Thread: Can a scanf function see a number input "1234" as 4 separate numbers?

Originally Posted by laserlight

No, zacs7 just noted that it is wrong, and it is indeed is wrong, even though it is syntactically correct.


I can get your program to crash if I wanted to... again due to the buffer overflow problem that zacs7 mentioned.

If you wanted to describe your suggested algorithm without providing a correct C program as an example, then use pseudocode, or describe your algorithm in prose (or use a flowchart if you can be bothered to do so).

Hey if you want to ! is that a joke ?! becuz its not funny
I ran the program with a number (string) of 18 digits and it works fine (limited by the int data type )at last I think you didn't understand the code
by the way alot of books start i=0
then for (i==0)
so im not taking you as C role models or sth
you cant understand the code thats it

Originally Posted by 8.brahim

I ran the program with a number (string) of 18 digits and it works fine (limited by the int data type )

How did you get 18 characters to fit into a 10-character array?

Ok, my understanding of this has been greatly improved, thank you very much.

Unfortunately the program itself has changed little.

Code:

#include<stdio.h>
#include <windows.h>

int main()
#define MAX 10

     /*  fill the first aray with their space numbers, 
     then print it to make sure*/
{
    char isbn_string[10];
    int a, b, c, x, y, z, isbn_dec[10];

    for (a=0; a < 10; a++)
    {
    isbn_dec[a]=a;
    printf("%d\n",isbn_dec[a]);
    }
    {
        isbn_dec[a]=a;
    } 
        printf("\nPlease enter ISBN: ");
        scanf("%s",isbn_string);
        printf("\nisbn is now the string: %s",isbn_string);
        
        /* So, right now, isbn is not an array of ints yet, its a
        string.  It needs to be copied 1 by 1 into an array of ints*/
  
            for (b=0; b<10; ++b)
        {   
            for (a=0; a<10; ++a)
            for (y=1; y<=10; ++y)

           /* string[b] should start as string[0], which is should be the first
           char in the string.*/
           
            isbn_string[b]= 0 - isbn_dec[a]; 
            isbn_dec[a] = y;  

          /*  on first loop, isbn_dec[0] is assigned to 1*/

            z = isbn_string[b] * isbn_dec[a] ;   
            printf("\n%d", z);
        }
    
    getch();
    return 0;
}

I'm still getting strange numbers back at the end. I think it has something to do with the way I'm handling the string and the string conversion. I'm still new to handling strings. Do i need a "*" in front of it or something here? I'm pretty stumped.

One of the values of indentation is that we can see what you believe your program to be doing, even if it does not match reality. Unfortunately, this:

Code:

            for (b=0; b<10; ++b)
        {   
            for (a=0; a<10; ++a)
            for (y=1; y<=10; ++y)

           /* string[b] should start as string[0], which is should be the first
           char in the string.*/
           
            isbn_string[b]= 0 - isbn_dec[a]; 
            isbn_dec[a] = y;  

          /*  on first loop, isbn_dec[0] is assigned to 1*/

            z = isbn_string[b] * isbn_dec[a] ;   
            printf("\n%d", z);
        }

doesn't give us much help. If you allow your editor/IDE to indent it, you should come up with something like this:

Code:

for (b=0; b<10; ++b)
{   
    for (a=0; a<10; ++a)
        for (y=1; y<=10; ++y)

            /* string[b] should start as string[0], which is should be the first
            char in the string.*/
           
            isbn_string[b]= 0 - isbn_dec[a]; 
    isbn_dec[a] = y;  

    /*  on first loop, isbn_dec[0] is assigned to 1*/

    z = isbn_string[b] * isbn_dec[a] ;   
    printf("\n%d", z);
}

which may give you some insight. (For instance, notice that 2/3 of your usage of the letter a is invalid.) Also I don't know what you intend by this line:

Code:

isbn_string[b]= 0 - isbn_dec[a];

but whatever it is, you're not doing it.

Thanks for the indentation tip, I haven't gotten to the "Indentation" chapter of the tutorial sets yet (and there actually is one, so I can see its importance). This is only my 4th or 5th program. I knew indenting was important, but wasn't sure on exactly how the style should look. So, thanks for the example.

Let me be more clear. I'm really confused about how to get each character is a string, say: "12345" to be assigned to to an integer array. Some on this forum hinted that doing this will involve subracting zero.

Code:

 isbn_string[b]= 0 - isbn_dec[a];

This was one of the many pathetic trial and error attempts I made before realizing I must be missing some key information and knowledge here and came back to the forum. The above code is simply a relic of my experiments (which has been extremely helpful, if odd looking to a veteran)

The entire last section of the program are my attempts to get each character of the string, into its own int, then multiply each char by its placement number in the string. i.e. string: 5467, will have 5*1, 4*2, 6*3 etc.

