What does an asterisk in a scanf format specifier mean? [duplicate]

Question

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• What is the difference between %*c%c and %c as a format specifier to scanf? 3 answers

So I stumbled across this code and I haven't been able to figure out what the purpose of it is, or how it works:

int word_count;
scanf("%d%*c", &word_count);

My first thought was that %*d was referencing a char pointer or disallowing word_count from taking char variables.

Can someone please shed some light on this?

Answer 1

*c means, that a char will be read but won't be assigned, for example for the input "30a" it will assign 30 to word_count, but 'a' will be ignored.

Answer 2

The * in "%*c" stands for assignment-suppressing character *: If this option is present, the function does not assign the result of the conversion to any receiving argument.¹ So the character will be read but not assigned to any variable.

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_Footnotes:

_{1. fscanf}

Answer 3

To quote the C11 standard, chapter §7.21.6.2, fscanf()

[...] Each conversion specification is introduced by the character %. After the %, the following appear in sequence:

— An optional assignment-suppressing character *.
— [...]
— A conversion specifier character

and regarding the behavior,

[..] Unless assignment suppression was indicated by a *, the result of the conversion is placed in the object pointed to by the first argument following the format argument that has not already received a conversion result. [...]

That means, in case of a format specifier like "%*c", a char will be read from the stdin but the scanned value won't get stored or assigned to anything. So, you don't need to supply a corresponding parameter.

So, in this case,

scanf("%d%*c", &word_count);

is a perfectly valid statement.

For example, What it does in a *nix environment is to clear the input buffer from the newline which is stored due to pressing ENTER key after the input.