Creating an atoi function in C - c

Question

I try to create atoi function, and think I made right code, but when I run it, it shows wrong one. I'm trying to figure it out, but don't know what I made it wrong please check the code and give some help
My code is
#include <stdio.h>
int my_atoi(char *str)
{
int i;
int res;
int sign;
i = 0;
res = 0;
sign = 1;//sign of '-' or '+'
while(str[i] == ' ' || (str[i] >= 9 && str[i] <= 13))
{
i++;
}
if(str[i] == '-')
{
sign = -1;
i++;
}
else if(str[i] == '+')
{
sign = 1;
i++;
}
while(str[i] >= '0' && str[i] <= '9')
{
res = res * 10 + str[i] + '0';
i++;
}
return(res * sign);// to make integer which has value of '-' or '+'
}
int main(void)
{
char str[] = "-2018shiba";
printf("%d\n", my_atoi(str));
return(0);
}
When I run it, it shows -108674

Answer 1

I am seeing multiple mistakes here.
If you want to convert a ASCII character into the corresponding integer you need to subtract '0'. Take a look at the ASCII table: for instance '7' is mapped by decimal value 55. Hence if you want to get 7 then you need to subtract the ASCII of '0' which is 48 (55 - 48 = 7):
int foo = str[i] - '0';
In the very last while loop of my_atoi. The value of an indexed numeral string representation is calculated by multiplying the value of str[i] with the numerical base to the power of the index starting from behind.
For example lets take a look at "1337":
7*10^0 + 3*10^1 + 3*10^2 + 1*10^3 = 7 + 30 + 300 + 1000 = 1337
As you can see, the 7 has the numerical index 0 and so on. Assuming you want to just ignore shiba your code be looking something like this:
#include <stdio.h>
#include <string.h>
// Return base^(exponent)
int my_pow(int base, unsigned int exponent)
{
if (exponent == 0) {
return 1;
} else {
int result = base;
for (int i = 1; i < exponent; i++) {
result *= base;
}
return result;
}
}
int my_atoi(char *str, size_t len)
{
int i;
int res;
int sign;
i = 0;
res = 0;
sign = 1;//sign of '-' or '+'
while(str[i] == ' ' || (str[i] >= 9 && str[i] <= 13))
{
i++;
}
if(str[i] == '-')
{
sign = -1;
i++;
}
else if(str[i] == '+')
{
sign = 1;
i++;
}
// Store the index where the number string starts
int j = i-1;
// Find the ending index of the number string
i = len;
while (str[i] < '0' || str[i] > '9') {
i--;
}
int num_end = i;
// Now start at the ending
while(i > j)
{
if (str[i] >= '0' && str[i] <= '9') {
res += my_pow(10, num_end-i) * (str[i] - '0');
} else {
// If a character unequal to a digit is found then skip it
num_end--;
}
i--;
}
return(res * sign);// to make integer which has value of '-' or '+'
}
int main(void)
{
char str[] = "-2018shiba";
printf("%d\n", my_atoi(str, strlen(str)));
char str2[] = "-20X18shiba";
printf("%d\n", my_atoi(str2, strlen(str2)));
return(0);
}