How to write the following code to get the same output without using any kind of loop?

Question

1.00/5 (1 vote)

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int main()
{
   int  count=0;
    char arr[7];
    printf("enter a string: ");
    scanf("%s",arr);
    for(int i=0;i<7;i++){
    switch(arr[i]){
    case 'A':
    count++;
    break;
    case 'E':
    count++;
    break;
    case 'I':
    count++;
    break;
    case 'O':
    count++;
    break;
    case 'U':
    count++;
    break;
    case 'a':
    count++;
    break;
    case 'e':
    count++;
    break;
    case 'i':
    count++;
    break;
    case 'o':
    count++;
    break;
    case 'u':
    count++;
    break;
    }
    }
    printf("\nthe number of vowels is: %d",count);

    return 0;
}

What I have tried:

I solved the question but i need to edit it without the loop

Posted 9-Mar-21 5:27am

Shadi Dwikat

Updated 9-Mar-21 11:21am

Add a Solution

3 solutions

Solution 3

Quote:
How to write the following code to get the same output without using any kind of loop?

Technically, it is impossible.
Even CPallini's clever Solution using recursion is a loop in disguise.
With this recursion, code do not show loop structure, but compiler transforms the code to a loop.

C++

// loopish version of CPallini solution
int vcount(const char * s)
{
  int ret=0;
  while (! *s) {
    ret += iv[(unsigned)*s]
    s++
  }
  return ret;
}

This particular recursion is named: Tail call - Wikipedia[^]
-----
Advice: Learn to indent properly your code, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.

C++

int main()
{
	int  count=0;
	char arr[7];
	printf("enter a string: ");
	scanf("%s",arr);
	for(int i=0;i<7;i++){
		switch(arr[i]){
		case 'A':
			count++;
			break;
		case 'E':
			count++;
			break;
		case 'I':
			count++;
			break;
		case 'O':
			count++;
			break;
		case 'U':
			count++;
			break;
		case 'a':
			count++;
			break;
		case 'e':
			count++;
			break;
		case 'i':
			count++;
			break;
		case 'o':
			count++;
			break;
		case 'u':
			count++;
			break;
		}
	}
	printf("\nthe number of vowels is: %d",count);

	return 0;
}

Indentation style - Wikipedia[^]

Professional programmer's editors have this feature and others ones such as parenthesis matching and syntax highlighting.
Notepad++ Home[^]
ultraedit[^]
Enabling Open Innovation & Collaboration | The Eclipse Foundation[^]

Posted 9-Mar-21 11:21am

Patrice T

Updated 9-Mar-21 14:20pm

v2

Comments

jeron1 9-Mar-21 20:15pm

Couldn't someone have 7 switch statements inline, one for every arr[i]? Yeah, it's awful, but possible I suppose.

Patrice T 9-Mar-21 20:20pm

Sure they can.

CPallini 10-Mar-21 2:05am

Indeed it is a loop in disguise.
Have my 5.

Patrice T 10-Mar-21 6:28am

Thank you

Solution 1

If you need to check every character in the string then you need the loop. But your switch block Could be made much simpler:

C++

switch(tolower(arr[i]))
{
    case 'a':
    case 'e':
    case 'i':
    case 'o':
    case 'u':
        count++;
        break;
    default: // all other charactgers
        break;
}

Posted 9-Mar-21 5:35am

Richard MacCutchan

Comments

CPallini 10-Mar-21 2:03am

My 5.

Richard MacCutchan 10-Mar-21 3:22am

Thanks, but your solution is far more elegant.

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CPallini · Accepted Answer · 2021-03-09T08:14:00

With C, you can unleash your creativity

C

#include <stdio.h>

const int iv[] =
{
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1,
  0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1,
  0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
};


int vcount(const char * s)
{
  if (! *s)
    return 0;
  return iv[(unsigned)*s] + vcount(s+1);
}

int main()
{
  char str[0x100];
  printf("please enter a string\n");
  fgets(str, 100, stdin);
  printf("%d vowels found.\n", vcount(str));
  return 0;
}