The aim of this program is to remove all vowels from an input. The function is working fine until I-CodePudding

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(void) {

        char array[1000];
        int x, y, len = 0;
        scanf("%s", array);
        len = strlen(array);
        
    for (x = 0; x < len; x  ) {
    
    if (array[x] == 'a' || array[x] == 'e' || array[x] == 'i' || array[x] == 'o' || array[x] == 'u' || array[x] == 'A' || array[x] == 'E' || array[x] == 'I' || array[x] == 'O' || array[x] == 'U') {

    for (y = x; y < len; y  ) {
        array[y] = array[y 1]; // Moving the non-vowels to a higher position to fill up the array.
    }
    x--; // Deleting those particular x's (The vowels in place).
    len--; // Deleting the length list.
    }
    
    array[len   1] = '\0';
}
    printf("%s", array);
    return 0;
}

The program is working fine until it uses an input that contains a space in the string, what can I do?

For example, "Hello World" prints "Hll" instead of "Hll Wrld".

CodePudding user response：

The scanf("%s", array); takes input till white space character. To take input till newline char can use scanf("%[^\n]", array);

And proper indentation is useful to understand code.

Just after this, the edited code works fine

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(void) {

    char array[1000];
    int x, y, len = 0;
    scanf("%[^\n]", array);
    len = strlen(array);
        
    for (x = 0; x < len; x  ) {
        if (array[x] == 'a' || array[x] == 'e' || array[x] == 'i' || array[x] == 'o' || array[x] == 'u' || array[x] == 'A' || array[x] == 'E' || array[x] == 'I' || array[x] == 'O' || array[x] == 'U') {
            for (y = x; y < len; y  ) {
                array[y] = array[y 1]; // Moving the non-vowels to a higher position to fill up the array.
            }
            x--; // Deleting those particular x's (The vowels in place).
            len--; // Deleting the length list.
        }
        array[len   1] = '\0';
    }
    printf("%s", array);
    return 0;
}

CodePudding user response：

The problem is that by default scanf accepts a space character as an input terminator, you can try using another function like fgets

Or set the conditions for the function using scanf("%[\n]", array);

CodePudding user response：

For starters the code is bad formatted.

And as any bad formatted code it has a bug.

This statement

array[len   1] = '\0';

can write outside the character array if the contained string does not contain a vowel.

For example consider the array

char array[] = "H';

The length of stored string is equal to 1. So this statement

array[len   1] = '\0';

is equivalent to

array[2] = '\0';

while the valid range of indices for this array is [0, 2).

The statement is just redundant and shall be removed because in this for loop

for (y = x; y < len; y  ) {
    array[y] = array[y 1]; // Moving the non-vowels to a higher position to fill up the array.
}

the terminating zero character '\0' is moved to left as it is required.

Also instead of this long if statement

if (array[x] == 'a' || array[x] == 'e' || array[x] == 'i' || array[x] == 'o' || array[x] == 'u' || array[x] == 'A' || array[x] == 'E' || array[x] == 'I' || array[x] == 'O' || array[x] == 'U') {

you could use the standard function strchr as for example

unsigned char c = array[x];
c = tolower( c );

uf ( strchr( "aeiou", c ) != NULL ) {
//...

Pay attention to that it is an inefficient approach to move a while sub-string to the left when a vowel is encountered.

As for you question then the conversion specifier s "matches a sequence of non-white-space characters.". That is as soon as a white space character after a sequence of non-white space characters is encountered that input is interrupted.

You should write

scanf( "