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Comments on Counting number of assignments that a `fscanf` format strings implies

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Counting number of assignments that a `fscanf` format strings implies

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I'm writing a function that counts the number of assignments for a fscanf format string. I studied the documentation in C standard 7.21.6.2

It looks like it works. It passes all test cases I have written and yields no warnings with -Wall -Wextra -pedantic -std=c17. While I appreciate design advises, my main concern is if the code is correct or not, so I would be grateful if you found any test case that breaks it. Minor things like variable naming and such is not really interesting.

The function requires the format string to be valid. If not, the behavior is undefined.

#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>

// Counts the number of assignments that should be made by scanf and alike.
//
// Assumes a valid format string. If it's not valid, behavior is undefined.

int count_assignments(const char *fmt) {
    int ret = 0;

    // Note that n is removed, because it suppresses assignments
    static const char specifiers[] = "diouxaefgcspAEFGX";
    static const char singlelength[] = "hljztL";
    static const char doublelength[] = "hl";

    while(*fmt) {
        if(*fmt == '%') { 
            fmt++;

            // Skip width modification
            while(isdigit(*fmt)) fmt++;

            // Check length modification
            if(strchr(singlelength, *fmt)) {
                fmt++;
                if(strchr(doublelength, *fmt)) {
                    fmt++;
                    goto READ_SPECIFIER;
                }
                goto READ_SPECIFIER;
            }
            if(*fmt == '[') {
                while(*fmt != ']') fmt++;
                ret++;
                goto END;
            }
            goto READ_SPECIFIER;
        }

        goto END;
        
    READ_SPECIFIER:
        if(strchr(specifiers, *fmt)) 
            ret++;
    END:
        fmt++;
    }

    return ret;
}

int main(void)
{
    struct test_case {
        const char *fmt;
        const int n;
    } test_cases[] = {
        { "foo", 0 },
        { "%s", 1 },
        { "%d%d", 2 },
        { "%lld", 1 },
        { "%%", 0 },
        { "%d%%%d", 2 },
        { "%2333d%c%33f", 3 },
        { "%[bar]", 1 }
    };

    for(size_t i=0; i<sizeof test_cases/sizeof test_cases[0]; i++) {
        struct test_case tc = test_cases[i];
        printf("%s %d %d\n", tc.fmt, tc.n, count_assignments(tc.fmt));
        assert(count_assignments(tc.fmt) == tc.n);
    }
}
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2 comment threads

Handling `%%` in a `scanf()` format string? (1 comment)
General comments (2 comments)
Handling `%%` in a `scanf()` format string?
Mythical Programmer‭ wrote almost 3 years ago

Does the code handle %%, which means a single % in the input? The POSIX specification for scanf() also provides some options your code does not consider — conversion specifiers S and C, the modifier m, and the %n$… notation which specifies which argument should be converted (so scanf("%2$d %1$d", &x, &y) assigns the first integer to y and the second to x). You can legitimately decide that the POSIX extras are not within remit.