Does sscanf touch pointers if no match has been found? [duplicate]

Question

If a line does not match a [fsv]scanf format, does scanf guarantee not to touch the provided pointers that are not matched?

For example, if

int int1 = 3;
int int2 = 5;
sscanf(line, "%d %d", &int1, &int2);

returns 0, are the integers guaranteed to be still 3 and 5, or can int1 have been changed?

Answer 1

The short answer is yes, in your case you can guarantee that int1 and int2 have not changed.

However, I would advise against relying on this behaviour, as it's likely to produce code that is difficult to read - and because:

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The long answer is it depends on your format string. Looking at the C11 standard for fscanf (s7.21.6.2.16), we have:

The fscanf function returns the value of the macro EOF if an input failure occurs before the first conversion (if any) has completed. Otherwise, the function returns the number of input items assigned, which can be fewer than provided for, or even zero, in the event of an early matching failure.

Critically important is this definition of input items from later in 7.21.6.2:

An input item is defined as the longest sequence of input characters which does not exceed any specified field width and which is, or is a prefix of, a matching input sequence

So. The number returned by scanf is the number of items read from the stream, not the number of pointers written to.

Additionally relevant is 7.21.6.2.2:

If the format is exhausted while arguments remain, the excess arguments are evaluated (as always) but are otherwise ignored.

The behaviour of ignoring arguments that aren't written to is also made explicit in an example at the end of that section:

In:

    #include <stdio.h>
    /* ... */
    int d1, d2, n1, n2, i;
    i = sscanf("123", "%d%n%n%d", &d1, &n1, &n2, &d2);

the value 123 is assigned to d1 and the value 3 to n1. Because %n can never get an input failure the value of 3 is also assigned to n2. The value of d2 is not affected. The value 1 is assigned to i.

In case you're not familiar with %n, it's "the number of characters read from the stream so far".

This is a great example to illustrate your question - here we have three pointers written to, and one pointer untouched. But, fscanf only returns 1 here - because it only assigned one "input item" from the stream.

So, in your example, yes, if you've got %d %d and you pass it something which causes 0 reads, then yes, the pointers will be untouched.

But, if you've got a %n in there, then your function could still return 0 or EOF while still consuming some input and writing to pointers. For example:

sscanf("aaa","aaa%n%d",&n1,&n2);

This writes 3 to n1, leaves n2 untouched, and returns EOF. And:

sscanf("aaa bbb","aaa%n%d",&n1,&n2);

This writes 3 to n1, leaves n2 untouched, and returns 0.