I made this program that passes a binary number into a function and prints the decimal value of that binary number. The problem is that if the binary number gets big(like 11 numbers), the function prints something completely different. I tried to solved for a couple of hours now but nothing worked.
So my question is: how can I change my program so that it prints the right decimal number even when the binary number gets big?
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdint.h>
int int_log2_64(uint64_t x) { return 63 ^ __builtin_clzll(x); }
#define K(T) (((sizeof(#T) - 1) << 32) - T)
int digit_count(uint32_t x)
{
static uint64_t table[] = {
K(0), K(0), K(0),
K(10), K(10), K(10), // 64
K(100), K(100), K(100), // 512
K(1000), K(1000), K(1000), // 4096
K(10000), K(10000), K(10000), // 32k
K(100000), K(100000), K(100000), // 256k
K(1000000), K(1000000), K(1000000), // 2048k
K(10000000), K(10000000), K(10000000), // 16M
K(100000000), K(100000000), K(100000000), // 128M
K(1000000000), K(1000000000), K(1000000000), // 1024M
K(1000000000), K(1000000000) // 4B
};
int lg2 = int_log2_64(x);
uint64_t n = (uint64_t)(x) + table[lg2];
return n >> 32;
}
void binaryToDecimal(long long int bin)
{
int l = digit_count(bin);
char str[l];
itoa(bin, str, 10);
float sum[l];
int x = l - 1;
float answer;
for (int i = 0; i < l; i++)
{
if (str[i] == '1')
{
sum[i] = pow(2, x);
}
x--;
}
for (int i = 0; i < l; i++)
{
answer = answer + sum[i];
}
printf("%.0f", answer);
}
int main()
{
long long int bin = 10101101101;
binaryToDecimal(bin);
}
P.S. I changed to code to this and it works
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
void binaryToDecimal(char *bin)
{
int l = strlen(bin);
int x = l - 1;
float sum[l];
float answer = 0;
for (int i = 0; i < l; i++)
{
if (bin[i] == '1')
answer += pow(2, x);
else
sum[i] = 0;
x--;
}
printf("%.0f", answer);
}
int main()
{
binaryToDecimal("010101101101");
}
How can I change my program so that it prints the right decimal number even when the binary number gets big?
By making your bin variable a string.
That is, you want
void binaryToDecimal(const char *str)
{
...
}
and then you can call things like
binaryToDecimal("101111000110000101001110");
When doing base conversions, I believe it is always a mistake to use an integer variable for the input. If I say
int x = 12;
it is not true that "x is a decimal integer". x is an integer, period — I just happened to use a decimal constant to get a value into it. Or if I say
int y = 0x7b;
then it's not meaningful to say that y is hexadecimal — again, y is just an integer.
The base in which an integer is represented only matters:
on input, when we read a number from the user using scanf with the %d, %o, or %x formats
when converting a string with the standard library atoi or strtol functions
on output, when we print a number using printf with the %d, %o, or %x formats
But in all of those cases, the representation where the base matters is a string of digit characters, not an integer.
There are two reasons not to write a "binary to ..." function that accepts an integer. One is that, as you've seen, it artificially and unnecessarily limits the range of the numbers you can convert. But the even bigger reason is that confuses the heck out of your readers, because it's just wrong. If I see a function call
f(1001)
I think to myself, "Okay, the constant value one thousand and one is being passed to function f." There is no circumstance under which I would imagine that it was actually trying to pass the binary number nine.
Your code is extremely overcomplicated. Never use floats to do integer calculations.
unsigned binaryToDecimal(long long int bin)
{
unsigned answer = 0;
int shift = 0;
while(bin)
{
answer += (bin % 10) << shift++;
bin /= 10;
}
return answer;
}
int main()
{
long long int bin = 10101101101;
printf("%u\n", binaryToDecimal(bin));
}
https://godbolt.org/z/5E8Gv6oqf
Or use strings to pass the binary number :
unsigned binaryToDecimal(char *str)
{
unsigned answer = 0;
while(*str)
{
answer <<= 1;
answer += *str++ == '1';
}
return answer;
}
int main()
{
printf("%u\n", binaryToDecimal("10101101101"));
}
https://godbolt.org/z/4vfnETY1f
Related
I working through a book on C on my own. This isn't homework to be turned in. I am writing a C program to determine the largest Fibonacci number my machine can produce. And instructed to use a nonrecursive method.
