What is the C program to check whether a given character is vowel or not using if else?

Answer 1

#include<stdio.h> main() { char key; clrscr(); printf("Enter the key"); scanf("\n %c",& key); if (key>='A' key<='Z' )&&(key>='a' key<='z') { printf("%c is an character \n",key); } else { printf("%c is not character",key); } getch(); }

Answer 2

bool is_vowel (char c) { c = tolower (c);

if (c=='a') return true;

else if (c=='e') return true;

else if (c=='i') return true;

else if (c=='o') return true;

else if (c=='u') return true;

else

return false;

}

This is rather verbose and inefficient however the 'shorthand' version doesn't improve efficiency:

bool is_vowel (char c) {

c = tolower (c);

if (c=='a' c=='e' c=='i' c=='o' c=='u')

return true;

else

return false;

}

It is in inefficient because for the letter 'u' and all non-vowels we must perform 5 comparisons whereas 'a' requires just one comparison. However, the first version of the function is really just a thinly-veiled switch:

bool is_vowel (char c) { switch (tolower (c)) { case 'a': case 'e':

case 'i':

case 'o':

case 'u': return true;

default: return false;

}

}

Here, the compiler can optimise the code to produce an efficient jump table thus creating a constant-time operation regardless of whether the letter is a vowel or not. To understand how this works, a switch is really just a goto statement that accepts an integral value rather than a label, such that we jump to the default label when there is no matching label:

bool is_vowel (char c) { // Note: not valid in C!

goto tolower(c);

'a':

'e':

'i':

'o':

'u': return true;

default: return false;

}