I have read this question ( Convert a string into an int ) before posting but I still have doubts.
I am very new to C and I'am trying to write a function which takes a string as parameter , something like this :
"99Hello" or "123ab4c"
And then returns the first the numbers it found into an int. So the result would be :
99 or 123
I'am trying to do this by finding the characters in the string(which correspond to numbers in the ASCII table) and then store them in the int I have created and returned by the function.
I'am having trouble though because when I find those character ( which correspond to numbers) I don't know how to save them to an int data type variable or how to convert them. I know there is a function called atoi which does this automatically but I would like to know how to do this by myself.
This is what I wrote so far :
int ReturnInt(char *str)
{
int i;
int new;
i = 0;
new = 0;
while (str[i] != '\0')
{
if( str[i] >= '0' && str[i] <= '9')
{
str[i] = new - '0';
}
i++;
}
return (new);
}
I think you should be able to form the desired integer from digits by multiplying them by 10 and adding them while iterating. Moreover, I think your loop's terminating condition should also include isDigit check
For example, for input 99Hello
i = 0, new = 0 -> new = 0 * 10 + 9 (9)
i = 1, new = 9 -> new = 9 * 10 + 9 (99)
i = 2, condition not satisfied as it's not a digit
int ReturnInt(char str[]) {
int i;
int new;
i = 0;
new = 0;
while (str[i] != '\0' && (str[i] >= '0' && str[i] <= '9'))
{
int currentDigit = str[i] - '0';
new = new * 10 + currentDigit;
i++;
}
return (new);
}
You just have to take a place holder , and that has to be multiplied by 10 for every iteration so that we get 10, 100, 1000 place holders and we can add that quantity to remaining numbers.
#include <stdio.h>
#include <ctype.h>
int str2num(char*strnum);
int main()
{
char *num1 = "122AB";
char *num2 = "2345AB9C";
char *num3 = "A23AB9C";
printf("num1 = %d\n", str2num(num1));
putchar('\n');
printf("num2 = %d\n", str2num(num2));
putchar('\n');
printf("num3 = %d\n", str2num(num3));
return 0;
}
int str2num(char*strnum)
{
int num = 0;
int ph = 10;// for getting tens,100s etc
while(strnum && *strnum && isdigit(*strnum))
{
num = num * ph + ( (*strnum) - '0');
strnum++;
}
return num;
}
Related
This is code to create a similar C library function atoi() without the use of any C runtime library routines.
I'm currently stuck on how to check for the first two digits of the char array s to see whether the input begins with "0x".
If it starts with 0x, this means that I can then convert it in to hexadecimal.
#include <stdio.h>
int checkforint(char x){
if (x>='0' && x<='9'){
return 1;
}
else{
return 0;
}
}
unsigned char converthex(char x){
//lets convert the character to lowercase, in the event its in uppercase
x = tolower(x);
if (x >= '0' && x<= '9') {
return ( x -'0');
}
if (x>='a' && x<='f'){
return (x - 'a' +10);
}
return 16;
}
int checkforhex(const char *a, const char *b){
if(a = '0' && b = 'x'){
return 1;
}else{
return 0;
}
}
//int checkforint
/* We read an ASCII character s and return the integer it represents*/
int atoi_ex(const char*s, int ishex)
{
int result = 0; //this is the result
int sign = 1; //this variable is to help us deal with negative numbers
//we initialise the sign as 1, as we will assume the input is positive, and change the sign accordingly if not
int i = 0; //iterative variable for the loop
int j = 2;
//we check if the input is a negative number
if (s[0] == '-') { //if the first digit is a negative symbol
sign = -1; //we set the sign as negative
i++; //also increment i, so that we can skip past the sign when we start the for loop
}
//now we can check whether the first characters start with 0x
if (ishex==1){
for (j=2; s[j]!='\0'; ++j)
result = result + converthex(s[j]);
return sign*result;
}
//iterate through all the characters
//we start from the first character of the input and then iterate through the whole input string
//for every iteration, we update the result accordingly
for (; s[i]!='\0'; ++i){
//this checks whether the current character is an integer or not
//if it is not an integer, we skip past it and go to the top of the loop and move to the next character
if (checkforint(s[i]) == 0){
continue;
} else {
result = result * 10 + s[i] -'0';
}
//result = s[i];
}
return sign * result;
}
int main(int argc)
{
int isithex;
char s[] = "-1";
char a = s[1];
char b = s[2];
isithex=checkforhex(a,b);
int val = atoi_ex(s,isithex);
printf("%d\n", val);
return 0;
}
There are several errors in your code. First, in C you start counting from zero. So in main(), you should write:
char a = s[0];
char b = s[1];
isithex = checkforhex(a, b);
Then, in checkforhex(), you should use == (two equal signs) to do comparisons, not =. So:
if (a == '0' && b == 'x')
However, as pointed out by kaylum, why not write the function to pass a pointer to the string instead of two characters? Like so:
int checkforhex(const char *str) {
if (str[0] == '0' && str[1] == 'x') {
...
