CODEWITHSUNDEEP
c++logic
user4375224
user4375224

Reputation:

C++ : Checking the contents of a string

Sorry , I am using an ancient compiler

#include <iostream.h>
#include <conio.h>
#include <ctype.h>
void main()
{
   char s[6] = "mOh1t*";
   int u = 0 , l=0 , d=0 , sp=0 , t = 0;
   for (int x = 0 ; s[x] ; x++)
   {
      if (isupper(s[x]))
        u++;
      else if(islower(s[x]))
        l++;
      else if (isdigit(s[x]))
        d++;
      t++;
   }
   sp = t - (u+l+d);
   cout<<"t:"<<t;
   cout<<"u:"<<u;
   cout<<"l:"<<l;
   cout<<"d:"<<d;
   cout<<"sp:"<<sp;
   getch();
}

The above code clearly counts the number of alphabets(uppercase and lowercase) , number of digits and number of special characters in a string.

I was wondering if the same is possible without using if statements/ternary operator/switch case. If yes , could I get a lead ?

Kudos to HoosierEE for the answer

UPDATE: Would it be possible to do the same as hoosierEE's answer without using inbuilt functions?

- A C++ newbie looking to learn

Upvotes: 2

Views: 110

Answers (1)

Alex Shroyer
Alex Shroyer

Reputation: 3829

If you just want to avoid if statements, you can treat booleans as 0 and 1 and sum up the characters like so:

for (int x = 0; s[x]; x++)
{
    u += isupper(s[x]);
    l += islower(s[x]);
    d += isdigit(s[x]);
    t++;
}

...But as @Angew mentions, you can't rely on the result to be just a 0 or 1 with these functions.

And to answer the UPDATE, here's how you could achieve what you want without stdlib functions:

for (int x = 0; s[x]; x++)
{
    u += ((s[x] >= 'A') && (s[x] <= 'Z'));
    l += ((s[x] >= 'a') && (s[x] <= 'z'));
    d += ((s[x] >= '0') && (s[x] <= '9'));
    t++;
}

You could look up the ASCII values and put numbers in there, but I think using character literals is more readable. Personal preference.

You might also consider being a little more strict with your termination condition:

for (int x = 0; s[x] != '\0'; x++)

Upvotes: 4

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