
unsigned long long pollard_rho(const unsigned long long N) {
// Require : N is a composite number, not a square.
// Ensure : you already performed trial division up to 23.
// Option : change the timeout, change the rand function.
static const int timeout = 18;
static unsigned long long rand_val = 2994439072U;
rand_val = (rand_val * 1025416097U + 286824428U) % 4294967291LLU;
unsigned long long gcd = 1, a, b, c, i = 0, j = 1, x = 1, y = 1 + rand_val % (N - 1);
for (; gcd == 1; ++i) {
if (i == j) {
if (j >> timeout)
break;
j <<= 1;
x = y; // "x" takes the previous value of "y" when "i" is a power of 2.
}
a = y, b = y; // computes y = f(y)
for (y = 0; a; a & 1 ? b >= N - y ? y -= N : 0, y += b : 0, a >>= 1, (c = b) >= N - b ? c -= N : 0, b += c);
y = (1 + y) % N; // function f performed f(y) = (y * y + 1) % N
for (a = y > x ? y - x : x - y, b = N; (a %= b) && (b %= a););
gcd = a | b; // the GCD(abs(y - x), N) was computed
// it continues until "gcd" is a non-trivial factor of N.
}
return gcd;
}
Usually you performed some trial division before calling the algorithm
The algorithm isn't designed to receive a prime number as input
Two consecutive calls may not result in the same answer
Alternately, there is a pure C quadratic sieve which factors numbers from 0 to 300-bit.
If in doubt about the primality of N you can use a C99 primality checker :
typedef unsigned long long int ulong;
ulong mul_mod(ulong a, ulong b, const ulong mod) {
ulong res = 0, c; // return (a * b) % mod, avoiding overflow errors while doing modular multiplication.
for (b %= mod; a; a & 1 ? b >= mod - res ? res -= mod : 0, res += b : 0, a >>= 1, (c = b) >= mod - b ? c -= mod : 0, b += c);
return res % mod;
}
ulong pow_mod(ulong n, ulong exp, const ulong mod) {
ulong res = 1; // return (n ^ exp) % mod
for (n %= mod; exp; exp & 1 ? res = mul_mod(res, n, mod) : 0, n = mul_mod(n, n, mod), exp >>= 1);
return res;
}
int is_prime(ulong N) {
// Perform a Miller-Rabin test, it should be a deterministic version.
const ulong n_primes = 9, primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23};
for (ulong i = 0; i < n_primes; ++i)
if (N % primes[i] == 0) return N == primes[i];
if (N < primes[n_primes - 1]) return 0;
int primality = 1, a = 0;
ulong b;
for (b = N - 1; ~b & 1; b >>= 1, ++a);
for (ulong i = 0; i < n_primes && primality; ++i) {
ulong c = pow_mod(primes[i], b, N);
if (c != 1) {
for (int j = a; j-- && (primality = c + 1 != N);)
c = mul_mod(c, c, N);
primality = !primality;
}
}
return primality;
}
To try it there is a factor function :
// return the number that was multiplied by itself to reach N.
ulong square_root(const ulong num) {
ulong res = 0, rem = num, a, b;
for (a = 1LLU << 62 ; a; a >>= 2) {
b = res + a;
res >>= 1;
if (rem >= b)
rem -= b, res += a;
}
return res;
}
ulong factor(ulong num){
const ulong root = square_root(num);
if (root * root == num) return root ;
const ulong n_primes = 9, primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23};
for (ulong i = 0; i < n_primes && primes[i] <= root; ++i)
if (num % primes[i] == 0) return primes[i];
if (is_prime(num))
return 1 ;
return pollard_rho(num);
}
Which is completed by the main function :
#include <assert.h>
int main(void){
for(ulong i = 2; i < 63; ++i){
ulong f = factor(i);
assert(f <= 1 || f >= i ? is_prime(i) : i % f == 0);
ulong j = (1LLU << i) - 1 ;
f = factor(j);
assert(f <= 1 || f >= j ? is_prime(j) : j % f == 0);
j = 1 | pow_mod((ulong) &main, i, -5);
f = factor(j);
assert(f <= 1 || f >= j ? is_prime(j) : j % f == 0);
}
}
There are some problems in your code:
you do not check the return value of scanf, so you cannot detect invalid or missing input and will have undefined behavior in those cases.
you only test divisors up to 1009, so composite numbers with larger prime factors do not produce any output.
prime numbers larger than 1009 do not produce any output.
you should probably output a newline after the factors.
