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I'm going a bit mad implementing this function. I think I have narrow the algorithm down but I have some strange behavior for some value. It seems to work for most of the values and bases, but for string "1000" it simply returns 0.
Below the code:
// A C program for
// implementation of atoi
#include <stdio.h>
#include <stdint.h>
int val(char c)
{
if (c >= '0' && c <= '9')
return (int)c - '0';
else
return (int)c - 'A' + 10;
}
int32_t my_atoi(uint8_t * ptr, uint8_t digits, uint32_t base){
// Initialize result
digits = 0;
// Initialize sign as positive
int sign = 1;
// Initialize index of first digit
int i = 0;
// If number is negative,
// then update sign
if (*ptr == '-') {
sign = -1;
i++;
}
// Iterate through all digits and update the result
for (; *(ptr+i) != '\0'; ++i){
digits = (digits * base) + val(*(ptr+i));
printf("Digits of %d is:%d\n",i,digits);
}
// Return result with sign
return sign * digits;
}
// Driver program to test above functions
int main()
{
uint8_t str[] = "1000";
uint8_t val2=0;
int val = my_atoi(str, val2, 16);
printf("%d \n", val);
return 0;
}
For the above code the output is:
Digits of 0 is:1
Digits of 1 is:16
Digits of 2 is:0
Digits of 3 is:0
0
I simply cannot understand why digits becomes 0 after it has the value 16.
Any help will be greatly appreciated.
Your digits is a uint8_t so it just overflows to 0. You should make it a larger, signed, data type like int32_t or int64_t so the sign works properly and it doesn't overflow.
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Below is the program for my assignment. All of it was given except the printing and logic. I searched many videos about it and did what they did, but 15's binary is 1111 and the result was 0000
#include <stdio.h>
#include <math.h>
// Function to convert decimal to binary
// The function is void type so, print the result inside the function
void decimal2Binary(int dec) {
// array to store binary number
int binaryNumber[32];
// Your logic goes here
// fill out the binaryNumber
// i to increment the loop and terminate
int i = 0;
// We are interested in positive numbers for now
while (dec > 0) {
// Please add your logic to build the binary array
binaryNumber[i] = dec%2;
dec = dec/2;
i++;
}
// Print the binary array.
printf("Binary: ");
for (int j = i - 1; j >= 0; j--) {
printf("%d",binaryNumber[i]);
}
}
// The main function goes here
// The main function calls the decimal2Binary
int main() {
int decimal = 15;
printf("Decimal: %d equal to ", decimal);
decimal2Binary(decimal);
return 0;
}
DECIMAL TO BINARY (RECURSIVE):
#include <stdio.h>
int find(int dec_num)
{
if (dec_num == 0)
return 0;
else
return (dec_num % 2 + 10 *
find(dec_num / 2));
}
int main()
{
int decimal_number;
scanf("%d", &decimal_number);
printf("%d", find(decimal_number));
return 0;
}
DECIMAL TO BINARY (ITERATIVE):
#include <stdio.h>
long dec_to_bin(int decimalnum){
int binarynum = 0;
int mod, place = 1;
while(decimalnum != 0){
mod = decimalnum % 2;
decimalnum = decimalnum/2;
binarynum = binarynum + (mod*place);
place = place*10;
}
return binarynum;
}
int main() {
int decimalnum;
scanf("%d", &decimalnum);
printf("enter a decimal number:\n %ld", dec_to_bin(decimalnum));
return 0;
}
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I need to print my characters BEFORE the previously printed character using printf. For example:
printf("1")
printf("0")
would need to output:
01
Is there a way to do this? I cannot use arrays. To be clear I'm printing in base two (binary) representation using a divide by two algorithm:
for(int i = 0; i < 16; i++){ // 16 bit int
tmp = num % 2;
if(tmp == 1){
printf("1");
} else {
printf("0");
}
num /= 2;
}
The above code prints the binary representation backwards.
Here is my solution to print binary without using array. you can ignore initial zeros by adding another if condition.
#include<stdio.h>
int main(){
int num =50;
int i;
for(i=15;i>=0;i--){
if( (1<<i) & num){
printf("1");
}
else printf("0");
}
return 0;
}
output:
0000000000110010
Recursion can perform like a "print before". #rici
The below does not always print 16 digits, just the decimal digits needed.
void print_binary(unsigned n) {
// Break the digits into 2 groups: 1) the one least digit and 2) the rest.
unsigned last_digit = n/10;
unsigned all_the_other_more_significant_digits = n/10;
// Let us print those more significant digits first;
if (all_the_other_more_significant_digits > 0) {
print_binary(all_the_other_more_significant_digits);
}
// Now print the last digit
printf("%u", last_digit);
}
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I do the sum of the digits like that:
while(number>0)
{
sum+=number%TEN;
number=number/TEN;
}
but I need that if the number is (for example) 123444 so it'll include only one 4 in the sum. how can I do that?
