为什么这个非静态变量通过函数调用保留其值？

Question

void testFunc(int);

int main(int argc, char** argv) {

    testFunc(1);
    testFunc(2);
    testFunc(3);

    return (EXIT_SUCCESS);
}

void testFunc(int another)
{
    int num;
    printf("num: %i\n", num);
    num = another;
}

output: num: 127383283 num: 1 num: 2

If I am printing the variable before I assign it to something each time, shouldn't I always get garbage values without a static keyword?

Answer 1

You are getting garbage values - it just so happens that in this case those garbage values happen to be the value that you assigned in the previous invocation of the function.

If you call another function in between the calls to testFunc() , or compile with higher optimisation settings, or with a different compiler, you'll probably see something completely different.

Answer 2

Because it ends up using the same memory each time for that variable, which hasn't changed.

This behavior is considered undefined. Although it may work reliably in your particular case it is not to be relied on in all cases.

And why don't you start accepting some answers for hell's sake?

Answer 3

The value of num without initialization is Indeterminate .
It can be anything. You cannot rely on it to be anything specific.
Note that using this indeterminate value results in Undefined Behavior , So this program has Undefined behavior.

Answer 4

In C/C++, your automatic variables are not guaranteed to have any particular value unless you initialize them to value you want. So this is by design

Answer 5

It's an accident. Add another function:

int use_stack(void)
{
    int a[4] = { rand(), rand(), rand(), rand() };
    int j = 0;
    for (int i = 0; i < 4; i++)
        j += a[i];
    return j;
}

Call it after the first two testFunc() calls. You're likely to see a random value as the 'saved value' in testFunc() .

Formally, the behaviour you are seeing is undefined behaviour and anything may happen.

Answer 6

Static just means that it can't change after compile time. Once you declare the value, it will remain that through the life of the code. It works the second and thirds times through since you are setting the value at the end of the function.

If you set it to static, you would not be able to assign the value to is in the function.