Does For-Loop counter stay?

Simple Question. Imagine this in ANSI-C:

int i;

for(i=0 ; i<5 ; i++){
   //Something...
}

printf("i is %d\n", i);

Will thi开发者_JS百科s output "i is 5" ?

Is i preserved or is the value of i undefined after the loop?

Yes. If i is declared outside of the for loop it remains in scope after the loop exits. It retains whatever value it had at the point the loop exited.

If you declatred I in the loop:

for (int i = 0 ; i < 5 ; i++)
{

}

Then i is undefined after the loop exit.

Variable i is defined outside of the scope of the loop (which is great, or you wouldn't be able to print it in that case).

And it is post-icnremented for every-turn of the loop, for which the end condition is "stop when i is bigger or equal to 5".

So it really makes perfect sense for i to be equal to 5 at this point.

A block scope is not exactly the same as a function scope in C. The variable i doesn't "get back" magically to its previous value when you step out of the loop's scope.

i's value will be 5 after your loop. Unless you did something like

i = 50000;

inside of it.

It's also generally recommended against using "i" after you exit the loop in most coding standards I have ever read. In particular do NOT do:

for(i = 0; i < num_elements; i++)
{
    if(element[i].id == id)
    {
        /* Do something to element here. */
        break;
    }
}

if(i == num_elements)
{
    fprintf(stderr, "Failed to find element %d.", id);
    succeeded == false;
}

While this will work it is poor coding. It is less readable and maintainable than the alternatives. E.g.

succeeded = false;

for(i = 0; i < num_elements; i++)
{
    if(element[i].id == id)
    {
        /* Do something to element here. */
        succeeded = true;
        break;
    }
}

if(false == succeeded)
{
    fprintf(stderr, "Failed to find element %d.", id);
}

Yes, variables are valid only inside the block in which they are declared. Here's an example:

#include <stdio.h>

void main(int argc, char *argv[])
{
    if(argc == 2) {
        int x;
        x = 7;
    }

    x = 1;
}

That's the compiler:

gcc ex.c
ex.c: In function ‘main’:
ex.c:10: error: ‘x’ undeclared (first use in this function)
ex.c:10: error: (Each undeclared identifier is reported only once
ex.c:10: error: for each function it appears in.)