#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int myatoi(const char* string) {
int i = 0;
while (*string) {
i = (i << 3) + (i<<1) + (*string -'0');
string++;
}
return i;
}
void decimal2binary(char *decimal, int *binary) {
decimal = malloc(sizeof(char) * 32);
long int dec = myatoi(decimal);
long int fraction;
long int remainder;
long int factor = 1;
long int fractionfactor = .1;
long int wholenum;
long int bin;
long int onechecker;
wholenum = (int) dec;
fraction = dec - wholenum;
while (wholenum != 0 ) {
remainder = wholenum % 2; // get remainder
bin = bin + remainder * factor; // store the binary as you get remainder
wholenum /= 2; // divide by 2
factor *= 10; // times by 10 so it goes to the next digit
}
long int binaryfrac = 0;
int i;
for (i = 0; i < 10; i++) {
fraction *= 2; // times by two first
onechecker = fraction; // onechecker is for checking if greater than one
binaryfrac += fractionfactor * onechecker; // store into binary as you go
if (onechecker == 1) {
fraction -= onechecker; // if greater than 1 subtract the 1
}
fractionfactor /= 10;
}
bin += binaryfrac;
*binary = bin;
free(decimal);
}
int main(int argc, char **argv) {
char *data;
data = malloc(sizeof(char) * 32);
int datai = 1;
if (argc != 4) {
printf("invalid number of arguments\n");
return 1;
}
if (strcmp(argv[1], "-d")) {
if (strcmp(argv[3], "-b")) {
decimal2binary(argv[2], &datai);
printf("output is : %d" , datai);
} else {
printf("invalid parameter");
}
} else {
printf("invalid parameter");
}
free(data);
return 0;
}
In this problem, myatoi works fine and the decimal2binary algorithm is correct, but every time I run the code it gives my output as 0. I do not know why. Is it a problem with pointers? I already set the address of variable data but the output still doesn't change.
./dec2bin "-d" "23" "-b"
The line:
long int fractionfactor = .1;
will set fractionfactor to 0 because the variable is defined as an integer. Try using a float or double instead.
Similarly,
long int dec = myatoi(decimal);
stores an integer value, so wholenum is unnecessary.
Instead of
i = (i << 3) + (i<<1) + (*string -'0');
the code will be much more readable as
i = i * 10 + (*string - '0');
and, with today's optimizing compilers, both versions will likely generate the same object code. In general, especially when your code isn't working, favor readability over optimization.
fraction *= 2; // times by two first
Comments like this, that simply translate code to English, are unnecessary unless you're using the language in an unusual way. You can assume the reader is familiar with the language; it's far more helpful to explain your reasoning instead.
Another coding tip: instead of writing
if (strcmp(argv[1], "-d")) {
if (strcmp(argv[3], "-b")) {
decimal2binary(argv[2], &datai);
printf("output is : %d" , datai);
} else {
printf("invalid parameter");
}
} else {
printf("invalid parameter");
}
you can refactor the nested if blocks to make them simpler and easier to understand. In general it's a good idea to check for error conditions early, to separate the error-checking from the core processing, and to explain errors as specifically as possible so the user will know how to correct them.
If you do this, it may also be easier to realize that both of the original conditions should be negated:
if (strcmp(argv[1], "-d") != 0) {
printf("Error: first parameter must be -d\n");
else if (strcmp(argv[3], "-b") != 0) {
printf("Error: third parameter must be -b\n");
} else {
decimal2binary(argv[2], &datai);
printf("Output is: %d\n" , datai);
}
void decimal2binary(char *decimal, int *binary) {
decimal = malloc(sizeof(char) * 32);
...
}
The above lines of code allocate a new block of memory to decimal, which will then no longer point to the input data. Then the line
long int dec = myatoi(decimal);
assigns the (random values in the) newly-allocated memory to dec.
So remove the line
decimal = malloc(sizeof(char) * 32);
and you will get the correct answer.
if(!strcmp(argv[3] , "-b"))
if(!strcmp(argv[3] , "-d"))
The result of the string compare function should be negated so that you can proceed. Else it will print invalid parameter. Because the strcmp returns '0' when the string is equal.
In the 'decimal2binary' function you are allocating a new memory block inside the function for the input parameter 'decimal',
decimal = malloc(sizeof(char) * 32);
This would actually overwrite your input parameter data.