Testing and reporting invalid input such as floating point numbers can be done more easily by reading the input as a full line and parsing it with strtol().
Here is a modified version:
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
int main() {
char input[120];
char ch;
char *p;
long x, i;
int last_errno;
printf("Enter an integer: ");
if (!fgets(input, sizeof input, stdin)) {
fprintf(stderr, "missing input\n");
return 1;
}
errno = 0;
x = strtol(input, &p, 0);
last_errno = errno;
if (p == input || sscanf(p, " %c", &ch) == 1) {
fprintf(stderr, "invalid input: %s", input);
return 1;
}
if (last_errno == ERANGE) {
fprintf(stderr, "number too large: %s", input);
return 1;
}
if (x < 0) {
fprintf(stderr, "number is negative: %ld\n", x);
return 1;
}
if (x <= 1) {
return 1;
}
printf("The prime factorization of %ld is", x);
while (x % 2 == 0) {
printf(" 2");
x = x / 2;
}
for (i = 3; x / i >= i;) {
if (x % i == 0) {
printf(" %ld", i);
x = x / i;
} else {
i = i + 2;
}
}
if (x > 1) {
printf(" %ld", x);
}
printf("\n");
return 0;
}
I am writing a program that converts user input to either
ASCII values or binary values. If the string contains letters,
each letter should be converted to ASCII. Strings of numbers will
be converted to binary value of entire string. If both letters and
numbers are entered, each letter will be converted to ASCII and the numbers can/will only be separated by letters, for example "32" will print the binary value "00100000", but "3a2" should be converted to "00000011", "97", "00000010".
The way the program is currently written, strings of numbers convert to binary perfectly. However, strings of letters add a decimal "0" to the end. The output converts each letter to its ASCII value, then converts the "0" to binary. I am unsure as to where this "0" is coming from. Additionally, strings beginning
and ending with digits (for example "6j3") will print the ASCII value of j, then the binary value of "6", skipping the "3" entirely and printing the "j" before the "6". I would like to print each ASCII/binary value in the exact order of the user input.
I am posting my entire code for any necessary clarification, but I believe the issue is in the determineChars() function. I am also looking to use the char* letters and char* numbers functions to efficiently handle the appropriate data and store the final num[] and let[] arrays, but I am unsure of how to do this.
I am quite a beginner to C, so excuse the messiness. Corrections, as well as any further optimizations would be greatly appreciated.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <unistd.h>
#include <string.h>
#define EIGHT_BITS 255
#define SIXTEEN_BITS 65535
#define THIRTY_TWO_BITS 4294967295UL
#define SIXTY_FOUR_BITS 18446744073709551615ULL
// defined in case of booleans
typedef enum { false, true } bool;
// GET # OF ELEMENTS IN STRING
size_t getSize(char* input) {
size_t size;
size = strlen(input);
printf("Number of Characters..... %d", size);
//printf("\n----------------------------------");
return size;
}
// DETERMINE NUMBER OF BITS TO OUTPUT
int getBitLength(unsigned long long d) {
int l;
if (d <= EIGHT_BITS) {
l = 8;
}
else if (d > EIGHT_BITS && d <= SIXTEEN_BITS) {
l = 16;
}
else if (d > SIXTEEN_BITS && d <= THIRTY_TWO_BITS) {
l = 32;
}
else if (d > THIRTY_TWO_BITS && d <= SIXTY_FOUR_BITS) {
l = 64;
}
printf("\nBits..................... %d", l);
return l;
}
// CONVERT INPUT TO BINARY VALUE
void convertToBinary(char* input) {
static int b[64];
int i, j, length, r;
unsigned long long decimal;
char* pEnd;
// converts input to ull
decimal = strtoull(input, &pEnd, 0);
printf("\n\n---------------- %I64u ------------------", decimal);
printf("\nULL...................... %I64u", decimal);
length = getBitLength(decimal);
// creates array
for (i = 0; i < length; i++) {
r = decimal % 2;
decimal /= 2;
b[i] = r;
}