Code:

#include<stdio.h>
#include <windows.h>

int main()
#define MAX 10

     /*  fill the first aray with their space numbers, 
     then print it to make sure*/
     
{
    char isbn_string[10];
    int a, b, c, x, y, z, isbn_dec[10];

    for (a=0; a < 10; a++)
    {
        isbn_dec[a]=a;
        printf("%d\n",isbn_dec[a]);
    }

    printf("\nPlease enter ISBN: ");
    scanf("%s",isbn_string);
    printf("\nisbn is now the string: %s",isbn_string);
        
    /* So, right now, isbn is not an array of ints yet, its a
    string.  It needs to be copied 1 by 1 into an array of ints*/
    
    for (b=0; b<10; ++b)
    {   
        for (a=0; a<10; ++a)
            for (y=1; y<=10; ++y)
            
                /* Code below will not attempt to attach each string char to 
                an array of ints*/
           
                isbn_string[b]= 0 - isbn_dec[a]; 
                isbn_dec[a] = y;  
            
                /* Code below attemps to multiply the string char, now turned int
                by the its placement number in the string, then printf the result.
                i.e. string: 5467, will have 5*1, 4*2, 6*3 etc */
            
                z = isbn_string[b] * isbn_dec[a] ;   
                printf("\n%d", z);
    }
    
    getch();
    return 0;
}

Thank you for you patience. I know from teaching other subjects that a noob's ignorance and lack of knowledge can make you want to pull your hair out.

You should read around in the forums a bit -- if so, you'll have seen this

Code:

value_of_char = digit - '0';

which does indeed look like subtracting zero.

Wow, I think I actually got it now. I thought it would take me much longer to understand all this, let alone get the code right.

Code:

#include<stdio.h>
#include <windows.h>

int main()
#define MAX 10

     
{
    char isbn_string[10];
    int a, b, x[10], y, z, isbn_dec[10];

    printf("\nPlease enter ISBN: ");
    scanf("%s",isbn_string);
    printf("\nisbn is now the string: %s",isbn_string);
        
    /* So, right now, isbn is not an array of ints yet, its a
    string.  It needs to be copied 1 by 1 into an array of ints*/
    
    for (b=0, y=1; b<9, y<=9; ++b,++y)
    {   
        /* the string char is assigned to an array int*/
        
        isbn_dec[b]= isbn_string[b] - '0'; 
        
        /* the new int is multiplied by y.  y represents the int's 
        digit placement*/
        
        x[b] = isbn_dec[b] * y;  
        printf("\n%d", x[b]);         
    }
    
    z = x[0] + x[1] + x[2] + x[3] + x[4] + x[5] + x[6] + x[7] + x[8];
    printf("\nSum of the first 9 digits: %d", z); 
    
    a = z%11;
    printf("\nRemainder of %d/11 = %d", z, a);
    
    if(a = isbn_dec[10]);
    {
         printf("\nYes, the ISBN is legit!");
    }
         
    getch();
    return 0;
}

Simplification of the code, especially in the loop helped big time.

I still have a question though.

Code:

isbn_dec[b]= isbn_string[b] - '0';

All my online searches in this yield info about both '0' and '/0'. As i understand it, '\0' is a character constant and called nul, and '0' is an integer constant. This sites FAQ is clear about this. And it seems C uses null termination strings, and '/0' signifies the end of a string.

I am confused as to why it subtracting '0' was necessary in the above code though. When I remove that part, the program doesn't work. I don't understand why.

'\0' is the end-of-string marker. '0' is just the character 0. You should look up ASCII.

Say I have a char, and that char is '1'.

Now I want to change that char, into a number. The ascii number for '1' is 49, so that's not what I want.

But all the ascii numbers (and letters), run in a continuous manner: 0 is 48, 1 is 49, etc.

So, if I subtract 48, or '0', from any char which is a digit, I will have changed it into a number - presto, chango!

0, '0' and '\0' have nothing to do with one another, conceptually. First one is a number, second one is a char, and third one is the end of string char.

Nice! bookmarked it! i will check it later. Thanks very much for sharing
__________
kite sale
kite lesson

I found this ..and i used this code to split and convert decimal to hex(change 10->16)

Code:

#include<stdio.h>
void main()
{
  int n  , i ;
  int stk[15] , top = 0;
  printf("Enter the number :");
  scanf("%d" , &n);
  while( n > 0 )
  {
        i = n % 10;
        stk[++top] = i;
        n = n / 10;

  }
  while(top)printf("%d " , stk[top--]);
}