GCC -Wuninitialized / -Wmaybe-uninitialized issues

Question

I am experiencing a very strange issue using gcc-4.7 (Ubuntu/Linaro 4.7.2-11precise2) 4.7.2. I am unable to compile the following valid code without a warning:

extern void dostuff(void);

int test(int arg1, int arg2)
{
    int ret;

    if (arg1) ret = arg2 ? 1 : 2;

    dostuff();

    if (arg1) return ret;

    return 0;
}

Compile options and output:

$ gcc-4.7 -o test.o -c -Os test.c -Wall
test.c: In function ‘test’:
test.c:5:6: warning: ‘ret’ may be used uninitialized in this function [-Wmaybe-uninitialized]

However, the following code compiles with no warning (albeit to slightly less efficient assembly):

extern void dostuff(void);

int test(int arg1, int arg2)
{
    int ret;

    if (arg1 && arg2) ret = 1;
    if (arg1 && !arg2) ret = 2;

    dostuff();

    if (arg1) return ret;

    return 0;
}

I am somewhat stuck and am considering this a compiler bug. Any thoughts?

No, the syntax is fine. I mean return 0 if `arg1=0, arg2=0`, return 1 if `arg1=1, arg2=1`, return 2 if `arg1=1, arg2=0`. This snippet is a simplified case of a much larger issue I'm having. — user593062, Jan 03 '13 at 04:14
Making ret `volatile` also resolves the issue, but isn't really ideal. — user593062, Jan 03 '13 at 04:15
`int ret;` does not mean `ret == 0`, If you change `int ret = 0;` that should solve your issue and fit correctly in your logic as well. The code what you have at present, doesn't initialize `ret` to `0`.`ret` has an Indeterminate value since it is local/auto variable which is not explicitly initialized.But that doesn't answer the underlying anomaly. — Alok Save, Jan 03 '13 at 04:17
Thanks Alok, but possibly this constructed example will produce misguided solutions. Code density is extremely important here, and in the actual function `ret` is a large array which I don't want to initialize if it is not used. Looking at my first program, ret is indeed never used uninitialized, so the warning is incorrect, no? — user593062, Jan 03 '13 at 04:23
Yes, seems to be problem with gcc. A little bit of research and I could find it.I added an answer.Hth. — Alok Save, Jan 03 '13 at 04:32
@AlokSave The code above will either initialize and return `ret` or it will not initialize but then also never use `ret` at all. Your objection is incorrect, the code as shown above will for sure never access `ret` before it is initialized, an explicit zero initializing is not needed, and if the compiler claims anything else, it's broken. — Mecki, Dec 24 '20 at 20:41

Alok Save · Accepted Answer · 2013-01-03T04:37:07.233

Indeed this is a known problem in gcc.
gcc is notorious for reporting incorrect uninitialized variables.
The shortcomings have been duly noted and there is a initiative to overcome the shortcomings:
Better Uninitialized Warnings:

The GNU Compiler Collection warns about the use of uninitialized variables with the option -Wuninitialized. However, the current implementation has some perceived shortcomings. On one hand, some users would like more verbose and consistent warnings. On the other hand, some users would like to get as few warnings as possible. The goal of this project is to implement both possibilities while at the same time improving the current capabilities.

The initiative aims at providing better warnings and it quotes a example case similar as your case. The relevant portion being:

What an user understands as a false positive may be different for the particular user. Some users are interested in cases that are hidden because of actions of the optimizers combined with the current environment. However, many users aren't, since that case is hidden because it cannot arise in the compiled code. The canonical example is

int x;
if (f ())
     x = 3;
return x;

where 'f' always return non-zero for the current environment, and thus, it may be optimized away. Here, a group of users would like to get an uninitialized warning since 'f' may return zero when compiled elsewhere. Yet, other group of users would consider spurious a warning about a situation that cannot arise in the executable being compiled.

On one hand, some users would like `-pedantic` and `-Werror`. So these users have to turn off `Wuninitialized` for gcc 4, 5, 6, 7. It is still broken in 2018. — puchu, Jan 20 '19 at 19:21

score 2 · Answer 2 · answered Dec 24 '20 at 20:54

Not sure if gcc has been fixed in the meantime. If not, you may want to try clang. It's the far better compiler IMHO and it does much better code analyzes.

Just because some comments claim the compiler is right, ret may be used uninitialized, here's the proof of the contrary. The code

int test(int arg1, int arg2)
{
    int ret;
    if (arg1) ret = arg2 ? 1 : 2;
    dostuff();
    if (arg1) return ret;
    return 0;
}

can easily be transformed to the following code by just combining the two identical if statements into one:

int test(int arg1, int arg2)
{
    if (arg1) {
        int ret = arg2 ? 1 : 2;
        dostuff();
        return ret;
    }
    dostuff();
    return 0;
}

This is equivalent code and now it should be obvious, that ret can never be used uninitialized. The compiler is wrong, the warning is pointless.

But then again, the code can be further simplified:

int test(int arg1, int arg2)
{
    dostuff();
    return (arg1 ? (arg2 ? 1 : 2) : 0);
}

Problem solved, ret is gone.

GCC -Wuninitialized / -Wmaybe-uninitialized issues

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