}
}
And in main() call it like so:
isithex = checkforhex(s);
I have a problem with one of the test for my solution for challenge in codewars. I have to write a function that returns alphabet position of characters in input string. My solution is below. I pass all my test and also tests from codewars but fail on this one (I did not implement this test code it was pat of the test code implemented by code wars):
Test(number_tests, should_pass) {
srand(time(NULL));
char in[11] = {0};
char *ptr;
for (int i = 0; i < 15; i++) {
for (int j = 0; j < 10; j++) {
char c = rand() % 10;
in[j] = c + '0';
}
ptr = alphabet_position(in);
cr_assert_eq(strcmp(ptr, ""), 0);
free(ptr);
}
}
The error I receive is following: The expression (strcmp(ptr, "")) == (0) is false. Thanks for the help!
p.s Also I noticed that I am leaking memory (I don't know how to solve this so I suppose I would use array to keep track of string and don't use malloc) --> I suppose this is not an issue I would just free(ptr) in main function.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char *alphabet_position(char *text);
// test
int main()
{
if (!strcmp("1 2 3", alphabet_position("abc")))
{
printf("success...\n");
}
else
{
printf("fail...\n");
}
if (!strcmp("", alphabet_position("..")))
{
printf("success...\n");
}
else
{
printf("fail...\n");
}
if (!strcmp("20 8 5 19 21 14 19 5 20 19 5 20 19 1 20 20 23 5 12 22 5 15 3 12 15 3 11", alphabet_position("The sunset sets at twelve o' clock.")))
{
printf("success...\n");
}
else
{
printf("fail...\n");
}
}
char *alphabet_position(char *text)
{
// signature: string -> string
// purpose: extact alphabet position of letters in input string and
// return string of alphabet positions
// return "123"; // stub
// track numerical value of each letter according to it's alphabet position
char *alph = "abcdefghijklmnopqrstuvwxyz";
// allocate maximum possible space for return string
// each char maps to two digit number + trailing space after number
char *s = malloc(sizeof(char) * (3 * strlen(text) + 1));
// keep track of the begining of return string
char *head = s;
int index = 0;
int flag = 0;
while(*text != '\0')
{
if ( ((*text > 64) && (*text < 91)) || ((*text > 96) && (*text < 123)))
{
flag = 1;
index = (int)(strchr(alph, tolower(*text)) - alph) + 1;
if (index > 9)
{
int n = index / 10;
int m = index % 10;
*s = n + '0';
s++;
*s = m + '0';
s++;
*s = ' ';
s++;
}
else
{
*s = index + '0';
s++;
*s = ' ';
s++;
}
}
text++;
}
if (flag != 0) // if string contains at least one letter
{
*(s -1) = '\0'; // remove the trailing space and insert string termination
}
return head;
}
Here is what I think is happening:
In the cases where none of the characters in the input string is an alphabet character, s is never used, and therefore the memory allocated by malloc() could be anything. malloc() does not clear / zero-out memory.
The fact that your input case of ".." passes is just coincidence. The codewars test case does many such non-alphabetical tests in a row, each of which causes a malloc(), and if any one of them fails, the whole thing fails.
I tried recreating this situation, but it's (as I say) unpredictable. To test this, add a debugging line to output the value of s when flag is still 0:
if (flag != 0) { // if string contains at least one letter
*(s -1) = '\0'; // remove the trailing space and insert string termination
}
else {
printf("flag is still 0 : %s\n", s);
}
I'll wager that sometimes you get a garbage / random string that is not "".
