Question

Say if I have a string like this

char foo[10] = "%r1%r2";

I want to take out the 1 and 2 and convert them into ints. How can I go about doing this?

Answer 1

if (sscanf(foo, "%%r%d%%r%d", &i1, &i2) != 2)
    ...format error...

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When you do the format for sscanf() I understand %d is a decimal int but why do you have %%r?

If you’re looking for a literal % in the source string, you use %% to specify that in the format string (and in printf(), you use %% in the format string to generate a % in the output); the r represents itself.

There are other ways to specify the conversion, such as %*[^0-9]%d%*[^0-9]%d; that uses assignment suppression (the *) and a scanset ([^0-9], anything that’s not a digit). This information should be obtainable from the manual page for sscanf().