// reverses array for binary value
printf("\nBinary Value............. ");
for (j = length - 1; j >= 0; j--) {
printf("%d", b[j]);
}
}
char* numbers(char* input) {
char* num = (char*) malloc(sizeof(char) * 25);
return num;
}
char* letters(char* input) {
char* let = (char*) malloc(sizeof(char) * 25);
return let;
}
void determineChars(char* input) {
int i;
char* num = numbers(input);
char* let = letters(input);
size_t inputSize = getSize(input);
// FOR EACH CHARACTER IN INPUT
for (i = 0; i < inputSize; i++) {
if (isdigit(input[i])) {
// stores number values from input into separate array
num[i] = input[i];
printf("\nNumbers: %c", num[i]);
}
if (!isdigit(input[i])) {
// stores letter values from input into separate array
let[i] = input[i];
printf("\nLetters: %c", let[i]);
// prints separator line + ASCII value
printf("\n\n---------------- %c ------------------", let[i]);
printf("\nASCII Value of %c......... %d", let[i], let[i]);
// removes char from input array
input[i] = ' ';
}
}
// char array must consist of digits only
convertToBinary(num);
}
int main() {
// GET INPUT
char input[50];
scanf("%s", input);
determineChars(input);
return 0;
}
I would like to print each ASCII/binary value in the exact order of the user input.
In that case, you would have to restructure your code a bit. This is because if the input contains only digits you will have to print binary and alternate being chars and digits if the string contains both. I have tried to do this with the following code, cleaned it up a bit, removed the warnings and memory leaks.
See if this is what you want:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <unistd.h>
#include <string.h>
#define EIGHT_BITS 255
#define SIXTEEN_BITS 65535
#define THIRTY_TWO_BITS 4294967295UL
#define SIXTY_FOUR_BITS 18446744073709551615ULL
// GET # OF ELEMENTS IN STRING
size_t getSize(char* input) {
size_t size;
size = strlen(input);
printf("Number of Characters..... %d\n", size);
//printf("\n----------------------------------");
return size;
}
// DETERMINE NUMBER OF BITS TO OUTPUT
int getBitLength(unsigned long long d) {
int l;
if (d <= EIGHT_BITS) {
l = 8;
}
else if (d > EIGHT_BITS && d <= SIXTEEN_BITS) {
l = 16;
}
else if (d > SIXTEEN_BITS && d <= THIRTY_TWO_BITS) {
l = 32;
}
else if (d > THIRTY_TWO_BITS && d <= SIXTY_FOUR_BITS) {
l = 64;
}
printf("\nBits..................... %d", l);
return l;
}
// CONVERT INPUT TO BINARY VALUE
void convertToBinary(char* input) {
static int b[64];
int i, j, length, r;
unsigned long long decimal;
char* pEnd;
// converts input to ull
decimal = strtoull(input, &pEnd, 0);
printf("\n---------------- %I64u ------------------", decimal);
printf("\nULL...................... %I64u", decimal);
length = getBitLength(decimal);
// creates array
for (i = 0; i < length; i++) {
r = decimal % 2;
decimal /= 2;
b[i] = r;
}
// reverses array for binary value
printf("\nBinary Value............. ");
for (j = length - 1; j >= 0; j--) {
printf("%d", b[j]);
}
printf("\n");
}
void determineChars(char* input) {
int i;
long ret;
char* ptr;
char c;
size_t inputSize = getSize(input);
ret = strtol(input, &ptr, 10);
if((ret == 0) || ((strlen(ptr) != 0) && (strlen(input) != strlen(ptr))))
{
for (i = 0; i < inputSize; i++) {
if (isdigit(input[i])) {
c = input[i];
printf("\nNumber: %c", c);
convertToBinary(&c);
}
if (!isdigit(input[i])) {
// stores letter values from input into separate array
printf("\nLetter: %c", input[i]);
// prints separator line + ASCII value
printf("\n---------------- %c ------------------\n", input[i]);
printf("ASCII Value of %c......... %d\n", input[i], input[i]);
// removes char from input array
}
}
}
else
convertToBinary(input);
}
int main() {
// GET INPUT
char input[50];
scanf("%s", input);
determineChars(input);
}
I also tried out the test cases you mentioned in the question along with few others and it seems to work fine.