Indian Currency format is 12345 → "12,345" (for odd length) and 123456 → "1,23,456" (for even length). I have included all possibilities such as
1. Minus sign: "-12,345".
2. Decimal Point: "-12,345.345" or "12,345.123".
3. Zero Condition 000000.123 → "0.123".
4.Minus and Zero Condition '-000000.123' -> "-0.123"
int currencyFormatter(char av_currency[], int av_strLen, char *ap_formattedNumber)
{
char flag = 'N'; //Taking a Flag to know whether thier is a decimal Point in Currency or not
int lengthOf = 0, index = 0, i = 0, j = 0;
char *decAr = NULL;
char *tmpCurrency = NULL;//Taking two Pointers one for Array with Commas(tmpCurrency) and decAr pointer for decimal Point array
char *s = NULL;
s = strstr(av_currency, ".");//Checking for decimal Point in array
if (s > 0)
{
flag = 'D'; // Changing Flag to show Decimal Point is Present in Array
s = strchr(av_currency, '.');
index = s - av_currency; //Index at which Decimal Point is present
av_strLen = strlen(av_currency) - index; // calculated formula to know length of an array needed to contain decimal point and Numbers after that
decAr = (char*)malloc(av_strLen*sizeof(char*));//allocated Memory using malloc
decAr[av_strLen] = '\0';
memmove(decAr, &av_currency[index], av_strLen); //memmove from decimal till end of array.
av_currency[index] = '\0';
if (!decAr)//Handled Null Condition for Pointer
{
return -1;//All errors for Negative Number
}
}
lengthOf = strlen(av_currency) + (strlen(av_currency) / 2); // Derived Formula(It Works for Indian Currency Format) to know the length of an array is needed to contain numbers and Commas Together.
tmpCurrency = (char*)malloc(lengthOf*sizeof(char*));
strrev(av_currency); //Reversed Array as commas comes at multiple of 3. eg=12345 reverse=54321 wdComma=543,21 index is 3 if number would had been bigger commas would had come at 3,6.
while (av_currency[i] != '\0')
{
if (j % 3 == 0 && j >= 3 && av_currency[i] != '-')//all Commas come at multiple of 3 when you reverse an amount
{
tmpCurrency[j] = ',';//If an , is found Increment only J as
is used as index number to store in tmpcurrency
j++;
continue;
}
tmpCurrency[j] = av_currency[i];//storing the Value in tmpCurrency
i++;//Incrementing
j++;//Incrementing
}
tmpCurrency[j] = '\0';//Null Condition
if (!tmpCurrency) // Checking for NULL Pointer
{
return -2; //all errors for Negative value
}
flag == 'D' ? strcpy(av_currency, (strcat(strrev(tmpCurrency), decAr))) : strcpy(av_currency, (strrev(tmpCurrency)));//Ternary Operator
strcpy(ap_formattedNumber,av_currency);//Copying formated number into original array
free(tmpCurrency);//Releasing the memory
free(decAr);//Releasing the Memory
return 0;
}
I have solution for above question.
Please try this code.