Related
I'm passing almost all leetCode tests with this, but not understanding why the output is wrong ("/0") when the input is:
a = "10100000100100110110010000010101111011011001101110111111111101000000101111001110001111100001101"
b = "110101001011101110001111100110001010100001101011101010000011011011001011101111001100000011011110011"
Anyone has an idea to what is not working ?
Thanks
#include <stdio.h>
#include <stdlib.h>
char * sumBinary(long int binary1, long int binary2, char * result);
char * addBinary(char * a, char * b)
{
char * result;
long int a_int;
long int b_int;
a_int = atoi(a);
b_int = atoi(b);
result = malloc(sizeof(*result) * 1000);
if (!result)
return (NULL);
sumBinary(a_int, b_int, result);
return (result);
}
char * sumBinary(long int binary1, long int binary2, char * result)
{
int i;
int t;
int rem;
int sum[1000];
i = 0;
t = 0;
rem = 0;
if ((binary1 == 0) && (binary2 == 0))
{
result[0] = '0';
result[1] = '\0';
}
else
{
while (binary1 != 0 || binary2 != 0)
{
sum[i++] = (binary1 %10 + binary2 % 10 + rem) % 2;
rem = (binary1 %10 + binary2 % 10 + rem) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (rem != 0)
sum[i++] = rem;
--i;
while (i >= 0)
{
result[t] = sum[i] + '0';
t++;
i--;
}
result[t] = '\0';
}
return (result);
}
For a start, you should be using atol(3), not atoi(3) if you're using long int. But that's not the main issue here.
atol(3) and atoi(3) expect strings containing decimal numbers, not binary, so that's not going to work well for you. You would need strtol(3), which you can tell to expect a string in ASCII binary. But again, this is not the main issue.
You don't give the question text, but I'm guessing they want you to add two arbitrarily-long ASCII-binary strings, resulting in an ASCII-binary string.
I imagine their expectation, given it's arbitrarily-long, is that you would be working entirely in the string domain. So you'd allocate for a string whose length is two greater than the longer of the two you get as parameters (+1 for the terminal NUL, the other +1 for a potential overflow digit).
Then you start from the end, working back to the start, adding the corresponding digits of the parameter strings, placing the results into the result string starting from its end (allowing for that terminal NUL), adding as if you were doing it by hand.
Don't forget to add a leading zero to the result string, if you don't overflow into that position.
Note that I'm not going to write the code for you. This is either a learning exercise or a test: either way, you need to do the coding so you can learn from it.
It is possible to convert integer to string in C without sprintf?
There's a nonstandard function:
char *string = itoa(numberToConvert, 10); // assuming you want a base-10 representation
Edit: it seems you want some algorithm to do this. Here's how in base-10:
#include <stdio.h>
#define STRINGIFY(x) #x
#define INTMIN_STR STRINGIFY(INT_MIN)
int main() {
int anInteger = -13765; // or whatever
if (anInteger == INT_MIN) { // handle corner case
puts(INTMIN_STR);
return 0;
}
int flag = 0;
char str[128] = { 0 }; // large enough for an int even on 64-bit
int i = 126;
if (anInteger < 0) {
flag = 1;
anInteger = -anInteger;
}
while (anInteger != 0) {
str[i--] = (anInteger % 10) + '0';
anInteger /= 10;
}
if (flag) str[i--] = '-';
printf("The number was: %s\n", str + i + 1);
return 0;
}
Here's an example of how it might work. Given a buffer and a size, we'll keep dividing by 10 and fill the buffer with digits. We'll return -1 if there is not enough space in the buffer.
int
integer_to_string(char *buf, size_t bufsize, int n)
{
char *start;
// Handle negative numbers.
//
if (n < 0)
{
if (!bufsize)
return -1;
*buf++ = '-';
bufsize--;
}
// Remember the start of the string... This will come into play
// at the end.
//
start = buf;
do
{
// Handle the current digit.
//
int digit;
if (!bufsize)
return -1;
digit = n % 10;
if (digit < 0)
digit *= -1;
*buf++ = digit + '0';
bufsize--;
n /= 10;
} while (n);
// Terminate the string.
//
if (!bufsize)
return -1;
*buf = 0;
// We wrote the string backwards, i.e. with least significant digits first.
// Now reverse the string.