My Code:
#include<stdio.h>
double fibo(int n);
int main(void)
{
int n = 0; // The number input by the user
double value; // Value of the series for the number input
while (n >= 0)
{
// Call fibo function
value = fibo(n);
// Output the value
printf("For %d the value of the fibonacci series = %.0f\n", n,
value);
n++;
}
return 0;
}
double fibo(int n)
{
int i; // For loop control variable
double one = 0; // First term
double two = 1; // Second term
double sum = 0; // placeholder
if (n == 0)
return 0;
else if (n == 1)
return 1;
else
{
for (i = 2; i <= n; i++)
{
sum = one + two;
one = two;
two = sum;
}
}
return sum;
Code works fine but I want to to break when the output gives me the fist instance of :
For 17127 the value of the fibonacci series = inf
Is there way to us an if statement like:
if (value == inf)
break;
The simplest is to use INFINITY or isinf().
Just did a little search and found this nice trick:
...
double value, temp; // Value of the series for the number input
while (n >= 0)
{
// Call fibo function
temp = fibo(n);
if (temp - temp != 0)
break;
else
value=temp;
...
well it turns out that whats happening is when temp hits Inf the if condition temp - temp produces Nan which equals nothing and the rest is just executing break; to exit the process.
I want to to break when the output gives me the first instance of : inf
Simply test against INFINITY from <math.h>. The output will not be an exact Fibonacci number.
#include <math.h>
#include <stdio.h>
int main(void) {
double a;
double b = 0;
double c = 1;
do {
a = b;
b = c;
c = a + b;
} while (c < INFINITY);
printf("%e\n", b);
return 0;
}
Output
1.306989e+308
long double
Use the widest floating point type and look for an inexact addition.
#include <fenv.h>
#include <stdio.h>
int main(void) {
long double a;
long double b = 0;
long double c = 1;
do {
a = b;
b = c;
c = a + b;
} while (fetestexcept(FE_INEXACT) == 0);
printf("%.0Lf\n", b);
return 0;
}
Output
12200160415121876738
Integers
Use the widest type available. This is akin to #Syed.Waris unsigned long long approach. Although common that unsigned long long and uintmax_t have the same range, using uintmax_t insures the widest.
uintmax_t: The following type designates an unsigned integer type capable of representing any value of any unsigned integer type:
#include <stdint.h>
#include <stdio.h>
uintmax_t a;
uintmax_t b = 0;
uintmax_t c = 1;
do {
a = b;
b = c;
c = a + b;
} while(c >= b);
printf("%ju\n", b);
Output
12200160415121876738
String
An alternative to double or some int type, is to create a simple string add function str_add(), then quite easy to form large Fibonacci numbers.
int main(void) {
char fib[3][4000];
strcpy(fib[0], "0");
strcpy(fib[1], "1");
int i;
for (i = 2; i <= 17127 && strlen(fib[1]) < sizeof fib[1] - 1; i++) {
printf("Fib(%3d) %s.\n", i, str_add(fib[2], fib[1], fib[0]));
strcpy(fib[0], fib[1]);
strcpy(fib[1], fib[2]);
}
printf("%zu\n", strlen(fib[2]));
return 0;
}
Output
Fib(1476) 13069...(299 digits)....71632. // Exact max `double`
Fib(17127) 95902...(3569 digits)...90818.
largest Fibonacci number my machine can produce
This question is not concerned with any data type but it is concerned with machine.
The basic rule of fibonacci is this:
n = (n-1) + (n-2)
You can take a big sized unsigned long long variable and you can keep on adding. But what if that datatype is overflowed? You are not concerned with data type. Your machine may produce a number even bigger than the long long. What would that number be ? Entire bits on RAM? Hard Disk ?
Since you are required to use an iterative method and not recursive method, your teacher/book/instructor might be testing you on loops (and not any standard API). Below is sample code using unsigned long long:
#include <stdio.h>
int main ()
{
unsigned long long a = 0;
unsigned long long b = 1;
unsigned long long c = a + b;
while(c >= b)
{
a = c;
c = b + c;
b = a;
}
printf("\n%llu\n", b);
return 0;
}
Output:
12200160415121876738
I've been trying to print out the Binary representation of a long long integer using C Programming
My code is
#include<stdio.h>
#include <stdlib.h>
#include<limits.h>
int main()
{
long long number, binaryRepresentation = 0, baseOfOne = 1, remainder;
scanf("%lld", &number);
while(number > 0) {
remainder = number % 2;
binaryRepresentation = binaryRepresentation + remainder * baseOfOne;
baseOfOne *= 10;
number = number / 2;
}
printf("%lld\n", binaryRepresentation);
}
The above code works fine when I provide an input of 5 and fails when the number is 9223372036854775807 (0x7FFFFFFFFFFFFFFF).