32
Number of Characters..... 2
---------------- 32 ------------------
ULL...................... 32
Bits..................... 8
Binary Value............. 00100000
3a2
Number of Characters..... 3
Number: 3
---------------- 3 ------------------
ULL...................... 3
Bits..................... 8
Binary Value............. 00000011
Letter: a
---------------- a ------------------
ASCII Value of a......... 97
Number: 2
---------------- 2 ------------------
ULL...................... 2
Bits..................... 8
Binary Value............. 00000010
6j3
Number of Characters..... 3
Number: 6
---------------- 6 ------------------
ULL...................... 6
Bits..................... 8
Binary Value............. 00000110
Letter: j
---------------- j ------------------
ASCII Value of j......... 106
Number: 3
---------------- 3 ------------------
ULL...................... 3
Bits..................... 8
Binary Value............. 00000011
I pass in a hex number into hex2bin and it prints out the binary number correctly but I don't want it to print out the number I want to return the number so I can use it to find the cardinality of the number. How would I store the number instead of printing it out?
int hex2bin (int n){
int i,k,mask;
for(i = sizeof(int) * 8 - 1; i >= 0; i--){
mask = 1 << i;
k = n & mask;
k == 0 ? printf("0"):printf("1");
}
return 0;
}
Perhaps something like this?
int result = 0;
int i, k...
...
result = result | (((k == 0) ? 0 : 1) << i;
...
return result;
Instead of being clever with an int, you could of course also simply use an array of variables instead.
Store the number in a string whose space is provided by a compound literal (Available since C99).
It works like OP's flow: Loop up to sizeof(int) * 8 times, finding the value of 1 bit and print/save it.
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
// Maximum buffer size needed
#define UTOA_BASE_2 (sizeof(unsigned)*CHAR_BIT + 1)
char *utoa_base2(char *s, unsigned x) {
s += UTOA_BASE_2 - 1;
*s = '\0';
do {
*(--s) = "01"[x % 2];
x /= 2;
} while (x);
return s;
}
#define TO_BASE2(x) utoa_base2((char [UTOA_BASE_2]){0} , (x))
void test(unsigned x) {
printf("base10:%10u base2:%5s ", x, TO_BASE2(x));
char *s = TO_BASE2(x);
// do stuff with `s`, it is valid for until the end of this block
printf("%s\n", s);
}
int main(void) {
test(0);
test(25);
test(UINT_MAX);
}
Sample output
base10: 0 base2: 0 0
base10: 25 base2:11001 11001
base10:4294967295 base2:11111111111111111111111111111111 11111111111111111111111111111111
This is a variation of this base-n answer.
You can use the strcat function to do that.
Note that the new hex2bin function in this answer assumes that the parameter char *buf has already been allocated and can hold at least 1+sizeof(int)*8 bytes including the null terminator:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// assume: buf is at least length 33
int hex2bin (int n, char *buf)
{
int i,k,mask;
for(i = sizeof(int) * 8 - 1; i >= 0; i--){
mask = 1 << i;
k = n & mask;
k == 0 ? strcat(buf, "0") : strcat(buf, "1");
}
return 0;
}
int main()
{
int n = 66555;
char buffer[1+sizeof(int)*8] = { 0 } ;
hex2bin(n, buffer);
printf("%s\n", buffer);
return 0;
}
I hope you will find this helpful :)
bool convertDecimalBNR(INT32 nDecimalValue, UINT32 * punFieldValue, INT32 nBitCount, DecimalBNRType * pDecimalSpecification)
{
bool bBNRConverted = false;
INT32 nBitIndex = nBitCount - 1;
INT32 nBitValue = anTwoExponents[nBitIndex];
*punFieldValue = 0;
if ((nDecimalValue >= pDecimalSpecification->nMinValue) && (nDecimalValue <= pDecimalSpecification->nMaxValue))
{
// if the value is negative, then add (-1 * (2 ^ (nBitCount - 1))) on itself and go on just like a positive value calculation.
