遍历C语言中的数组

Question

My code looks like this:

void main()
{
    int vect[10], i;

    for (i=0; i<5; i++)
        vect[i] = i*2;

    printf("Vector: ");

    for (i=0; i<10; i++)
        printf("%d ", vect[i]);

    printf("\n");

When executing it, it will always show me this kind of output:

To make it just show the elements I entered (the first five; 0 2 4 6 8) I must use a counter or is there a way of telling it to only show me those elements?

Is there a reason why the elements 5, 6, 7 and 9 are always the same but the 8th changes every time? I rewrote the program to change how it shows the elements and it is the same way: it shows the five elements I entered, then three that remain always constant, then one that changes every time the program is executed and then a last constant one. Why is this?

Answer 1

int vect[10] indices 5-9 are not initialised- you need to assign something to them otherwise they will (probably) return garbage, as this is undefined behaviour ( C99 standard, section 5.1.2 "Execution environments" ). You can also define vect as static, ie static int vect[10] , since static variables will be automatically initialised to 0, and a static int array will have all elements automatically initialised to 0.

Answer 2

In your code, vect is an automatic local array which is not initialized explicitly upon definition. So, at that point, all the element values are indeterminate.

To quote C11 , chapter §6.7.9

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

In the first for loop, you only initialize 5 elements, the rest elements are uninitialized and they contains indeterminate values.

Trying to read indeterminate values invoke undefined behavior .

Related, Annex J, same standard, for undefined behaviour

The value of an object with automatic storage duration is used while it is indeterminate.

Once your program exhibits UB, nothing is guaranteed.

FWIW, void main() should at least be int main(void) .