Why is my function atoi not working? [closed] - c

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I am trying to create my personal atoi function but i doesn't work properly. I don't know why.
void ft_putchar(char c)
{
write(1, &c, 1);
}
int ft_atoi(const char *str)
{
int i;
i = 0;
while (str[i] == 9 || str[i] == 32 || str[i] == 43 || str[i] == 45)
i++;
while (str[i] > 48 && str[i] < 57)
{
ft_putchar(str[i]);
i++;
}
return (0);
}
int main(void)
{
printf("%d", atoi(" 2794ffafsaasf"));
printf("\n%d",ft_atoi(" 2794fsffsf"));
return(0);
}
I kinda works but it doesn't. It gives me a weird result where it even ignores '\n'.
The result it gives me is this.
272794 and a new line.
with only my function it gives me only the number 27.
EDIT:
I have created a new program. But it still doesn't work. It simply can't see digits.
int ft_atoi(const char *str)
{
int i;
int n;
i = 0;
while (str[i] == '\t' || str[i] == ' ' || str[i] == '+' || str[i] == '-')
++i;
n = i;
while (str[n] >= '0' && str[n] <= '9')
{
++n;
}
return(str[n]);
}

Answer 1

Your test for digits has the bounds set wrong. You're testing str[i] > 48 && str[i] < 57, but 48 is the ordinal for the 0 character, and 57 is the ordinal for 9. This means you only consider 1 through 8 inclusive to be numeric digits, and you stop parsing 2794fsffsf at the 9, not at the first f.
Changing the test to str[i] >= 48 && str[i] <= 57 would fix the bounds issue, but would still be less than self-documenting. To make it obvious what you're doing to people who don't have the ASCII tables memorized, you could do:
while ('0' <= str[i] && str[i] <= '9')
or possibly slightly slower but even more obviously with ctype.h:
while (isdigit(str[i]))
You could similarly replace the many non-obvious tests for whitespace ordinal values with isspace(str[i]).

Answer 2

First thing is to ignore leading whitespace. You can do that with i as you're doing, but it's easier to do it by shifting the start of the string forward. I'm guessing that you're reimplementing the standard C library as an exercise. I'm not, so I'm going to make use of the ctype.h functions. Adjust as you like, the basic code remains the same. You should probably write your own ft_isspace and ft_isdigit rather than hard coding the logic.
/* Move str to point at the first non-space character */
while( isspace(str[0]) ) {
str++;
}
Now there's no need for an i to hold your place, you can just work with str[0] and it will be the first non-whitespace character.
Next we need to find and add up our digits. Again, I'll use isdigit. Follow the advice in ShadowRanger's answer if you can't use that function.
The technique is to convert each character to a digit and add it to the total. Since we're reading a number left to right, each new digit multiplies the existing sum by 10. For example, if we had 1234 it would be... 1 then 10 + 2 then 120 + 3 then 1230 + 4.
int num = 0;
while( isdigit(str[0]) ) {
int digit = str[0] - '0';
num *= 10;
num += digit;
str++;
}
Once again I'm manipulating the pointer rather than using an index. This is just to save an integer, and to avoid mixing up two string iteration techniques in a single function.
And, finally, don't forget to return the number!
return num;
What about negative numbers? We need to look for a - before we check for digits, if it's there flip the sign. We also need to handle a + and ignore it.
/* Assume it's positive. */
short sign = 1;
/* Check for a sign. */
switch( str[0] ) {
case '-':
sign = -1;
str++;
break;
case '+':
str++;
break;
}
And then multiply by the sign to flip the result.
return sign * num;

Answer 3

Your code only writes out the numbers - and not all, as pointed out before me (because it excludes '0' and '9' and they should be included!) - to the file identified by 1 (what is maybe the standard output, I don't remember exactly...), and returns 0 as the parameter of the printf. It seems to me a little strange, you only had 27 printed out with your function, but it might be environment(?) specific. (I have not tested it, so it possibly not exactly work as I suppose...)
I've made my version of atoi, hoping I can show where to start / the way how you should do yours:
int nvi9_atoi(const char *s) {
int n = 0, valid = 0, prev = 0;
while(1) { // infinite loop
if (*s == '\0') break; // if it is the end of the string, finish further processing
if (*s >= '0' && *s <= '9') { // check whether the current character is an ascii number
n *= 10; // "move" all previous digits (eg. 1 -> 10, 43 -> 430); if n is 0, this has no effect
if (n >= 0) n += (*s - '0'); // if n is not negative, add the value of the last digit (here is ascii-number "conversion"!)
else n -= (*s - '0'); // if it is negative, the number should be substracted (note eg. if there was -10 so far, and the new number is 2, the new value should be -12, not -8)
if (n > 0 && !valid) { // if n not 0, and there was no digits before it, check if there was a minus sign before
valid = 1; // preventing more check
if (prev == '-') n *= -1; // makes n negative
}
} else if (valid) break; // if there was numbers processed, but the current character is not a number, finish loop
prev = *s; // store current character for minus sign checking of next loop
s++; // move string pointer to the next character
}
return n; // return the converted value
}
It is also untested, so fixes / corrections / improvements are welcome!
This function can even process strings like " sfg afg-65sd1g". In that case the returned value will be -65, because it looks for the first number, then returns.