//
--buf;
while (start < buf)
{
char a = *start;
*start = *buf;
*buf = a;
++start;
--buf;
}
return 0;
}
Unfortunately none of the answers above can really work out in a clean way in a situation where you need to concoct a string of alphanumeric characters.There are really weird cases I've seen, especially in interviews and at work.
The only bad part of the code is that you need to know the bounds of the integer so you can allocate "string" properly.
In spite of C being hailed predictable, it can have weird behaviour in a large system if you get lost in the coding.
The solution below returns a string of the integer representation with a null terminating character. This does not rely on any outer functions and works on negative integers as well!!
#include <stdio.h>
#include <stdlib.h>
void IntegertoString(char * string, int number) {
if(number == 0) { string[0] = '0'; return; };
int divide = 0;
int modResult;
int length = 0;
int isNegative = 0;
int copyOfNumber;
int offset = 0;
copyOfNumber = number;
if( number < 0 ) {
isNegative = 1;
number = 0 - number;
length++;
}
while(copyOfNumber != 0)
{
length++;
copyOfNumber /= 10;
}
for(divide = 0; divide < length; divide++) {
modResult = number % 10;
number = number / 10;
string[length - (divide + 1)] = modResult + '0';
}
if(isNegative) {
string[0] = '-';
}
string[length] = '\0';
}
int main(void) {
char string[10];
int number = -131230;
IntegertoString(string, number);
printf("%s\n", string);
return 0;
}
You can use itoa where available. If it is not available on your platform, the following implementation may be of interest:
https://web.archive.org/web/20130722203238/https://www.student.cs.uwaterloo.ca/~cs350/common/os161-src-html/atoi_8c-source.html
Usage:
char *numberAsString = itoa(integerValue);
UPDATE
Based on the R..'s comments, it may be worth modifying an existing itoa implementation to accept a result buffer from the caller, rather than having itoa allocate and return a buffer.
Such an implementation should accept both a buffer and the length of the buffer, taking care not to write past the end of the caller-provided buffer.
int i = 24344; /*integer*/
char *str = itoa(i);
/*allocates required memory and
then converts integer to string and the address of first byte of memory is returned to str pointer.*/
I'm trying to do a simple function to convert a number codified in a string form to a float, using the function below (attached also), without using the function strtof.
I'm getting wrong values, could somebody solve this? Why is this giving me incorrect values?
For example if the input string is "123456789.123" the function will return a float value of 123456789.123...
Im using DevC++
if string == "12345678"
output = 12345678.000000 (CORRECT)
if string == "-12345678"
output = -12345678.000000 (CORRECT)
if string == "123456789"
output = 123456792.000000 (INCORRECT)
if string == "-123456789"
output = -123456792.000000 (INCORRECT)
if string == "1000.1"
output = 999.799988 (INCORRECT)
if string == "-1000.1"
-output = -999.799988 (INCORRECT)
My code so far is:
#include <stdio.h>
#include <stdbool.h>
#include <stdint.h>
#include <string.h>
#include <math.h>
float StringToFloat (uint8_t *var);
int main()
{
uint8_t string[64] = "-1000.1";
float value = StringToFloat (string);
printf("%f", value);
return 0;
}
float StringToFloat (uint8_t *var)
{
float multiplier;
float result = 0;
bool negative_num = false;
bool found_comma_or_dot = false;
uint16_t numbers_before_dot = 0;
uint16_t numbers_after_dot = 0;
uint16_t i = 0;
while (*(var+i) != 0)
{
if (*(var+i) == '-') { negative_num = true; }
else if (*(var+i) == '.' || *(var+i) == ',') { found_comma_or_dot = true; }
else
{
if (found_comma_or_dot == false) { numbers_before_dot++; }
if (found_comma_or_dot == true) { numbers_after_dot++; }
}
i++;
}
multiplier = pow (10, numbers_before_dot-1);
while (*var != 0)
{
if (*var == '-') { var++; }
if (numbers_before_dot > 0)
{
numbers_before_dot--;
result += ( (*var) - 0x30) * (multiplier);
multiplier /= 10;
}
else if (numbers_after_dot > 0)
{
numbers_after_dot--;
result += ( (*var) - 0x30) * (multiplier);
multiplier /= 10;
}
var++;
}
if (negative_num == true)
{
result *= (-1);
}
return result;
}
A problem is precision. A float is simply not very precise. In order to pass your test cases, change from float to double. This can be seen if you run this code:
float t = 123456789;
printf("%f\n", t);
Your code is very overly complex. Here is a much more slick solution.
