Check if char array contains digits - c++

Question

So I have a random char array, i.e "asdd1vnb24vnf63vbn,-5h-2kk", and I should check if it contains digits, and if it does, my task is to add them with opposite signs (S=-1-2-4-6-3+5+2) without using string functions.
But for now I'm stuck here and simply cannot understand why it doesn't work and "Sum" gives random numbers, though it is able to print digits in an array.
int main ()
{
char s [] = "asdd1vnb24vnf63vbn,-5h-2kk";
int sum = 0;
for (int i = 0; i < sizeof(s)/sizeof(s[0]); i++) {
if (isdigit(s[i]) !=0) {
sum += s[i];
cout << "Sum is " << sum << endl;
}
}
system ("pause");
return 0;
}
Can somebody tell me what's wrong?

Answer 1

you add the ASCII value to sum variable not the integer value
only need little correction in your code in sum += s[i] to sum += s[i] - '0'

Answer 2

You should add not s[i], instead it should be s[i] - '0'.
Why?
s[i] is a char code, for 'zero' it could be 48, then 49 for 'one', 50 for 'two' and so on. So, you increment ascii code value instead of char value
Once you decrement s[i] with '0' char, you receive exact digit value (in case it is really a digit, because we don't see your isdigit implementation)

Answer 3

Besides the obvious sum += s[i] - '0' I see another problem - topic starter wont to inverse digits sign. So you need to check preceding '-' key:
if (isdigit(s[i])) {
if (i > 0 && s[i - 1] == '-')
sum += s[i] - '0';
else
sum -= s[i] - '0';

Answer 4

my task is to add them with opposite signs
If that's the case, in addition to the advice given in the other answers, you need to keep track of the polarity of the sign, whether it is negative or positive.
#include <iostream>
#include <cctype>
int main()
{
char ptr[] = "asdd1vnb24vnf63vbn,-5h-2kk";
int multiplier = 1; // this starts out at 1
char *p = ptr;
int value = 0;
while ( *p ) // keep going until the end of the string
{
if (isdigit(*p))
{
value += multiplier * (*p - '0');
multiplier *= -1; // switch polarity
}
++p;
}
std::cout << value;
}
Live Example