why does the variable become 0 suddenly?(c language)

i was practicing basic so i write a code to ask about user name and then i wanted to print user name character by character , then the whole name and the length of it and at the end the even numbers characters of the user name like the 2nd,4th and so on
i got two problems first the variable holding name length become zero at the last for cycle
my second problem was the even characters as i get odd ones too

i entered the user name : james
i expected the out put to be :

j
5
a 
5
m
5
e
5
s
5
hello james , your name has 5 characters
ae

#include <stdio.h>
#include <cs50.h>
#include <string.h>

int main(void)
{
    string name = get_string("whats your name ? ");
    int x = strlen(name);
    char new[4] = "";
    for (int i = 0 ; i < x ; i++)
    {
        printf("%c\n" ,name[i] );
        if ( (i+1) % 2  == 0)
        {
            strcat(new,&name[i]);
        }
            printf("%i\n",x);
    }

    printf("hello %s ,your name has %i characters \n", name,x);
    printf("%s\n",new);
}

whats your name ? james
j
5
a
0
hello james ,your name is 0 characters long
ames

The main issue is that the new buffer is not large enough to hold the data you are writing to it. I suspect the primary confusion is the incorrect usage of strcat. When you write strcat(new,&name[i]);, it does not copy a single character to the end of new. If name is "james" and i is 2, then &name[i] is the address of the 'm' and strcat attempts to copy the string "mes" to the end of new. If you just want to copy a single character, strcat is the wrong tool. IMO, this confusion is exacerbated by the use of the string typedef, whose only purpose is to confuse students. Stop using string and use char * instead. You must understand this detail if you are ever to understand C.

If you want to iterate through the string and copy every other character into an array, perhaps you want something like below. Note that you must check the array bounds if you are using unknown input.

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

int
main(int argc, char **argv)
{
	char *name = argc > 1 ? argv[1] : "james";
	int length = strlen(name);
	char new[128] = "";
	char *d = new;
	/*
	Checking the parity of i inside the loop is a code smell, but
	we leave this as dead code for comparison to the original.  Simply
	skipping the unwanted indices is done below the dead code.
	for( int i = 0; i < length && i < sizeof new - 1; i += 1 ){
		if( (i+1) % 2 == 0 ){
			*d++ = name[i];
		}
	}
	*/
	for( int i = 1; i < length && i < sizeof new - 1; i += 2 ){
		*d++ = name[i];
	}
	*d = '
#include <stdio.h>
#include <string.h>
int
main(int argc, char **argv)
{
char *name = argc > 1 ? argv[1] : "james";
int length = strlen(name);
char new[128] = "";
char *d = new;
/*
Checking the parity of i inside the loop is a code smell, but
we leave this as dead code for comparison to the original.  Simply
skipping the unwanted indices is done below the dead code.
for( int i = 0; i < length && i < sizeof new - 1; i += 1 ){
if( (i+1) % 2 == 0 ){
*d++ = name[i];
}
}
*/
for( int i = 1; i < length && i < sizeof new - 1; i += 2 ){
*d++ = name[i];
}
*d = '\0';
printf("hello %s, your name has %i characters \n", name, length);
printf("%s\n", new);
return 0;
}
';
	printf("hello %s, your name has %i characters \n", name, length);
	printf("%s\n", new);
	return 0;
}

Many would think it more natural to write the loop as something like:

char *end = new + min(length, sizeof new);
for( name += 1; d < end; name += 2 ){                                      
        *d++ = *name;                                                      
} 

and get rid of all of the indexing. Note, this requires a definition of min to compute the minimum, and it loses track of the beginning of the original name, so you would need to keep track of the start of the name or use a different variable for the loop. Instead of indexing, you just keep track of three pointers; one to the character you want to copy, one to the destination to which you want to copy it, and one to the end of the destination buffer so you don't go past it.

Your entered a string "james" that contains 6 characters including the terminating zero character '\0'.

whats your name ? james

And you declared a character array that has only 4 elements.

char new[4] = "";

In this if statement

    if ( (i+1) % 2  == 0)
    {
        strcat(new,&name[i]);
    }

when for example i is equal to 1 the string pointed to by the pointer expression name + 1 is appended to the character array new.

        strcat(new,&name[i]);

This is the following string (represented as a character array) { 'a', 'm', 'e', 's', '\0' }. As it is seen the string contains 5 characters including the terminating zero character '\0'. As a result this call of strcat results in overwritting memory outside the array new that invokes undefined behavior.

There is no need to use the function strcat because instead of storing only one character in the array new the function tries to append a whole substring.

You could write for example

for (int i = 0 ; i < x ; i++)
{
    printf("%c\n" ,name[i] );
    if ( (i+1) % 2  == 0)
    {
        new[i / 2] = name[i];
    }
        printf("%i\n",x);
}

Pay attention to that the array new will contain a string because all its elements were initially initialized by zeroes in the array declaration

char new[4] = "";

On the other hand, using the magic number 4 in the array declaration is unsafe. Instead you could declare for example a variable length array like

int x = strlen(name);
char new[x / 2 + 1];
memset( new, 0, x / 2 + 1 );

In this case your program will work even if the user will enter a more long name.