double StringToFloat(uint8_t *var)
{
// First check if negative. If it is, just step one step forward
// and treat the rest as a positive number. But remember the sign.
double sign = 1;
if(*var=='-') {
sign = -1;
var++;
}
// Read until either the string terminates or we hit a dot
uint32_t integer_part = 0;
while(*var != 0) {
if(*var == '.' || *var == ',') {
var++;
break;
}
integer_part = 10*integer_part + (*var - '0');
var++;
}
// If we hit the string terminator in previous loop, we will do so
// in the beginning of this loop too. If you think it makes things
// clearer, you can add the boolean found_comma_or_dot to explicitly
// skip this loop.
uint32_t decimal_part = 0;
uint32_t decimal_size = 0;
while(*var != 0) {
decimal_part = 10*decimal_part + (*var - '0');
var++;
decimal_size++;
}
return sign * (integer_part + decimal_part/pow(10, decimal_size));
}
Note that I changed uint16_t to uint32_t because I'm using them in another way. If this is not a viable option for you, you can change them to a floating type, but that may cause loss of precision.
There are reasons for using uint16_t and float, but the you will have to live with the limitations. That's just the way it is.
The problem is that float doesn't have enough precision to represent the number 123456789. We have 123456789 ≥ 2^26; in that range float can only represent multiples of 8.
The error for numbers with decimal point is caused by you treating the decimal point like an additional digit instead of ignoring it. Since the ASCII code for "." is two less than the code for "0", that "." is treated like a digit "-2".
PS. Using 0x30 instead of '0' is very bad style.
PS. Use double instead of float unless you have a very good reason that you can explain clearly why you wouldn't.
I am working on a method in C where I am trying to convert a decimal to its base. I am having trouble with returning a Char*. I still am unsure how to return a pointer. When I compile this code, I get a warning saying
"warning: function returns address of local variable [-Wreturn-local-addr]".
and it has to do with my res character. I am uncertain with why I cannot return res, if it's a char. I don't understand what I am suppose to return if I can't return res. Please help.
//return res;
char reVal(int num)
{
if (num >= 0 && num <= 9)
return (char)(num + '0');
else if(num = 10)
{
return (char)(num - 10 + 'A');
}
else if(num = 11)
{
return (char)(num - 11 + 'B');
}
else if(num = 12)
{
return (char)(num - 12 + 'C');
}
else if(num = 13)
{
return (char)(num - 13 + 'D');
}
else if(num = 14)
{
return (char)(num - 14 + 'E');
}
else if(num = 15)
{
return (char)(num - 15 + 'F');
}
}
// Utility function to reverse a string
void strev(char *str)
{
int len = strlen(str);
int i;
for (i = 0; i < len/2; i++)
{
char temp = str[i];
str[i] = str[len-i-1];
str[len-i-1] = temp;
}
}
char* decToBase(int base, int dec)
{
int index = 0; // Initialize index of result
char res[100]; // Convert input number is given base by repeatedly
// dividing it by base and taking remainder
while (dec > 0)
{
res[index++] = reVal(dec % base);
dec /= base;
}
res[index] = '\0';
// Reverse the result
strev(res);
return res;
int main()
{
char* base = decToBase(16, 248);
}
Regardless, the outcome I want is to have the method return "f8" as the outcome.
In your decToBase() function, the issue it's warning about is the use of char res[500];, which is an array that's allocated on the stack as a local variable. These are all thrown away when the function returns, so if you return a pointer to (or: the address of) the res array, this pointer points to junk on the stack.
You have to find some other way to manage this allocation, and though some might suggest using malloc() to allocate memory from the system, this is probably a bad idea because it's asking for problems with memory leaks.
Better is to pass in the buffer you want it to fill, and use that. Then the caller does the allocation and you don't worry about memory leaks.
char *decToBase(int base, int dec, char *outbuf)
{
int index = 0; // Initialize index of result
// Convert input number is given base by repeatedly
// dividing it by base and taking remainder
while (dec > 0)
{
outbuf[index++] = reVal(dec % base);
dec /= base;
}
outbuf[index] = '\0';
// Reverse the result
strev(outbuf);
return outbuf;
}
and then your main function would look like:
int main()
{
char decbuf[500];
decToBase(16, 248, decbuf);
printf("Buffer is %s\n", decbuf);
}
This is still not super ideal, because your decToBase() function doesn't know how big outbuf is, and overflows are possible, so the experience and/or paranoid programmer will also pass in the size of outbuf so your function knows how much to use.
