printf 中的“%.#s”是什么意思？

Question

I have got a question in hackerearth to check if there is an error in this printf statement.Can you tell me if there is any meaning for this line below? Here

str="qwertyABC"

printf("%.#s",str)

Answer 1

This statement will invoke undefined behavior because format specification is invalid since # is not defined as a conversion specifier.

To use # as a flag, it must be placed before . , which specifies the precision.

Moreover, using # as a flag, like printf("%#.s",str) will also invoke undefined behavior because the # ("alternative form") flag is not defined for the %s conversion specifier.

Answer 2

There are two errors in your printf statement:

1. The # is not a valid field width or precision specifier.
2. The statement is missing a terminating semicolon.