Declaration difference and scope?

Question

In question Declaration difference? it was asked what the difference is between

int i;

for (i=0; i<100; i++) {
    //some loop....
}

and:

for (int i=0; i<100; i++) {
    //some loop....
}

The answers are clear; the 2nd is C99 and the scope of i is limited to the loop. I don't have C99 so I can't test and hence ask it as a question: what would be the resolution in the following case:

int i = 32;

for (int i=i; i<100; i++) {
    // some loop
}

Would the "new" i be initialized with the "old" i? Or would the old i already be inaccessible becaus a new i has already been declared?

Answer 1

In this for loop statement

int i = 32;

for (int i = i; i < 100; i++) {
    // some loop
}

variable i declared in the for statement has indeterminate value. The problem is that as soon as a declarator is defined (in this case it consists from identifier i) it hides an entity with the same name in the given scope. So in this declaration

int i = i;

variable i refers to itself in the right side of the = .

Another similar example. Let's assume that you have a typedef.

typedef int Int;

You may write after its definition

Int Int;

In this case the name Int of the object of type Int hides the typedef definition and you may not already write

Int Another_Int;

because the compiler will issue an error.

According to the C Standard (6.2.1 Scopes of identifiers)

4 ...If the declarator or type specifier that declares the identifier appears inside a block or within the list of parameter declarations in a function definition, the identifier has block scope, which terminates at the end of the associated block.

A more clear is written in the C++ Standard (3.3.2 Point of declaration)

1 The point of declaration for a name is immediately after its complete declarator (Clause 8) and before its initializer (if any), except as noted below. [ Example:

int x = 12;
{ int x = x; }

Here the second x is initialized with its own (indeterminate) value. —end example ]

Take into account that in this code snippet

int i = 10;

{
    int i[i];
}

inside the compound statement there is declared array int i[10]; that is outer variable i is used as the array size because inner variable i will be declared only when its declarator will be completed.

Answer 2

See C11 6.8.5.3: "If clause-1 is a declaration, the scope of any identifiers it declares is the remainder of the declaration and the entire loop, including the other two expressions".

The 2nd i refers to the i being defined, not the old one.

The whole thing is UB because you are using the value of i (the i being defined inside the loop) without a previous assignment (or initialization).

---

Edit with working (but different) example

You can still use the old value through the use of a pointer

int i = 42;
int *old_i = &i;
for (int i = *old_i; i < 50; i++) printf("%d ", i);