C++ : Checking the contents of a string

Question

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

If you are really looking to learn, get a newer compiler and learn the modern standards. — crashmstr, Dec 09 '14 at 14:44
@Mohit_Bhasi if you are using something like Turbo C++, the C++ language has changed significantly (as a note, I was using Turbo C++ over 20 years ago). — crashmstr, Dec 09 '14 at 14:47
@crashmstr yup and i am using it now , sure will upgrade . Thanks for advice :) — , Dec 09 '14 at 14:50
Unrelated: Your `s` array is too small. The string has 6 characters, so there's no space left for the nul terminator. Never specify the size of a `char[]` if you want to initialise it with a string literal. — Angew is no longer proud of SO, Dec 09 '14 at 14:53
@Angew what effect would it have(if size of array = number of elements)? just curious :) — , Dec 09 '14 at 14:55
@Mohit_Bhasi Think about the terminating condition of your `for` loop: `s[x] != 0` (the `!= 0` part is implied). Is this ever guaranteed to be true when the nul terminator is missing, or will you happily run over the buffer and start reading and processing random memory? — Angew is no longer proud of SO, Dec 09 '14 at 15:03

Alex Shroyer · Accepted Answer · 2014-12-09T15:17:43.353

4

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++)

edited Dec 09 '14 at 15:17

answered Dec 09 '14 at 14:41

Alex Shroyer

3,499
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Yes, C and C++ (and [others](https://en.wikipedia.org/wiki/Boolean_data_type#C.2C_C.2B.2B.2C_Objective-C.2C_AWK.2C_Perl)) say `true == 1` and `false == 0`. – Alex Shroyer Dec 09 '14 at 14:45
More like `0` is `false` and anything other than `0` is `true`. – crashmstr Dec 09 '14 at 14:48
2

However, the isupper, islower and isdigit functions return an int where non-zero denotes true and 0 denotes false. – George Houpis Dec 09 '14 at 14:50
1

This will fail spectacularly unless you cast the result to a `bool` first - `std::isupper` and friends return arbitrary non-zero numbers as true values. – Angew is no longer proud of SO Dec 09 '14 at 14:52
@Angew Better to fail early... ;) I'll edit my answer. – Alex Shroyer Dec 09 '14 at 14:54
2

@hoosierEE what is the difference between s[x]!='\0' and s[x]? always thought they meant the same – Dec 09 '14 at 15:19
@Mohit_Bhasi Yes, they are strictly equivalent. – Angew is no longer proud of SO Dec 09 '14 at 15:31
@hoosierEE then why is it suggested to use s[x]!='\0'? – Dec 09 '14 at 15:32
Comparing to a null terminator *character* [better conveys your intent](http://stackoverflow.com/a/7578643/2037637), both for your own benefit (when you come back to this code months later) and your coworkers'. – Alex Shroyer Dec 09 '14 at 15:42

C++ : Checking the contents of a string

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