局部变量在C中初始化为零

Question

I thought that the local variable in C is not initialized. But when I compiled this code with gcc.

void f() {
    static int s;
    int n;

    printf("static s = %d\n", s++);
    printf("local  n = %d\n", n++);

    f();
}

main() {
    f();
}

And run this code, the partial result is:

static s = 0
local  n = 0
static s = 1
local  n = 0
static s = 2
local  n = 0
static s = 3
local  n = 0
static s = 4
local  n = 0
static s = 5
local  n = 0
...
static s = 261974
local  n = 0
static s = 261975
local  n = 0
static s = 261976
local  n = 0
static s = 261977
local  n = 0
static s = 261978
local  n = 0
static s = 261979
local  n = 0
static s = 261980
local  n = 0
static s = 261981
local  n = 0
Segmentation fault: 11

Could anyone please explain this? Or refer to a standard reference that C won't init local vars?

Answer 1

ISO/IEC 9899:TC3 WG14/N1256 (C99 standard) section 6.7.8 clause 10:

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.

If an object that has static storage duration is not initialized explicitly, then:

• if it has pointer type, it is initialized to a null pointer;
• if it has arithmetic type, it is initialized to (positive or unsigned) zero;
• if it is an aggregate, every member is initialized (recursively) according to these rules;
• if it is a union, the first named member is initialized (recursively) according to these rules.

Your variable fits the first category. Indeterminate means it can be anything (including 0). Just because it is zero in the tests you have performed, does not mean it will always be or that you can rely on that behaivour. The behaivour might change even with the same compiler depending on the compilation options and optimization level.

Answer 2

In my experience, it might or might not be initialized to 0 depending on the compiler and the flags used during compile.

• Is there a gcc flag to initialise local variable storage?
• Does gcc automatically initialize static variables to zero?

Answer 3

Local non- static variables are not initialized -- which typically means they contain garbage.

0 is just as valid a garbage value as any other. But the initial value could just as easily have been 42 or -12345 .

Just don't do that. You need to ensure that you don't read the value of any variable unless it's been initialized. Reading an uninitialized variable has undefined behavior, which means that the possible behavior is not limited to printing some arbitrary value.

A good way to do to do that is to use an explicit initializer:

int n = 0;

(Incidentally, you're missing the required #include <stdio.h> , and the correct declaration for main is int main(void) .)

Answer 4

What you're writing exhibits undefined behavior (and I assume gets you a scolding from your compiler). The fact that this compiler produced a program with this output today doesn't particularly mean anything. The compiler may just be setting all stack memory to zero. Or the stack may be advancing through previously zeroed-memory. Or the stack is staying exactly where it is (the compiler couldn't "unroll" your main, after all) and the location of n happened to be a zero word. Today.

Answer 5

If the variable is initialized inside a function it is not initialized automatically. When it's declared outside any function, it's initialized with 0.