#include <stdio.h>
void main()
{
    int n,i;
    int arr[5]={5,4,3,2,1};
    int *ptr;
    printf("input the number you want to find:\n");
    scanf("%d",&n);
    for(i=0;i<5;i  )
        if(arr[i]==n)
        {
            ptr=&arr[i];
        printf("number '%d' is present in the array and it is the %d st/nd/rd/th term in the array.\n its address is: %d",n,i 1,ptr);
        }
}

**>i added else printf("number not found"); here but its also looping and printing a lot **

CodePudding user response：

Keep it simple... just add the print after the loop and add a return when you have a match.

Like:

#include <stdio.h>
int main(void)
{
    int n,i;
    int arr[5]={5,4,3,2,1};
    int *ptr;
    printf("input the number you want to find:\n");
    if (scanf("%d",&n) != 1)
    {
        // Input error
        exit(1);
    }
    for(i=0;i<5;i  )
    {
        if(arr[i]==n)
        {
            ptr=&arr[i];
            printf("number '%d' is present in the array and it is the %d st/nd/rd/th term in the array.\n its address is: %p",n,i 1,(void*)ptr);
            return 0;  // Done... just end the program
        }
    }
    puts("not found");
}

BTW:

Notice that it shall be int main(void)

Notice that pointers are to be printed using %p with a cast to (void*)

CodePudding user response：

1. Always check the return value of the scanf!!
2. Initialize pointer to NULL. If nothing found it will remain NULL
3. Use the correct format for the address (%p), and cast pointer to (void *) before passing to printf
4. main return type is int.

int main()
{
    int n;
    int arr[]={5,4,3,5,2,5,1};
    int *ptr = NULL;
    printf("input the number you want to find:\n");
    if(scanf("%d",&n) == 1)
    {
        for(size_t i=0;i<sizeof(arr)/ sizeof(arr[0]);i  )
            if(arr[i]==n)
            {
                ptr=&arr[i];
                printf("number '%d' is present in the array and it is the %zu st/nd/rd/th term in the array.\n its address is: %p",n,i 1, (void *)ptr);
            }
        if(!ptr) printf("NUMBER NOT FOUND!!!!!\n");
    }
    else
    {
        printf("SCANF ERROR\n");
    }
}

https://godbolt.org/z/6zcE4nnz8

CodePudding user response：

Just put the call of printf outside the for loop. For example

int *ptr = NULL;

printf("input the number you want to find:\n");
scanf("%d",&n);

for( i = 0; ptr == NULL && i < 5; i   )
{
    if ( arr[i] == n )
    {
        ptr = arr   i;
    }
}

if ( ptr != NULL )
{
    printf("number '%d' is present in the array and it is the %td st/nd/rd/th term in the array.\n its address is: %p\n", 
           n, ptr - arr   1, ( void * )ptr );
}
else
{
    printf( "number %d is not present in the array.\n", n );
}

Also do not use magic numbers like 5. Instead use named constants. And declare variables in minimal scopes where they are used. And using the conversion specifier %d with a pointers has undefined behavior. You need to use the conversion specifier %p.

Here is a demonstrative program.

#include <stdio.h>

int main(void) 
{
    int arr[] = { 5, 4, 3, 2, 1 };
    const size_t N = sizeof( arr ) / sizeof( *arr );
    
    int n = 0;
    
    printf( "input the number you want to find (default is %d): ", n );
    scanf( "%d", &n );
    
    int *ptr = NULL;
    
    for ( size_t i = 0; ptr == NULL && i < N; i   )
    {
        if ( arr[i] == n )
        {
            ptr = arr   i;
        }
    }

    if ( ptr != NULL )
    {
        printf( "number '%d' is present in the array and it is the %td st/nd/rd/th term in the array.\n its address is: %p\n", 
                n, ptr - arr   1, ( void * )ptr);
    }
    else
    {
        printf( "number %d is not present in the array.\n", n );
    }
}

The program output might look like

input the number you want to find (default is 0): 3
number '3' is present in the array and it is the 3 st/nd/rd/th term in the array.
 its address is: 0x7ffefecb1778