GCC 7, -Wimplicit-fallthrough warnings, and portable way to clear them?

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We are catching warnings from GCC 7 for implicit fall through in a switch statement. Previously, we cleared them under Clang (that's the reason for the comment seen below):

g++ -DNDEBUG -g2 -O3 -std=c++17 -Wall -Wextra -fPIC -c authenc.cpp
asn.cpp: In member function ‘void EncodedObjectFilter::Put(const byte*, size_t)’:
asn.cpp:359:18: warning: this statement may fall through [-Wimplicit-fallthrough=]
    m_state = BODY;  // fall through
                  ^
asn.cpp:361:3: note: here
   case BODY:
   ^~~~

The GCC manual states to use __attribute__ ((fallthrough)), but its not portable. The manual also states "... it is also possible to add a fallthrough comment to silence the warning", but it only offer FALLTHRU (is this really the only choice?):

switch (cond)
  {
  case 1:
    bar (0);
    /* FALLTHRU */
  default:
    …
  }

Is there a portable way to clear the fall through warning for both Clang and GCC? If so, then what is it?

switch-statement gcc warnings c++
6 Answers

C++17 [[fallthrough]]

Example:

int main(int argc, char **argv) {
    switch (argc) {
        case 0:
            argc = 1;
            [[fallthrough]];
        case 1:
            argc = 2;
    };
}

Compile with:

g++ -std=c++17 -Wimplicit-fallthrough main.cpp

If you remove the [[fallthrough]];, GCC warns:

main.cpp: In function ‘int main()’:
main.cpp:5:15: warning: this statement may fall through [-Wimplicit-fallthrough=]
             argc = 1;
             ~~^~~
main.cpp:6:9: note: here
         case 1:
         ^~~~

Also note from the example that the warning only happens if you fall beacross two cases: the last case statement (case 1 here) generates no warnings even though it has no break.

The following constructs don't generate the warning either:

#include <cstdlib>

[[noreturn]] void my_noreturn_func() {
    exit(1);
}

int main(int argc, char **argv) {
    // Erm, an actual break
    switch (argc) {
        case 0:
            argc = 1;
            break;
        case 1:
            argc = 2;
    }

    // Return also works.
    switch (argc) {
        case 0:
            argc = 1;
            return 0;
        case 1:
            argc = 2;
    }

    // noreturn functions are also work.
    // https://stackoverflow.com/questions/10538291/what-is-the-point-of-noreturn/47444782#47444782
    switch (argc) {
        case 0:
            argc = 1;
            my_noreturn_func();
        case 1:
            argc = 2;
    }

    // Empty case synonyms are fine.
    switch (argc) {
        case 0:
        case 1:
            argc = 2;
    }

    // Magic comment mentioned at:
    // https://stackoverflow.com/a/45137452/895245
    switch (argc) {
        case 0:
            argc = 1;
            // fall through
        case 1:
            argc = 2;
    }

    switch (argc) {
        // GCC extension for pre C++17.
        case 0:
            argc = 1;
            __attribute__ ((fallthrough));
        case 1:
            argc = 2;
    }

    switch (argc) {
        // GCC examines all braches.
        case 0:
            if (argv[0][0] == 'm') {
                [[fallthrough]];
            } else {
                return 0;
            }
        case 1:
            argc = 2;
    }
}

We can see from the last one that GCC examines all possible branches, and warns if any of them don't have [[fallthrough]]; or break or return.

You might also want to check for feature availability with macros as in this GEM5 inspired snippet:

#if defined __has_cpp_attribute
    #if __has_cpp_attribute(fallthrough)
        #define MY_FALLTHROUGH [[fallthrough]]
    #else
        #define MY_FALLTHROUGH
    #endif
#else
    #define MY_FALLTHROUGH
#endif

See also: https://en.cppreference.com/w/cpp/language/attributes/fallthrough

Tested on GCC 7.4.0, Ubuntu 18.04.

See also

C version of this question: How to do an explicit fall-through in C

user_8952450

Another example: The Linux kernel provides a fallthrough pseudo-keyword macro. Which can be used as:

switch (cond) {
case 1:
    foo();
    fallthrough;
case 2:
    bar();
    break;
default:
    baz();
}

In kernel v5.10 its implemented like this:

#if __has_attribute(__fallthrough__)
# define fallthrough                    __attribute__((__fallthrough__))
#else
# define fallthrough                    do {} while (0)  /* fallthrough */
#endif
user_20208270

Clean C solution:

int r(int a) {
    switch(a) {
    case 0:
        a += 3;
    case 1:
        a += 2;
    default:
        a += a;
    }
    return a;
}

becomes:

int h(int a) {
    switch(a) {
    case 0:
        a += 3;
        goto one;
    case 1:
    one:
        a += 2;
        goto others;
    default:
    others:
        a += a;
    }
    return a;
}

EDIT: Moved the labels after case statements, as suggested by Stéphane Gourichon in comments, to see the fallthrough more easily.

user_65576210

Nobody mentioned disabling the warning altogether and this might not be the answer the OP was looking for but I think it should be included for completeness since it also works for both compilers:

-Wno-implicit-fallthrough

If for some reason you are not able to change the source code, this keeps the compile output clean allowing for a clear view on other problems (but of course one has to be aware of what one loses).

user_19054910

You can also try:

#pragma GCC diagnostic push
#pragma GCC diagnostic ignored "-Wimplicit-fallthrough"
// your code
#pragma GCC diagnostic pop
user_50933080
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