why does this not print the string?

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

int main(){
  
  const char *adj[] = {"swag","nice","asparagus","ugly"};

  for (int i=0;i<sizeof(adj); i++)
    printf("you are extremely %s\n", *adj[i]);

}

i'm actually new to C and i'm trying test programs to learn, in this case i was hoping for it to print the name a few times with all the adjectives, but it doesn't when using %s in the last print, if i use %c it will only print the first letter of the string

tried swapping %c and %s, but neither work, %c prints only the 1st letter, %s prints nothing

For starters sizeof( adj ) is equal to 4 * sizeof( char * ) where 4 is the number of initializers. If sizeof( char * ) is equal to for example 8 then the for loop looks in fact like

for (int i=0;i < 32; i++)

while the array contains only 4 elements.

To get the number of elements in the array you need to use the expression sizeof( adj ) / sizeof( *adj ) that is the same as sizeof( adj ) / sizeof( char * ).

And as the expression has the type size_t it is much better to declare the variable i also as having the type size_t

for ( size_t i = 0; i < sizeof( adj ) / sizeof( *adj ); i++ )

And the argument of the call of printf shall be a pointer to a string while the expression *adj[i] has the type const char.

That is you need to write

for ( size_t i = 0; i < sizeof( adj ) / sizeof( *adj ); i++ )
{
    printf( "you are extremely %s\n", adj[i] );
}

Here is the correct way to do it

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

int main(){
  
  const char *adj[] = {"swag","nice","asparagus","ugly"};

  for (int i=0;i<sizeof(adj)/sizeof(*adj); i++)
    printf("you are extremely %s\n", adj[i]);
}

outputs

you are extremely swag
you are extremely nice
you are extremely asparagus
you are extremely ugly

sizeof(adj) returns the total size of the pointers. There are 4 pointers each is of 32bit or 64bit, so the number of elements have to be obtained by sizeof(adj)/sizeof(*adj) i.e., total size in bytes/ size of one pointer.

The second error in your code is in the printf, you have to print the string pointed by the char pointer not the single char.

The expression

sizeof(adj)

or simply

sizeof adj

will give you the size of the array in bytes. The size of the array in bytes is not necessarily identical to the number of elements in the array. If the size of an element of the array is larger than a single byte, then the size of the array in bytes will be larger than the number of elements in the array.

Therefore, in order to determine the number of elements in an array, you must divide the total size of the array by the size of a single element of the array. You can do this using the following expression:

sizeof adj / sizeof *adj

For the reasons stated above, you should change the line

for (int i=0;i<sizeof(adj); i++)

to:

for ( int i=0; i < sizeof adj / sizeof *adj; i++ )

Another issue is that the line

printf("you are extremely %s\n", *adj[i]);

is wrong. The %s printf format specifier requires that you pass as a function argument a char * that points to the null-terminated string that is to be printed. However, that is not what you are doing. You are instead passing as the function argument the first character that is to be printed, which is of type char. Because you are passing a char instead of a char * as a function argument to printf, your program has undefined behavior.

For this reason, you should remove the *, so that the line looks like this:

printf( "you are extremely %s\n", adj[i] );