Formatted string in scanf() in C

Question

Just like printf(), I was trying to use optional specifiers in scanf() format string. I tried to use the width and precision specifier. Now in printf() it simply reserves the columns and print according to these specifiers but what happens with scanf()? Like what is meaning of %3d and %3.3f here? Is this even relevant in case of scanf()? I have a little idea that width in this case represents the number of characters that are to be read for some particular format but not sure. Below code explains this further:

#include<stdio.h>
int main()
{
int a;
float b;
printf("Enter Numbers:\n");
scanf("%3d %3.3f",&a,&b);
printf("Entered Numbers are\n");
printf("%d %f",a,b);
return 0;
}

Answer 1

Since you specified in the comments that what you really want to know is 'what if i forcefully try to do it' ... Here are the results (with Clang)

warning: invalid conversion specifier '.'

and

warning: data argument not used by format string

The program compiles , however, since these are just warnings.

Upon executing the binary, and entering the variables asked for:

1. The "%d" for a gets stored properly.
2. Regardless of what value is entered, the " %3.3f " for b always stores 0.000000

In short, the it does what almost any other code that compiles with warnings does - not behave as intended. This is neither undefined, nor unspecified behaviour, but it is wrong.

Suggestion : Refrain from asking questions that are of the nature ' what happens if I try to compile this '. Just try and see for yourself !