Why we need to put space before %c? [duplicate]

The following code gives the bizarre o/p as soon as I compile it.

main() {
    char name[3];
    float price[3];
    int pages[3], i;

    printf ( "\nEnter names, prices and no. of pages of 3 books\n" ) ;

    for ( i = 0 ; i <= 2 ; i++ )
        scanf ("%c %f %d", &name[i], &price[i], &pages[i] );

    printf ( "\nAnd this is what you entered\n" ) ;

    for ( i = 0 ; i <= 2 ; i++ )
        printf ( "%c %f %d\n", name[i], price[i], pages[i] );
}

But if we give the space in the scanf statement before %c, it gives proper o/p.

Can anyone please explain me why is it so?

Update:-

If I am providing the input like this-

then my question is why not we are giving space before %f and space before %d? Why we need to give space before %c only?

Solution 1:

Adding the space to the format string enables scanf to consume the newline character from the input that happens everytime you press return. Without the space, name[i] will receive the char '\n', and the real char is left to be misinterpreted by %f.

So, say your input is

a 1.0 2
b 3.0 4
c 5.0 6

The program sees it more like this:

a 1.0 2\nb 3.0 4\nc 5.0 6\n\377

That is, the line-breaks are actual characters in the file (and the \377 here indicates "end of file").

The first scanf will appear to work fine, consuming a char, a float, and an integer. But it leaves the input like this:

\nb 3.0 4\nc 5.0 6\n\377

So the second scanf will read the '\n' as its %c, unless you get rid of it first.

Adding the space to the format string instructs scanf to discard any whitespace characters (any of space ' ', tab '\t', or newline '\n').

A directive is one of the following:

A sequence of white-space characters (space, tab, newline, etc.; see isspace(3)). This directive matches any amount of white space, including none, in the input.

...

from http://linux.die.net/man/3/scanf

This sort of problem arises whenever you use scanf with %c in a loop. Because, assuming free-form input, newlines can happen anywhere. So, it's common to try to avoid the whole issue by using a two-tiered approach. You read lines of input into a buffer (using fgets); chop-off the silly newline characters; then use sscanf instead of scanf to read from the buffer (string) instead of straight from the file.

Solution 2:

Incorrect input using %c

Consider the following snippet of code:

int main( ){
int x;
char y;
scanf("%d", &x);
scanf("%c", &y);
printf("%d %c", x, y);
}

Behavior:
When you run the above program, the first scanf is called which will wait for you to enter a decimal number. Say you enter 12(That’s ASCII ‘1’ and ASCII ‘2’). Then you hit the "enter" key to signal the end of the input. Next the program will execute the second scanf, except you will notice that the program doesn't wait for you to input a character and instead goes right ahead to output 12 followed by a ‘\n’.

Explanation:
Why does that happen? Let’s look at the behavior of the program step-bystep. Initially there is nothing in the buffer. When the first scanf() is called, it has nothing to read, and so it waits. It keeps waiting until you type 1,2, then "enter". Now what's in the buffer are the character 1, the character 2, and the character ‘\n’. Remember that ‘\n’ signifies the end of input, once all fields have been entered, but it is also stored in the buffer as an ASCII character. At this point scanf will read the largest decimal input from the buffer and convert that to an integer. In this example, it finds the string "12" and converts it to the decimal value twelve and puts it in x. Then scanf returns control back to the main function and returns the value 1, for being able to convert one entry successfully. In our example, we do not catch the return value of the scanf in a variable. For robust code, it is important to check the return value of scanf( ) to make sure that the user inputted the correct data.

What is now left in the buffer is ‘\n’. Next, the second scanf is called and it's expecting a character. Since the buffer already has the ‘\n’ character in it, scanf sees that, takes it from the buffer, and puts it in y. That's why when you execute the printf afterwards, 12 and “enter” are printed to the screen.

Solution:
Moral of the story is, enter is a character like any other, and is inputted to the buffer, and consumed from the buffer by %c just like any other ASCII character. To fix this, try using this code instead:

int main( ){
int x;
char y;
scanf("%d", &x);
scanf("\n%c", &y); /* note the \n !! */
printf("%d %c", x, y);
}

How does this fix the problem?

** So you again type ‘1’,’2’,’\n’. The first scanf reads "1" and "2", converts that to the decimal twelve and leaves the ‘\n’ in the buffer. The next scanf now expects a ‘\n’ at the beginning of the next input. It finds the ‘\n’ in the buffer, reads it, and discards it. Now the buffer is empty and scanf is waiting on the user to input a character. Say the user inputs ‘c’, followed by the enter key. ‘c’ is now assigned to y, NOT "enter". Therefore printf will output "12 c" to the screen. NOTE: there is again a ‘\n’ sitting in the queue now. So if you need to do another scanf for a single character, you will have to "consume" that '\n' before taking another character in from the user.

This is not an issue for any other format specifier, as they all ignore white spaces before the input.

Why we need to put space before %c? [duplicate]

Solution 1:

Solution 2:

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