#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <stdlib.h>
char *printComma(double input_number,char *demo,char ap_it[],char ap_type[])
{
char *result = NULL;
char *lp_decimal_number = NULL;
char *main_number = NULL;
char *decimal_pos = NULL;
char zero[1] = {0};
int i = 0;
int j = 0;
int z = 0;
int cnt = 0;
int decimal_index = 0;
int lp_decimal_numberLen = 0 ;
int flag_dec = 0;
int flag_minus = 0 ;
int length_main;
int k = 0;
int length_demo=0;
sprintf(demo,"%lf",input_number);
if(strcmp(ap_type,"P") == 0)
{
if(strcmp(ap_it,"A") == 0 || strcmp(ap_it,"B") == 0)
{
sprintf_s(demo,40,"%0.4lf",input_number);
}
else
{
sprintf_s(demo,40,"%0.2lf",input_number);
}
}
else
{
sprintf_s(demo,40,"%.0lf",input_number);
}
length_demo = strlen(demo); // finds the length of original string
result = (char *)malloc((length_demo+10)*sizeof(char));
main_number = (char *)malloc((length_demo+50)*sizeof(char));
z = strspn(demo[0] == '-' ? (demo + 1) : demo , "0");
if(z != 0)
{
if(demo[0] != '-')
{
memcpy(main_number,&demo[z],length_demo);
main_number[length_demo]='\0';
}
else
{
puts(main_number);
main_number[length_demo]='\0';
flag_minus=1;
}
}
else
{
memcpy(main_number,&demo[0],length_demo);
main_number[length_demo]='\0';
}
length_main=strlen(main_number);
decimal_pos = strstr(main_number,".");
if(decimal_pos > 0)
{
decimal_index = decimal_pos - main_number ; // Getting postion of decimal
lp_decimal_numberLen = length_main - decimal_index; // Calculating the endpoint for decimal number
if(length_main > 3) //Copying the decimal part to a separate array
{
lp_decimal_number = (char *) malloc(lp_decimal_numberLen+1);
memcpy( lp_decimal_number, &main_number[decimal_index], lp_decimal_numberLen );
lp_decimal_number[lp_decimal_numberLen] = '\0';
flag_dec=1;
main_number[decimal_index]='\0';
}
}
//logic for comma starts here
strrev(main_number);
i = 0;
while(main_number[i] != '\0')
{
if (j%3 == 0 && j>=3 && main_number[i]!='-' && main_number[i]!='$')
{
result[j] = ',';
cnt++;
j++;
continue;
}
else if(cnt==1 || cnt==2)
{
result[j] = main_number[i];
}
else
{
result[j] = main_number[i];
}
i++;
j++;
}
result[j] = '\0';
if(flag_dec==0)
{
if(flag_minus==0)
return(strrev(result));
else
{
strcat(result,"-");
return(strrev(result));
}
}
else
{
if(flag_minus==0)
return(strcat(strrev(result),lp_decimal_number));
else
{
strcat(result,"-");
return(strcat(strrev(result),lp_decimal_number));
}
}
}
int main()
{
double number;
char num[25] = {0};
char it_type[] = "A";
char ap_type[] = "P";
char *formattedNumber = NULL;
printf("\n Enter the number n: ");
scanf("%lf",&number);
formattedNumber=printComma(number,num,it_type,ap_type);
printf("\n Final Result = %s ",formattedNumber);
getch();
return 0;
}
Do only positive values with your function!
Check if the value is negative before calling the function; call the function with the positive value; if it was negative to start with, add the minus sign afterwards.
int needssign = 0;
if (val < 0) needssign = 1;
indianformat(res, abs(val));
if (needssign) sprintf(res, "-%s", res);
Or make your current function a helper function and use the code above for the new improved function for formatting in Indian format.
I'm creating a program that adds and subtracts 2 numbers. Then I have to output this answer into different bases.
My answer is in decimal format, of type long double, such as:
long double answer;
answer = numberOne + numberTwo;
I want to convert this answer into binary. Now I have code used earlier in my program that does this, but with a char pointer:
char * decimalBinary (char * decimalNumber)
{
bool zeroFront = true;
int i;
int z;
int j = 0;
int n = atoi(decimalNumber);
char * binaryNum = malloc(32+1);
binaryNum[32] = '\0';
int current_index=1;
int end_index = strlen(decimalNumber)-1;
//Error check for valid decimal input, needed error check for beginning of code
while(current_index <= end_index)
{
if(decimalNumber[current_index] != '0' &&decimalNumber[current_index] != '1' &&decimalNumber[current_index] != '2' &&decimalNumber[current_index] != '3' &&decimalNumber[current_index] != '4' &&decimalNumber[current_index] != '5' &&dec[current_index] != '6' &&dec[current_index] != '7' &&decimalNumber[current_index] != '8' &&decimalNumber[current_index] != '9')
{
binaryNum[0] = -8;
return binaryNum;
}
current_index++;
}
for (i = 31; i >= 0; i--) {
z = n >> i;
if (z & 1)
{
binaryNum[j] = '1';
j++;
zeroFront = false;
}
else if (!zeroFront)
{
binaryNum[j] = '0';
j++;
}
}
binaryNum[j] = '\0';
return binaryNum;
}
My preferred solution is to use the code I already have in my program to convert my answer into a binary format, but as you can see the parameters are conflicting, and I'm not sure how to go about doing that.