But that's a step you'll get to later.
I have the following working code; it accepts a string input as the function parameter and spits out the same string converted to a decimal.
I'm not going to bother accounting for negative inputs, although I understand that I can set a boolean flag to true when the first indexed character is a "-". If the flag switches to true, take the total output and multiply by -1.
Anyway, I'm pretty stuck on where to go from here; I'd like to adjust my code so that I can account for a decimal place. Multiplying by 10 and adding the next digit (after converting that digit from an ASCII value) yields an integer that is displayed in decimal in the output. This obviously won't work for numbers that are smaller than 1. I understand why (but not really how) to identify where the decimal point is and say that "for anything AFTER this string index containing a decimal point, do this differently"). Also, I know that instead of multiplying by a power of 10 and adding the next number, I have to multiply by a factor of -10, but I'm not sure how this fits into my existing code...
#include <stdio.h>
#include <string.h>
int num = 0;
int finalValue(char *string1) {
int i = 0;
if (string1[i] != '\0') {
if (string1[i]<'0' || string1[i]>'9') {
printf("Sorry, we can't convert this to an integer\n\n");
}
else {
num *= 10;
num += string1[i] - '0';
//don't bother using a 'for' loop because recursion is already sort-of a for loop
finalValue(&string1[i+1]);
}
}
return num;
}
int main(int argc, const char * argv[]) {
printf("string to integer conversion yields %i\n",(finalValue("99256")));
return 0;
}
I made some adjustments to the above code and it works, but it's a little ugly when it comes to the decimal part. For some reason, the actual integer output is always higher than the string put in...the math is wrong somewhere. I accounted for that by subtracting a static amount (and manually multiplying by another negative power of 10) from the final return value...I'd like to avoid doing that, so can anybody see where my math / control flow is going wrong?
#include <stdio.h>
#include <string.h>
//here we are setting up a boolean flag and two variables
#define TRUE 1
#define FALSE 0
double num = 0;
double dec = 0.0;
int flag = 0;
double final = 0.0;
double pow(double x, double y);
//we declare our function that will output a DOUBLE
double finalValue(char *string1) {
//we have a variable final that we will return, which is just a combination of the >1 and <1 parts of the float.
//i and j are counters
int i = 0;
int j = 0;
//this will go through the string until it comes across the null value at the very end of the string, which is always present in C.
if (string1[i] != '\0') {
//as long as the current value of i isn't 'null', this code will run. It tests to see if a flag is true. If it isn't true, skip this and keep going. Once the flag is set to TRUE in the else statement below, this code will continue to run so that we can properly convert the decimal characers to floats.
if (flag == TRUE) {
dec += ((string1[i] - '0') * pow(10,-j));
j++;
finalValue(&string1[i+1]);
}
//this will be the first code to execute. It converts the characters to the left of the decimal (greater than 1) to an integer. Then it adds it to the 'num' global variable.
else {
num *= 10;
num += string1[i] - '0';
// This else statement will continue to run until it comes across a decimal point. The code below has been written to detect the decimal point and change the boolean flag to TRUE when it finds it. This is so that we can isolate the right part of the decimal and treat it differently (mathematically speaking). The ASCII value of a '.' is 46.
//Once the flag has been set to true, this else statement will no longer execute. The control flow will return to the top of the function, and the if statement saying "if the flag is TRUE, execute this' will be the only code to run.
if (string1[i+1] == '.'){
flag = TRUE;
}
//while this code block is running (before the flag is set to true) use recursion to keep converting characters into integers
finalValue(&string1[i+1]);
}
}
else {
final = num + dec;
return final;
}
return final;
}
int main(int argc, const char * argv[]) {
printf("string to integer conversion yields %.2f\n",(finalValue("234.89")));
return 0;
}
I see that you have implemented it correctly using global variables. This works, but here is an idea on how to avoid global variables.
A pretty standard practice is adding parameters to your recursive function:
double finalValue_recursive(char *string, int flag1, int data2)
{
...