1.Test Case
5
101
2.Test Case
9223372036854775807
-1024819115206086201
Using a denary number to represent binary digits never ends particularly well: you'll be vulnerable to overflow for a surprisingly small input, and all subsequent arithmetic operations will be meaningless.
Another approach is to print the numbers out as you go, but using a recursive technique so you print the numbers in the reverse order to which they are processed:
#include <stdio.h>
unsigned long long output(unsigned long long n)
{
unsigned long long m = n ? output(n / 2) : 0;
printf("%d", (int)(n % 2));
return m;
}
int main()
{
unsigned long long number = 9223372036854775807;
output(number);
printf("\n");
}
Output:
0111111111111111111111111111111111111111111111111111111111111111
I've also changed the type to unsigned long long which has a better defined bit pattern, and % does strange things for negative numbers anyway.
Really though, all I'm doing here is abusing the stack as a way of storing what is really an array of zeros and ones.
As Bathsheba's answer states, you need more space than is
available if you use a decimal number to represent a bit sequence like that.
Since you intend to print the result, it's best to do that one bit at a time. We can do this by creating a mask with only the highest bit set. The magic to create this for any type is to complement a zero of that type to get an "all ones" number; we then subtract half of that (i.e. 1111.... - 0111....) to get only a single bit. We can then shift it rightwards along the number to determine the state of each bit in turn.
Here's a re-worked version using that logic, with the following other changes:
I use a separate function, returning (like printf) the number of characters printed.
I accept an unsigned value, as we were ignoring negative values anyway.
I process arguments from the command line - I tend to find that more convenient that having to type stuff on stdin.
#include <stdio.h>
#include <stdlib.h>
int print_binary(unsigned long long n)
{
int printed = 0;
/* ~ZERO - ~ZERO/2 is the value 1000... of ZERO's type */
for (unsigned long long mask = ~0ull - ~0ull/2; mask; mask /= 2) {
if (putc(n & mask ? '1' : '0', stdout) < 0)
return EOF;
else
++printed;
}
return printed;
}
int main(int argc, char **argv)
{
for (int i = 1; i < argc; ++i) {
print_binary(strtoull(argv[i], 0, 10));
puts("");
}
}
Exercises for the reader:
Avoid printing leading zeros (hint: either keep a boolean flag that indicates you've seen the first 1, or have a separate loop to shift the mask before printing). Don't forget to check that print_binary(0) still produces output!
Check for errors when using strtoull to convert the input values from decimal strings.
Adapt the function to write to a character array instead of stdout.
Just to spell out some of the comments, the simplest thing to do is use a char array to hold the binary digits. Also, when dealing with bits, the bit-wise operators are a little more clear. Otherwise, I've kept your basic code structure.
int main()
{
char bits[64];
int i = 0;
unsigned long long number; // note the "unsigned" type here which makes more sense
scanf("%lld", &number);
while (number > 0) {
bits[i++] = number & 1; // get the current bit
number >>= 1; // shift number right by 1 bit (divide by 2)
}
if ( i == 0 ) // The original number was 0!
