char array c++ vowels - c++

Question

I am trying to make a program that will use a switch statement and see if an element of a char array is a vowel and which one, but i am stuck at how to check the elements:
int prob2() {
char uName[25] = "";
int voCo = 0;
cout<<"Enter you first and last name, under 25 chars please: ";
cin>>uName;
int i = 0;
while(i <= 25){
switch(i){
case 1:
voCo++;
break;
case 2:
voCo++;
break;
case 3:
voCo++;
break;
case 4:
voCo++;
break;
case 5:
voCo++;
break;
default:
break;
}
i++;
}
cout<<"Your first and last name have: "<<voCo<<" vowels in them."<<endl;
return 0;
}

Answer 1

It seems you mean the following
#include <iostream>
#include <cctype>
using namespace std;
//...
size_t prob2()
{
const size_t N = 25;
char uName[N] = "";
size_t voCo = 0;
cout<<"Enter you first and last name, under " << N << " chars please: ";
cin.getline( uName, N );
for ( char *p = uName; *p != '\0'; ++p ) *p = toupper( ( unsigned char )*p );
for ( const char *p = uName; *p != '\0'; ++p )
{
switch( *p )
{
case 'A':
voCo++;
break;
case 'E':
voCo++;
break;
case 'I':
voCo++;
break;
case 'O':
voCo++;
break;
case 'U':
voCo++;
break;
default:
break;
}
}
cout<<"Your first and last name have: "<<voCo<<" vowels in them."<<endl;
return voCo;
}

Answer 2

You could try something like this:
const std::string vowels = "aeiou";
const std::string name = "martin luther king, jr.";
const unsigned int name_length = name.length();
unsigned int vowel_count = 0U;
for (unsigned int i = 0U; i < name_length; ++i)
{
if (vowels.find(name[i]) != std::string::npos)
{
++vowel_count;
}
}
No need for switch statement. This is one of many possible algorithms or implementations.
Edit 1: An array of counts
You could also use an array of counts:
unsigned int counts[26] = {0};
for (unsigned int i = 0U; i < name_length; ++i)
{
const c = std::tolower(name[i]);
if (isalpha(c))
{
counts[c - 'a']++;
}
}
const unsigned int vowel count =
counts['a'] + counts['e'] + counts['i']
+ counts['o'] + counts['u'];

Answer 3

First of all decouple user interaction from the logic solving your requirement. I think we can safely assume you can collect the input in this case and save it into an string. So we will not waste our time with that.
We will focus on developing and testing the code that solves the requirement. In a standard C++. Now here is the deep end of the pool. The code.
// mike.h
#pragma once
// std::string view requires C++17
#include <string_view>
// always use namespace,to avoid name clashes
namespace mike {
// make 'sv' the string_view literal available
using namespace std::string_view_literals;
// declare and define compile time
// string view literal
// 'constexpr' guarantees compile time
// notice the use of 'sv'
constexpr auto vowels = "eaiouEAIOU"sv;
// compile time function to count literals
// again 'constexpr' guarantees compile time
// inline gurantees we can include this header many times
// without making accidental duplicates of `count_vowels`
// 'in_' argument has 'std::string_view' passed by value
// pass by value is preferred standard C++ method
// of functions arguments passing
// 'std::string_view' is standard C++ preferred type
// to pass strings into functions
inline constexpr size_t
count_vowels(std::string_view in_)
{
// return type is size_t
// we can count very large number of vowels
// but all at compile time
size_t rezult{};
// this is C+17 'range for'
// we cast implicitly references to input elements
// from, `char const &` to `int const &`
// cost of that is very likely 0
for (int const & ch_ : in_)
for (int const & v_ : vowels)
// there is no if() here
// we simply add 0's or 1's, to the rezult
// false is 0, true is 1
// the correct by the book way of coding that is
// static cast from bool to int
// rezult += static_cast<int>( v_ == ch_ ) ;
rezult += v_ == ch_ ;
return rezult;
}
// runtime speed of this call is 0 (zero)
// all happens at compile time
// notice how we pass normal string literal
// no need to create string_view
constexpr size_t r1
= count_vowels("abra ca dabra");
// no runtime tests necessary
// `static_assert()` is compile time assert
// failure message is optional
static_assert(r1 == 5,
"compile time calculation failed, 'abra ca dabra', must contain 5 vowels");
} // mike ns
Hopefully there are a lots of comments. Solution does not use switch() statement or if() statements. Thanks to standard C++ constructs, code is very simple, resilient and probably very fast when compiled by modern optimizing compilers.
Solution works at compile time too. That is not stopping you to use it in your run-time scenario. Although, I would advise again using native char array. std::string might be a perfect match here.
std::string input_ = collect_user_input() ;
int rezult = count_vowels(input_);
Enjoy the standard C++ ...