CODEWITHSUNDEEP
c
danny
danny

Reputation: 1637

char array to integer array

I got a set of characters as input using scanf which is actually like this "1854?156X".

(using scanf("%c",&input[i]) , input is an array of 10 characters);

In further processing of code I want to multiply the (first digit *10) & (second digit *9) and so on.

So, while multiplying it is taking the ASCII value of 1 which is actually (49 *10), instead of (1*10).

input[i]*counter;

where counter is 

int counter=10;

How can I convert the char array to integer array where the exact digit should be multiplied instead of ascii value of that character?

Upvotes: 1

Views: 296

Answers (4)

Brian Cain
Brian Cain

Reputation: 14619

Use strtol to convert these character strings into integral types.

It's superior to atoi because the failure modes are not ambiguous.

Upvotes: 0

Sebi
Sebi

Reputation: 4532

Definitely use atoi(). It spares you a lot of manual character checking. 

int input_val = atoi(input);
int temp = input_val;

while(temp)
{
  int digit = temp % 10;
  temp /= 10;
}

This gives you each digit from right to left in "digit". You can construct your number using addition and multiplying by powers of 10.

Upvotes: 1

nullpotent
nullpotent

Reputation: 9260

Take the ascii value and subtract 48.

That is input[i] - '0'

...or rather use atoi(). It returns zero if it fails on conversion, so you don't need to bother with '?' and such characters in your array while you traverse it.

Upvotes: 0

cmh
cmh

Reputation: 10947

If you know ahead of time that you value is ascii then simply subtract the value of '0' from it. E.g. '9' - '0' = 9'

Upvotes: 3

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