atoi function doesn't work in for loop - c

Question

I am trying to get the output of this equation :
44 - 10 + 11 / 5
it work properly as follow :
char str[] = "44 - 10 + 11 / 5";
int sum = 0 ;
sum += atoi(str); // 0 + 44
sum += atoi(str+3); // 44 + 10
sum += atoi(str+5); // 54 + 11
sum += atoi(str+7); // 65 / 5 = 13
printf("%d/n",sum); // output = 13
but it doesn't work if I put the atoi() function in the for loop:
char str[] = "44 - 10 + 11 / 5";
int sum = 0;
int i;
sum += atoi(str);
for (i = 0; i < 100; i++) {
if (!(str[i] >= 0 && str[i] <= 9)) { //if str[i] is not a number
sum += atoi(str + i);
}
}
printf("%d/n", sum); // output = 0

Answer 1

You are comparing a char value that contains the ASCII representation of a number, to an actual number. Change
if(!(str[i]>=0 && str[i]<=9)){
to
if(!(str[i]>='0' && str[i]<='9')){
I did not check if the rest of the code is correct, but certainly, this is one issue.

Answer 2

if(!(str[i]>=0 && str[i]<=9)){ //if str[i] is not a number
This is incorrect. '0' and 0 are not same. Neither is 9 and '9'. The value of '9' will be integer that represents '9' in ASCII chart.
Rest of the logic seems suspicious too, for example what happens if you have "55", code will enter if twice, once for 55 and second time for 5, isn't it?

Answer 3

Because your code does not do the same calculations as your previous one. In the first one you are shifting the pointer to the next number and you are adding it to the number, so your code does the following 44 + 10 + 11 + 5.
In the second case you are checking whether the character code is between 0 and 9, but the numbers are between 48 and 57 - the characters '0' and '9'. So the if statement would be always wrong. Your intention was probably this, but it won't work.
for(i = 0; i < 100 ; i++){
if(!(str[i]>='0' && str[i]<='9')){ //if str[i] is not a number
sum += atoi(str+i);
}
}
This also does not work, because here you are doing the wrong shifts of the pointer by one, resulting in the reading of the following numbers: 44, 4, 10, 0, 11, 1, 5. Because of the atoi properties. To perform the exact same code you did before you`d have to write the following code:
int skip_non_diggits(int i, char *str) {
while (str[i] < '0' || str[i] > '9') {
if (str[i] == 0)
return -1; //end of string case
i++;
}
return i;
}
int skip_diggits(int i, char *str) {
while (str[i] >= '0' && str[i] <= '9') {
if (str[i] == 0)
return -1; //end of string case
i++;
}
return i;
}
These functions could be used as follows:
int i = 0, sum = 0;
while (i != -1) {
sum += atoi(str+i);
i = skip_diggits(i, str); // positions i after the current number
i = skip_non_diggits(i, str); // positions i at the begin of the next number
}
BTW
Don't use for (i = 0; i < 100; i++) to iterate through the strings you don't know the size of. If the string is shorter in length than 100 you're derefferencing memory which probably isn't yours creating undefined behavior. Instead check the strlen method or check whether the character is not equal to \0 which in C indicates the end of the string.

Answer 4

This seems a bit odd. What's the goal?
Agree with previous comment.
Why are you doing atoi() on what is NOT a number?
You will go past the end of the string.
Your spaces COUNT in the example
and so on.
you don't update i when you find a number, so you will count "44" as 44 first and 4 second.
If you're trying to learn, hook up a debugger, step through the code, and see how the data changes.