What does a # sign after a % sign in a scanf() function mean?

Question

What does the following code mean,in C

scanf("%d%#d%d",&a,&b,&c);

if given values 1 2 3 it gives output as 1 0 0

P.S- I know it is used with printf() statement but here in scanf() statement it gives random behaviour.

Answer 1

TL;DR; - A # after a % sign in the format string of scanf() function is wrong code.

Explanation:

The # here is a flag character, which is allowed in fprintf() and family, not in fscanf() and family.

In case of your code, the presence of # after % is treated as invalid conversion specifier. As per 7.21.6.2,

If a conversion specification is invalid, the behavior is undefined

So, your code produces undefined behaviour.

Hint: you can check the return value of scanf() to check how many elements were "scanned" successfully.

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However, FWIW, using # with %d in printf() also is undefined behaviour.

Just for reference: As per the C11 standard document , chapter §7.21.6.1, flag characters part, (emphasis mine)

#

The result is converted to an ‘‘alternative form’’. For o conversion, it increases the precision, if and only if necessary, to force the first digit of the result to be a zero (if the value and precision are both 0, a single 0 is printed). For x (or X) conversion, a nonzero result has 0x (or 0X) prefixed to it. For a, A, e, E, f, F, g, and G conversions, the result of converting a floating-point number always contains a decimal-point character, even if no digits follow it. (Normally, a decimal-point character appears in the result of these conversions only if a digit follows it.) For g and G conversions, trailing zeros are not removed from the result. For other conversions, the behavior is undefined.

Answer 2

According to the Standard, the use of # is illegal.

Its use makes your program invoke Undefined Behaviour.

Of course, if your implementation defines it, it is defined behaviour for your implementation and it does what your documentation says.