if (nDecimalValue < 0)
{
nDecimalValue += nBitValue;
nBitIndex--;
nBitValue /= 2;
*punFieldValue |= BIT_0_ONLY_ONE;
}
while (nBitIndex >= 0)
{
*punFieldValue = (*punFieldValue << 1);
if (nDecimalValue >= nBitValue)
{
nDecimalValue -= nBitValue;
*punFieldValue |= BIT_0_ONLY_ONE;
}
nBitIndex--;
nBitValue /= 2;
}
if (nDecimalValue <= nBitValue)
{
bBNRConverted = true;
}
}
return (bBNRConverted);
}
I can't understand the following atoi implementation code, specifically this line:
k = (k << 3) + (k << 1) + (*p) - '0';
Here is the code:
int my_atoi(char *p) {
int k = 0;
while (*p) {
k = (k << 3) + (k << 1) + (*p) - '0';
p++;
}
return k;
}
Can someone explain it to me ?
Another question: what should be the algorithm of atof implementation ?
<< is bit shift, (k<<3)+(k<<1) is k*10, written by someone who thought he was more clever than a compiler (well, he was wrong...)
(*p) - '0' is subtracting the value of character 0 from the character pointed by p, effectively converting the character to a number.
I hope you can figure out the rest... just remember how the decimal system works.
Here is a specification for the standard function atoi. Sorry for not quoting the standard, but this will work just as fine (from: http://www.cplusplus.com/reference/clibrary/cstdlib/atoi/ )
The function first discards as many whitespace characters (as in
isspace) as necessary until the first non-whitespace character is
found. Then, starting from this character, takes an optional initial
plus or minus sign followed by as many base-10 digits as possible, and
interprets them as a numerical value.
The string can contain additional characters after those that form the
integral number, which are ignored and have no effect on the behavior
of this function.
If the first sequence of non-whitespace characters in str is not a
valid integral number, or if no such sequence exists because either
str is empty or it contains only whitespace characters, no conversion
is performed and zero is returned.
k = (k << 3) + (k << 1);
means
k = k * 2³ + k * 2¹ = k * 8 + k * 2 = k * 10
Does that help?
The *p - '0' term adds the value of the next digit; this works because C requires that the digit characters have consecutive values, so that '1' == '0' + 1, '2' == '0' + 2, etc.
As for your second question (atof), that should be its own question, and it's the subject for a thesis, not something simple to answer...
#include <stdio.h>
#include <errno.h>
#include <limits.h>
double atof(const char *string);
int debug=1;
int main(int argc, char **argv)
{
char *str1="3.14159",*str2="3",*str3="0.707106",*str4="-5.2";
double f1,f2,f3,f4;
if (debug) printf("convert %s, %s, %s, %s\n",str1,str2,str3,str4);
f1=atof(str1);
f2=atof(str2);
f3=atof(str3);
f4=atof(str4);
if (debug) printf("converted values=%f, %f, %f, %f\n",f1,f2,f3,f4);
if (argc > 1)
{
printf("string %s is floating point %f\n",argv[1],atof(argv[1]));
}
}
double atof(const char *string)
{
double result=0.0;
double multiplier=1;
double divisor=1.0;
int integer_portion=0;
if (!string) return result;
integer_portion=atoi(string);
result = (double)integer_portion;
if (debug) printf("so far %s looks like %f\n",string,result);
/* capture whether string is negative, don't use "result" as it could be 0 */
if (*string == '-')
{
result *= -1; /* won't care if it was 0 in integer portion */
multiplier = -1;
}
while (*string && (*string != '.'))
{
string++;
}
if (debug) printf("fractional part=%s\n",string);
// if we haven't hit end of string, go past the decimal point
if (*string)
{
string++;
if (debug) printf("first char after decimal=%c\n",*string);
}
while (*string)
{
if (*string < '0' || *string > '9') return result;
divisor *= 10.0;
result += (double)(*string - '0')/divisor;
if (debug) printf("result so far=%f\n",result);
string++;
}
return result*multiplier;
}
Interestingly, the man page for atoi doesn't indicate setting of errno so if you're talking any number > (2^31)-1, you're out of luck and similarly for numbers less than -2^31 (assuming 32-bit int). You'll get back an answer but it won't be what you want. Here's one that could take a range of -((2^31)-1) to (2^31)-1, and return INT_MIN (-(2^31)) if in error. errno could then be checked to see if it overflowed.