printf("0");
for ( ; i > 0; i-- )
printf("%d", bits[i]); // or... putchar('0' + bits[i])
printf("\n");
}
I am not sure what you really want to achieve, but here is some code that prints the binary representation of a number (change the typedef to the integral type you want):
typedef int shift_t;
#define NBITS (sizeof(shift_t)*8)
void printnum(shift_t num, int nbits)
{
int k= (num&(1LL<<nbits))?1:0;
printf("%d",k);
if (nbits) printnum(num,nbits-1);
}
void test(void)
{
shift_t l;
l= -1;
printnum(l,NBITS-1);
printf("\n");
l= (1<<(NBITS-2));
printnum(l,NBITS-1);
printf("\n");
l= 5;
printnum(l,NBITS-1);
printf("\n");
}
If you don't mind to print the digits separately, you could use the following approach:
#include<stdio.h>
#include <stdlib.h>
#include<limits.h>
void bindigit(long long num);
int main()
{
long long number, binaryRepresentation = 0, baseOfOne = 1, remainder;
scanf("%lld", &number);
bindigit(number);
printf("\n");
}
void bindigit(long long num) {
int remainder;
if (num < 2LL) {
printf("%d",(int)num);
} else {
remainder = num % 2;
bindigit(num/2);
printf("%d",remainder);
}
}
Finally I tried a code myself with idea from your codes which worked,
#include<stdio.h>
#include<stdlib.h>
int main() {
unsigned long long number;
int binaryRepresentation[70], remainder, counter, count = 0;
scanf("%llu", &number);
while(number > 0) {
remainder = number % 2;
binaryRepresentation[count++] = remainder;
number = number / 2;
}
for(counter = count-1; counter >= 0; counter--) {
printf("%d", binaryRepresentation[counter]);
}
}
I have to write a C program for one of my classes that converts a given binary number to decimal. My program works for smaller inputs, but not for larger ones. I believe this may be due to the conversion specifier I am using for scanf() but I am not positive. My code is below
#include<stdio.h>
#include<math.h>
int main(void)
{
unsigned long inputNum = 0;
int currentBinary = 0;
int count = 0;
float decimalNumber = 0;
printf( "Input a binary number: " );
scanf( "%lu", &inputNum );
while (inputNum != 0)
{
currentBinary = inputNum % 10;
inputNum = inputNum / 10;
printf("%d\t%d\n", currentBinary, inputNum);
decimalNumber += currentBinary * pow(2, count);
++count;
}
printf("Decimal conversion: %.0f", decimalNumber);
return 0;
}
Running with a small binary number:
Input a binary number: 1011
1 101
1 10
0 1
1 0
Decimal conversion: 11
Running with a larger binary number:
Input a binary number: 1000100011111000
2 399133551
1 39913355
5 3991335
5 399133
3 39913
3 3991
1 399
9 39
9 3
3 0
Decimal conversion: 5264
"1000100011111000" is a 20 digit number. Certainly unsigned long is too small on your platform.
unsigned long is good - up to at least 10 digits.1
unsigned long long is better - up to at least 20 digits.1
To get past that:
Below is an any size conversion by reading 1 char at a time and forming an unbounded string.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
// Double the decimal form of string: "512" --> "1024"
char *sdouble(char *s, size_t *len, int carry) {
size_t i = *len;
while (i > 0) {
i--;
int sum = (s[i] - '0')*2 + carry;
s[i] = sum%10 + '0';
carry = sum/10;
}
if (carry) {
(*len)++;
s = realloc(s, *len + 1); // TBD OOM check
memmove(&s[1], s, *len);
s[0] = carry + '0';
}
return s;
}
int main(void) {
int ch;
size_t len = 1;
char *s = malloc(len + 1); // TBD OOM check
strcpy(s, "0");
while ((ch = fgetc(stdin)) >= '0' && ch <= '1') {
s = sdouble(s, &len, ch - '0');
}
puts(s);
free(s);
return 0;
}
100 digits
1111111111000000000011111111110000000000111111111100000000001111111111000000000011111111110000000000
1266413867935323811836706421760
1 When the lead digit is 0 or 1.
When you do this for a large number inputNum
currentBinary = inputNum % 10;
its top portion gets "sliced off" on conversion to int. If you would like to stay within the bounds of an unsigned long, switch currentBinary to unsigned long as well, and use an unsigned long format specifier in printf. Moreover, unsigned long may not be sufficiently large on many platforms, so you need to use unsigned long long.
Demo.
Better yet, switch to reading the input in a string, validating it to be zeros and ones (you have to do that anyway) and do the conversion in a cleaner character-by-character way. This would let you go beyond the 64-bit of 19 binary digits to have a full-scale int input.
unsigned long supports a maximum number of 4294967295, which means in the process of scanf( "%lu", &inputNum ); you've sliced the decimal number 1000100011111000 to a 32-bit unsigned long number.
I think scanf inputNum to a string would help a lot. In the while loop condition check if the string is empty now, and in the loop body get the last char of the string, detect if it's an '1' of a '0', and then calculate the binary number using this info.