Have an array of all digits initialized to zero
int digits[10] = { 0 };
Then before adding a digit you check if digits[that_digit] is zero, if yes you set it to 1 and add to sum, if no keep going ...
while(number>0)
{
int one = number%TEN;
if ( ! digits[one]) {
sum+=one;
digits[one] = 1;
}
number=number/TEN;
}
Edit, no array version
Add an int initialized to 0, the bit i indicates if that digit i has already been summed.
If 1 was added, bit 1 set to 1, if 2, bit 2 set to 1 etc...
int bits = 0;
while(number>0)
{
int one = number%TEN;
if (!(bits & (1<<one))) {
sum+=one;
bits |= 1<<one;
}
number=number/TEN;
}
First you should put some code here whatever you tried, to give you basic idea to solve your problem I am putting simple code below.
#include<stdio.h>
#include<malloc.h>
int main()
{
int input, digit, temp, sum = 0;
printf("Enter Input Number :\n");
scanf("%d",&input);
temp = input;
//first find how many digits are there
for(digit = 0 ; temp != 0 ;digit++, temp /= 10);
//create one array equal to no of digits, use dynamic array because once you find different digits you can re-allocate memory and save some memory
int *p = malloc(digit * sizeof(int));
//now store all the digits in dynamic array
p[0] = input % 10;//1
for(int i = 0; i < digit ;i++) {
input /= 10;
p[i+1] = input %10;
if(p[i] != p[i+1])
sum = sum + p[i];
}
printf("sum of different digits : = %d \n",sum);
free(p);
p = 0;
return 0;
}
Explanation of this code I mentioned in comments itself, it may not work for all test case, remaining try yourself.
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I need a function (in the C language) which can convert the binary contents of an array, in this case myArray[8] {*MSB* 1, 1, 1, 1, 1, 1, 1, 1 *LSB*} into the decimal equivalent = 255.
Any ideas on how to solve this, as efficiently as possible, as it is being done on a microcontroller?
Using bitwise shifting
#include <stdio.h>
int main(void) {
char myArray[8] = {1,1,1,1,1,1,1,1};
int i;
int result = 0;
for (i=0; i<sizeof(myArray); i++){
result += myArray[sizeof(myArray)-i-1] << i;
}
printf("%d\n", result);
return 0;
}
One solution is to add and shift in a Horner scheme.
result = (...((myArray[0])<<1 + myArray[1])<< 1+ ... ) << 1 ... ) + myArray[8];
or in multiple expressions:
result = myArray[0];
result <<= 1;
result += myArray[1];
result <<= 1;
...
result += myArray[8];
This solution is for MSB.
set up the for loop
unsigned int result = myarray[0];
for (int i = 1; i < 8, i++);
{
result <<= 1;
result += myarray[i];
}
#include <stdio.h>
int main()
{
int a[8] = {1,1,1,1,1,1,1,1};
int n=8,dec=0;
int j=0,f;
for(int i=(n-1);i>=0;i--)
{
dec=(a[i]*(int)pow(2,j))+dec;
j++;
}
printf("The converted Decimal number is: %d",dec);
return 0;
}
Output: The converted Decimal number is: 255
The result is also available at: http://ideone.com/NQK0Il
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This is my function and I need help....
I must try to catch the range of short int without error handler or try&catch.
I cant´t find my failure in this algorithm — I hope you can help me.
short int checkShortInt(char * myString)
{
short int i = 0;
short int len;
if((myString[i]=='+')||(myString[i]=='-')) i++;
for (len = i; myString[len] != '\0'; len++);
if(len-i>5) return(0);
if(myString[i+0]<'3') return(1);
if(myString[i+0]>'3') return(0);
if(myString[i+1]<'2') return(1);
if(myString[i+1]>'2') return(0);
if(myString[i+2]<'7') return(1);
if(myString[i+2]>'7') return(0);
if(myString[i+3]<'6') return(1);
if(myString[i+3]>'6') return(0);
if(myString[i+4]>'7') return(0);
return(1);
}
Note that the range covered by short is asymmetric when two's complement is used: it normally ranges from -32768 to 32767.
If the length of the value is smaller than 5, the string clearly fits into a short.
My two cents. Maybe overkill...
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <inttypes.h>
int checkShortInt(char *myString)
{
long long i = strtoll(myString, 0, 0);
return (i >= SHRT_MIN && i <= SHRT_MAX);
}
short int checkShortInt(char * myString)
{
short int i = 0;
int total = 0;
// Skip a leading +/- sign.
if((myString[i]=='+')||(myString[i]=='-')) i++;
while(myString[i] != '\0')
{
total = 10*total + myString[i]-'0';
++i;
}
if (SHRT_MIN <= total && total <= SHRT_MAX)
{
return total;
}
return 0;
}
Sample inputs and results:
checkShortInt("53") ==> 53
checkShortInt("+125") ==> 125
checkShortInt("0") ==> 0
checkShortInt("70345") ==> 0