can't assign character when while loop is done in C - c

Question

I am trying to find solution how to add '\0' at the end of string (array of characters in C ).
I already have array "number" which is generated from for loop.
int i=0;
int j=0;
char s[] = "123.45e-10";
while (s[i] != '\0') {
if (isdigit(s[i]) && eFound == false) {
number[j++] = s[i];
}
i++
}
I tried to do this:
i = 0;
while (isdigit(number[i])){
i++;
}
printf("%d", i);
number[i] = 'b';
I have no compile errors, but when running program visual studio says
"Debug Assertion Fail".
Why is this not possible to do in C?
EDIT (whole code added):
#pragma warning(disable:4996) //disables crt warnings
#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>
#include<string.h>
double stof(char s[]);
int main(){
char string[] = "123.45e-10";
stof(string);
}
double stof(char s[]) {
bool isNegative = false;
bool eFound = false;
int i = 0;
int j = 0;
int k = 0;
char number[10];
char potention[11];
int dotPosition;
while (s[i] != '\0') {
if (isdigit(s[i]) && eFound == false) {
number[j++] = s[i];
}
else if (s[i] == '.') {
dotPosition = i;
}
else if (s[i] == 'e' || s[i] == 'E')
eFound = true;
else if (s[i] == '-')
isNegative = true;
else if (eFound== true && isdigit(s[i])){
potention[k++] = s[i];
}
i++;
}
i = 0;
while (isdigit(number[i])){
i++;
}
printf("%d", i);
number[i] = 'b';
const int charSize = dotPosition;
int potentionN = atoi(potention);
char beforeDecimal[sizeof(number)];
char addedNulls[20];
int g = 0;
if (isNegative == true) {
strncpy(beforeDecimal, number, dotPosition);
int addNull = potentionN - sizeof(beforeDecimal);
printf("%d", addNull);
while (g < addNull) {
addedNulls[g] = '0';
g++;
}
printf("%d", strlen(addedNulls));
}
return 0.0;
}

Answer 1

Why is this not possible to do in C?
Every computable answer can be found with a Turing-complete language. Your question is more like, where is the error in my code?
In C, it's very important to ensure you read & write only to defined memory: those locations covered by your variables. You're testing your input locations with
while (s[i] != '\0')
but nowhere do you ensure that you stay within the bounds of number and potention. Especially here:
i = 0;
while (isdigit(number[i])){
i++;
}
At no point do you write a nondigit to number. Why should that loop ever end?
The instant fix is to initialize number:
char number[10] = {};
That causes the compiler to initialize the array with any elements inside {}, and the rest of it with zeros. Since the initializer is empty, the whole array is filled with zeros.
That way, provided you don't fill number with digits, your while loop will terterminate. A belt-and-suspenders solution is make double-sure:
i = 0;
while (i < sizeof(number) && isdigit(number[i])) {
i++;
}
I haven't debugged your whole program; I haven't even compiled it. The run-time error you're seeing suggests you're writing to undefined memory. If you check the array bounds in each loop, you can avoid that particular problem and move on to the next one. ;-)