I was tasked with writing a binary to decimal converted with taking larger binary inputs, but using embedded C programming in which we are not allowed to use library functions such as strlen. I found a simpler way to write this conversion tool using C, with both strlen, and also sizeof, as shown in the code below. Hope this helps. As you can see, strlen is commented out but either approach works fine. Sizeof just accounts for the 0 elecment in the array and that is why sizeof (number) -1 is used. Cheers!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
const char number[] = "100111111111111111111111";
int binToDec(char *);
int main()
{
printf("Output: %d", binToDec(&number));
}
int binToDec(char *n)
{
char *num = n;
int decimal_value = 0;
int base = 1;
int i;
int len = sizeof(number)-1;
//int len = strlen(number);
for (i=len-1; i>=0; i--)
{
if (num[i] == '1')
decimal_value += base;
base = base * 2;
}
return decimal_value;
}
I have this code to get from binary to decimal:
#include <stdio.h>
#include <math.h>
#include <stdint.h>
int main() {
printf("%lld\n", binaryToDecimal(11110000111100001111000011110000));
return 1;
}
long long binaryToDecimal(long long binary) {
int power = 0;
return binaryToDecimalHelper(binary, power);
}
long long binaryToDecimalHelper(long long binary, int power) {
if (binary != 0) {
long long i = binary % (double)10;
return (i * pow(2, power))
+ binaryToDecimalHelper(binary / 10, power + 1);
} else {
return 0;
}
}
It works fine for small values (up to 16 bits) , but for 32 bits (which is what I need) it just returns garbage.
The number 11110000111100001111000011110000 is of type int, which can't hold a number as big as 11110000111100001111000011110000 in your machine. It's better to use a string representation instead ("11110000111100001111000011110000") and adjust your algorithm, if you can.
if you are limited to 32 bits maximum this is one example:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void setStr(char *c, const char * x)
{
int i = 0;
while(x[i] != '\0')
{
c[i] = x[i];
i++;
}
}
void prepend(char* s, const char* t)
{
size_t len = strlen(t);
size_t i;
memmove(s + len, s, strlen(s) + 1);
for (i = 0; i < len; ++i)
{
s[i] = t[i];
}
}
int main(int argc, char const *argv[])
{
char *x = malloc(33*sizeof(char));
setStr(x, "111");
while (strlen(x) < 31) // not really necessary, but will help to 'cut' bytes if necessary
{
prepend(x,"0");
}
printf("%s\n", x);
int i = strtol(x,NULL,2);
printf("%d\n",i);
free(x);
return 0;
}
The first thing to be clear on is that your code does not convert anything to decimal, but rather to an int. Conversion to a decimal character string representation of that integer is performed by the printf() call.
The literal constant 11110000111100001111000011110000 is interpreted by the compiler (or would be if it were not so large) as a decimal value, and as such will require 104 bits to store;
i.e. log10(1111000011110000111100001111000010) / log10(210)
Representing a binary value with a decimal integer containing only 1 and 0 digits does not make much mathematical sense - though it may be convenient for small integers. A 64 bit unsigned long long is good for only 20 decimal digits (using just 1 and 0 - it can represent all 19 digit positive decimal integers, and some 20 digit values);
i.e. log10(210) * 6410
If you need longer binary values, then you should probably use a string representation. This is in fact simpler and more efficient in any case - you can use the fact that the machine representation of integers already is binary:
#include <stdio.h>
#include <stdint.h>
uint64_t binstrToInt( const char* binstr )
{
uint64_t result = 0 ;
int bit = 0;
while( binstr[bit] != '\0' )
{
if( binstr[bit] == '1' )
{
result |= 1 ;
}
bit++ ;
if( binstr[bit] != '\0' )
{
result <<= 1 ;
}
}
return result ;
}
int main()
{
printf("%llu\n", binstrToInt( "11110000111100001111000011110000" ) ) ;
return 0 ;
}
This would be easiest by far with a string as input instead of an int, and would allow longer numbers. Your problem is probably being caused by integer overflow.
String version:
#include <math.h>
#include <stdio.h>
#include <string.h>
int main() {
const char * const string_to_convert = "1010"
int result = 0;
for( int i = strlen(string_to_convert) - 1; i >= 0; --i ) {
if( string_to_convert[i] == '1' ) {
// Careful with pow() -- returns double, may round incorrectly
result += (int)pow( 10.0d, (double)i )
}
}
fprintf( stdout, "%d", result );
return 0;
}
Also, I'm not sure what the point of the return 1 is. Usually a non-zero return value from main indicates an error.