#include <stdio.h>
#include <errno.h> /* for errno */
#include <limits.h> /* for INT_MIN */
#include <string.h> /* for strerror */
extern int errno;
int debug=0;
int atoi(const char *c)
{
int previous_result=0, result=0;
int multiplier=1;
if (debug) printf("converting %s to integer\n",c?c:"");
if (c && *c == '-')
{
multiplier = -1;
c++;
}
else
{
multiplier = 1;
}
if (debug) printf("multiplier = %d\n",multiplier);
while (*c)
{
if (*c < '0' || *c > '9')
{
return result * multiplier;
}
result *= 10;
if (result < previous_result)
{
if (debug) printf("number overflowed - return INT_MIN, errno=%d\n",errno);
errno = EOVERFLOW;
return(INT_MIN);
}
else
{
previous_result *= 10;
}
if (debug) printf("%c\n",*c);
result += *c - '0';
if (result < previous_result)
{
if (debug) printf("number overflowed - return MIN_INT\n");
errno = EOVERFLOW;
return(INT_MIN);
}
else
{
previous_result += *c - '0';
}
c++;
}
return(result * multiplier);
}
int main(int argc,char **argv)
{
int result;
printf("INT_MIN=%d will be output when number too high or too low, and errno set\n",INT_MIN);
printf("string=%s, int=%d\n","563",atoi("563"));
printf("string=%s, int=%d\n","-563",atoi("-563"));
printf("string=%s, int=%d\n","-5a3",atoi("-5a3"));
if (argc > 1)
{
result=atoi(argv[1]);
printf("atoi(%s)=%d %s",argv[1],result,(result==INT_MIN)?", errno=":"",errno,strerror(errno));
if (errno) printf("%d - %s\n",errno,strerror(errno));
else printf("\n");
}
return(errno);
}
Here is my implementation(tested successfully with cases containing and starting with letters, +, - and zero's).
I tried to reverse-engineer atoi function in Visual Studio. If the input string only contained numerical characters, it could be implemented in one loop. but it gets complicated because you should take care of -,+ and letters.
int atoi(char *s)
{
int c=1, a=0, sign, start, end, base=1;
//Determine if the number is negative or positive
if (s[0] == '-')
sign = -1;
else if (s[0] <= '9' && s[0] >= '0')
sign = 1;
else if (s[0] == '+')
sign = 2;
//No further processing if it starts with a letter
else
return 0;
//Scanning the string to find the position of the last consecutive number
while (s[c] != '\n' && s[c] <= '9' && s[c] >= '0')
c++;
//Index of the last consecutive number from beginning
start = c - 1;
//Based on sign, index of the 1st number is set
if (sign==-1)
end = 1;
else if (sign==1)
end = 0;
//When it starts with +, it is actually positive but with a different index
//for the 1st number
else
{
end = 1;
sign = 1;
}
//This the main loop of algorithm which generates the absolute value of the
//number from consecutive numerical characters.
for (int i = start; i >=end ; i--)
{
a += (s[i]-'0') * base;
base *= 10;
}
//The correct sign of generated absolute value is applied
return sign*a;
}
about atoi() hint code from here:
and based on the atoi(), my implementation of atof():
[have same limitation of original code, doesn't check length, etc]
double atof(const char* s)
{
double value_h = 0;
double value_l = 0;
double sign = 1;
if (*s == '+' || *s == '-')
{
if (*s == '-') sign = -1;
++s;
}
while (*s >= 0x30 && *s <= 0x39)
{
value_h *= 10;
value_h += (double)(*s - 0x30);
++s;
}
// 0x2E == '.'
if (*s == 0x2E)
{
double divider = 1;
++s;
while (*s >= 0x30 && *s <= 0x39)
{
divider *= 10;
value_l *= 10;
value_l += (double)(*s - 0x30);
++s;
}
return (value_h + value_l/divider) * sign;
}
else
{
return value_h * sign;
}
}