}
Then you wrap your recursive function with additional parameters into another function:
double finalValue(char *string)
{
return finalValue_recursive(string, 0, 0);
}
Using this template for code, you can implement it this way (it appears that only one additional parameter is needed):
double finalValue_recursive(char *s, int pow10)
{
if (*s == '\0') // end of line
{
return 0;
}
else if (*s == '-') // leading minus sign; I assume pow10 is 0 here
{
return -finalValue_recursive(s + 1, 0);
}
else if (*s == '.')
{
return finalValue_recursive(s + 1, -1);
}
else if (pow10 == 0) // decoding the integer part
{
int digit = *s - '0';
return finalValue_recursive(s + 1, 0) * 10 + digit;
}
else // decoding the fractional part
{
int digit = *s - '0';
return finalValue_recursive(s + 1, pow10 - 1) + digit * pow(10.0, pow10);
}
}
double finalValue(char *string)
{
return finalValue_recursive(string, 0);
}
Also keep track of the occurrence of the decimal point.
int num = 0;
const char *dp = NULL;
int dp_offset = 0;
int finalValue(const char *string1) {
int i = 0;
if (string1[i] != '\0') {
if (string1[i]<'0' || string1[i]>'9') {
if (dp == NULL && string1[i] == '.') {
dp = string1;
finalValue(&string1[i+1]);
} else {
printf("Sorry, we can't convert this to an integer\n\n");
} else {
} else {
num *= 10;
num += string1[i] - '0';
finalValue(&string1[i+1]);
}
} else if (dp) {
dp_offset = string1 - dp;
}
return num;
}
After calling finalValue() code can use the value of dp_offset to adjust the return value. Since this effort may be the beginning of a of a complete floating-point conversion, the value of dp_offset can be added to the exponent before begin applied to the significand.
Consider simplification
//int i = 0;
//if (string1[i] ...
if (*string1 ...
Note: using recursion here to find to do string to int is a questionable approach especially as it uses global variables to get the job done. A simply function would suffice. Something like untested code:
#include <stdio.h>
#include <stdlib.h>
long long fp_parse(const char *s, int *dp_offset) {
int dp = '.';
const char *dp_ptr = NULL;
long long sum = 0;
for (;;) {
if (*s >= '0' && *s <= '9') {
sum = sum * 10 + *s - '0';
} else if (*s == dp) {
dp_ptr = s;
} else if (*s) {
perror("Unexpected character");
break;
} else {
break;
}
s++;
}
*dp_offset = dp_ptr ? (s - dp_ptr -1) : 0;
return sum;
}
Figured it out:
#include <stdio.h>
#include <string.h>
//here we are setting up a boolean flag and two variables
#define TRUE 1
#define FALSE 0
double num = 0;
double dec = 0.0;
int flag = 0;
double final = 0.0;
double pow(double x, double y);
int j = 1;
//we declare our function that will output a DOUBLE
double finalValue(char *string1) {
//i is a counter
int i = 0;
//this will go through the string until it comes across the null value at the very end of the string, which is always present in C.
if (string1[i] != '\0') {
double newGuy = string1[i] - 48;
//as long as the current value of i isn't 'null', this code will run. It tests to see if a flag is true. If it isn't true, skip this and keep going. Once the flag is set to TRUE in the else statement below, this code will continue to run so that we can properly convert the decimal characers to floats.
if (flag == TRUE) {
newGuy = newGuy * pow(10,(j)*-1);
dec += newGuy;
j++;
finalValue(&string1[i+1]);
}
//this will be the first code to execute. It converts the characters to the left of the decimal (greater than 1) to an integer. Then it adds it to the 'num' global variable.
else {
num *= 10;
num += string1[i] - '0';
// This else statement will continue to run until it comes across a decimal point. The code below has been written to detect the decimal point and change the boolean flag to TRUE when it finds it. This is so that we can isolate the right part of the decimal and treat it differently (mathematically speaking). The ASCII value of a '.' is 46.
//Once the flag has been set to true, this else statement will no longer execute. The control flow will return to the top of the function, and the if statement saying "if the flag is TRUE, execute this' will be the only code to run.
if (string1[i+1] == 46){
flag = TRUE;
finalValue(&string1[i+2]);
}
//while this code block is running (before the flag is set to true) use recursion to keep converting characters into integers
finalValue(&string1[i+1]);
}
}
else {
final = num + dec;
return final;
}
return final;
}
int main(int argc, const char * argv[]) {
printf("string to integer conversion yields %.2f\n",(finalValue("234.89")));
return 0;
}