I have a simple code to convert binary to decimal numbers. In my compiler, the decomposition works just fine for number less than 1000, beyond the output is always the same 1023. Anybody has an idea ?
#include <stdio.h>
#include <stdlib.h>
// how many power of ten is there in a number
// (I don't use the pow() function to avoid trouble with floating numbers)
int residu(int N)
{
int i=0;
while(N>=1){
N=N/10;
i++;
}
return i;
}
//exponentiating a number a by a number b
int power(int a, int b){
int i;
int res=1;
for (i=0;i<b;i++){res=a*res;}
return res;
}
//converting a number N
int main()
{
int i;
//the number to convert
int N;
scanf("%d",&N);
//the final decimal result
int res=0;
//we decompose N by descending powers of 10, and M is the rest
int M=0;
for(i=0;i<residu(N);i++){
// simple loop to look if there is a power of (residu(N)-1-i) in N,
// if yes we increment the binary decomposition by
// power(2,residu(N)-1-i)
if(M+ power(10,residu(N)-1-i) <= N)
{
M = M+power(10,residu(N)-1-i);
res=power(2,residu(N)-1-i)+res;
}
}
printf("%d\n",res);
}
Yes try this :
#include <stdio.h>
int main(void)
{
char bin; int dec = 0;
while (bin != '\n') {
scanf("%c",&bin);
if (bin == '1') dec = dec * 2 + 1;
else if (bin == '0') dec *= 2; }
printf("%d\n", dec);
return 0;
}
Most likely this is because you are using an int to store your binary number. An int will not store numbers above 2^31, which is 10 digits long, and 1023 is the largest number you can get with 10 binary digits.
It would be much easier for you to read your input number as a string, and then process each character of the string.
After a little experimentation, I think that your program is intended to accept a number consisting of 1's and 0's only as a base-10 number (the %d reads a decimal number). For example, given input 10, it outputs 2; given 1010, it outputs 10; given 10111001, it outputs 185.
So far, so good. Unfortunately, given 1234, it outputs 15, which is a little unexpected.
If you are running on a machine where int is a 32-bit signed value, then you can't enter a number with more than 10 digits, because you overflow the limit of a 32-bit int (which can handle ±2 billion, in round terms). The scanf() function doesn't handle overflows well.
You could help yourself by echoing your inputs; this is a standard debugging technique. Make sure the computer got the value you are expecting.
I'm not going to attempt to fix the code because I think you're going about the problem in completely the wrong way. (I'm not even sure whether it's best described as binary to decimal, or decimal to binary, or decimal to binary to decimal!) You would do better to read the input as a string of (up to 31) characters, then validate that each one is either a 0 or a 1. Assuming that's correct, then you can process the string very straight-forwardly to generate a value which can be formatted by printf() as a decimal.
Shift left is the same than multiply by 2 and is more efficient, so I think it is a more c-like answer:
#include <stdio.h>
#include <stdlib.h>
int bin2int(const char *bin)
{
int i, j;
j = sizeof(int)*8;
while ( (j--) && ((*bin=='0') || (*bin=='1')) ) {
i <<= 1;
if ( *bin=='1' ) i++;
bin++;
}
return i;
}
int main(void)
{
char* input = NULL;
size_t size = 0;
while ( getline(&input, &size, stdin) > 0 ) {
printf("%i\n", bin2int(input));
}
free(input);
}
#include <stdio.h> //printf
#include <string.h> //strlen
#include <stdint.h> //uintX_t or use int instead - depend on platform.
/* reverse string */
char *strrev(char *str){
int end = strlen(str)-1;
int start = 0;
while( start<end ){
str[start] ^= str[end];
str[end] ^= str[start];
str[start] ^= str[end];
++start;
--end;
}
return str;
}
/* transform binary string to integer */
uint32_t binstr2int(char *bs){
uint32_t ret = 0;
uint32_t val = 1;
while(*bs){
if (*bs++ == '1') ret = ret + val;
val = val*2;
}
return ret;
}
int main(void){
char binstr[] = "1010101001010101110100010011111"; //1428875423
printf("Binary: %s, Int: %d\n", binstr, binstr2int(strrev(binstr)));
return 0;
}