Another possible solution that detracts from having reusable code in my program, is to create a different function all together that converts a decimal to a binary, but accepting a parameter of type long double, which is a bit unclear to me as well.
Edit:
Instead of long double, my answer is of type int.
If you really want to reuse your function without modifications, you can transform answer into a decimal string and pass the string to your function.
char stringAnswer[20];
sprintf(stringAnswer, "%d", answer);
printf("the binary answer is %s\n", decimalBinary(stringAnswer));
But a better solution should be to split the function decimalBinary into two functions : the first one to check that all digits are ok, and the second one to convert a int into a binary string.
Then you'll be able to call directly this second function with answer as parameter.
Rather than use a magic number 32, better to let the compiler deduce the needed size as an int is not always 32 bits. Checking allocation results is a good habit.
#include <assert.h>
#include <stdlib.h>
#define INT_MAX_BIN_WIDTH (sizeof(int) * CHAR_BIT)
char * binaryNum = malloc(INT_MAX_BIN_WIDTH+1);
assert(binaryNum != NULL);
binaryNum[INT_MAX_BIN_WIDTH] = '\0'; // null character
Rather than checking against each digit, since '0' to '9' must be sequential:
// if(decimalNumber[current_index] != '0' &&decimalNumber[current_index] != '1' ...
if (decimalNumber[current_index] < '0' || decimalNumber[current_index] >= '9') ...
// or
if (!isdigit((unsigned char) decimalNumber[current_index])) ...
Problem does not address negative numbers. Better to state that they will not occur or better, make code handle them.
Code allocates memory, but does not free it. Consider letting the higher level code allocate/free and supply the needed buffer to decimalBinary(char *dest, size_t size, const char *src). Robust code would supply the size too.
char *binaryNum = malloc(INT_MAX_BIN_WIDTH+1);
assert(binaryNum != NULL);
decimalBinary(binaryNum, INT_MAX_BIN_WIDTH+1, "123");
do_something(binaryNum);
free(binaryNum);
Following is a solution that is not limited to 32 bits. It does not cope with negative numbers nor memory allocation - certainly it should provide some ideas for your eventual solution.
#include <stdio.h>
#include <string.h>
static void times10(char *binaryNumber, int carry) {
size_t length = strlen(binaryNumber);
size_t i = length;
while (i > 0) {
i--;
int sum = (binaryNumber[i] - '0') * 10 + carry;
binaryNumber[i] = sum % 2 + '0';
carry = sum / 2;
}
while (carry) {
memmove(&binaryNumber[1], &binaryNumber[0], ++length);
binaryNumber[0] = carry % 2 + '0';
carry /= 2;
}
}
char *decimalBinary(char *binaryNumber, const char *decimalNumber) {
strcpy(binaryNumber, "0");
int ch;
while ((ch = *decimalNumber++) >= '0' && (ch <= '9')) {
times10(binaryNumber, ch - '0');
}
return binaryNumber;
}
int main(void) {
char buf10[200];
puts(decimalBinary(buf10, "123"));
puts(decimalBinary(buf10, "123456"));
puts(decimalBinary(buf10, "123456789012345678901234567890"));
return 0;
}
So lets say I have a character array called str of size 12, and I input 1000110 in the array, with str[0] = 1, str[1] = 0 etc... I tested the array by printing it in a for loop, it works.
I then want to count how many integers are initiated in the array. In this case, the value should be: 7
Here is my code, for whatever reason, the output is 1, not 7.
int length;
length = 12;
int actual_length()
{
int act_length = 0;
int i;
for (i = 0; i < length; i++)
{
if ( str[i] == 0 || str[i] == 1 )
{
act_length++;
}
}
printf("TEST ACTUAL LENGTH: %d\n", act_length);
return act_length;
}
I also tried
(int)str[i]
for the comparisons, but that did not change the outcome.
You have a string that consists of characters.
You need to compare against the character value:
if ( str[i] == '0' || str[i] == '1' )