CODEWITHSUNDEEP
cswitch-statement
Talha
Talha

Reputation: 23

Why this switch case goes to the default case?

The code is supposed to check for a vowel input. If found, it would print "Vowel" and if there is none it should print "consonent". But the compiler is going to the default case regardless of the input and I can't find where the mistake is. Please Help.

Here's my code:

#include<stdio.h>

void main()
{
    char ch;
    printf("Insert a Char \n");
    scanf("%d", &ch);

    switch(ch)
    {
    case 'a':
    case 'e':
    case 'i':
    case 'o':
    case 'u':
        printf("Vowel");
        break;

    default:
        printf("Consonent");
    }
}

Upvotes: 0

Views: 83

Answers (4)

Talha
Talha

Reputation: 23

Thanks Everyone. I appreciate the help! I should have noticed the data type though.

include

void main() {

char ch;
printf("Insert a Char \n");
scanf("%c", &ch);

switch(ch)
{
case 'a':
case 'A':

case 'e':
case 'E':

case 'i':
case 'I':

case 'o':
case 'O':

case 'u':
case 'U':
    printf("Vowel");
    break;

default:
    printf("Consonant");
}

}

Upvotes: 0

Vlad from Moscow
Vlad from Moscow

Reputation: 310920

The conversion specifier %d in this call

scanf("%d", &ch);

is invalid. It tries to read a number. So entering a letter like 'A' results in input error.

The compiler can issue a warning or even an error reporting that for example

format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘char *’

Because using an invalid format can result in undefined behavior.

Here is a demonstrative program

#include <stdio.h>

int main(void) 
{
    char s[] = "ABCD";

    puts( s );

    scanf( "%d", ( int * )s );

    puts( s );

    return 0;
}

If for example even to enter a valid ASCII code as for example 65 of the letter 'A' then the program output might look like

ABCD
A

That is the memory occupied by the array was overwritten.

Instead use the following call.(if you want to skip white spaces)

scanf( " %c", &ch );

or the following call

scanf( "%c", &ch );

Pay attention to that according to the C Standard the function main without parameters shall be declared like

int main( void )

Upvotes: 1

Lewis Kelsey
Lewis Kelsey

Reputation: 4677

you should use "%c" as the format. when you use "%d", the value is 0 (because no integers have been detected in the input string). Refer to https://en.cppreference.com/w/c/io/fscanf#Example.

This issue, and void main() instead of int main(), should produce a compiler warning.

Upvotes: 0

Abhay Kumar
Abhay Kumar

Reputation: 343

First thing %d accepts an integer, not a character so correct this and run then place break after each case ends if you don't place break it will run all case till the end which is the default. Example:- 

char ch;
printf("Insert a Char \n");
scanf("%c", &ch);

switch(ch)
{
case 'a':printf("Vowel");
    break;

case 'e':printf("Vowel");
    break;

case 'i':printf("Vowel");
    break;

case 'o':printf("Vowel");
    break;

case 'u':
    printf("Vowel");
    break;

default:
    printf("Consonent");
}

Hope it will help. Happy coding :-)

Upvotes: 0

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