What are the signs of crosses initialization?

问题:

Consider following code:

#include <iostream>
using namespace std;

int main()
{
    int x, y, i;
    cin >> x >> y >> i;
    switch(i) {
        case 1:
            // int r = x + y; -- OK
            int r = 1; // Failed to Compile
            cout << r;
            break;
        case 2:
            r = x - y;
            cout << r;
            break;
    };
}

G++ complains crosses initialization of 'int r'.My questions are:

1. What is crosses initialization?
2. Why do the first initializer x + y pass the compilation,but the later failed?
3. What are the problems of so-called crosses initialization?

EDIT:
I know I should use brackets to specify the scope of r but I want to know why,for example why non-POD could not be defined in multi-case switch statement.

Thanks.

回答1:

The version with int r = x + y; won't compile either.

The problem is that it is possible for r to come to scope without its initializer being executed. The code would compile fine if you removed the initializer completely (i.e. the line would read int r;).

The best thing you can do is to limit the scope of the variable. That way you'll satisfy both the compiler and the reader.

switch(i)
{
case 1:
    {
        int r = 1;
        cout << r;
    }
    break;
case 2:
    {
        int r = x - y;
        cout << r;
    }
    break;
};

The Standard says (6.7/3):

It is possible to transfer into a block, but not in a way that bypasses declarations with initialization. A program that jumps from a point where a local variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has POD type (3.9) and is declared without an initializer (8.5).

回答2:

You should put the contents of the case in brackets to give it scope, that way you can declare local variables inside it:

switch(i) {
    case 1:
        {
            // int r = x + y; -- OK
            int r = 1; // Failed to Compile
            cout << r;
        }
        break;
    case 2:
        ...
        break;
};

回答3:

It is possible to transfer into a block, but not in a way that bypasses declarations with initialization. A program that jumps from a point where a local variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has POD type and is declared without an initializer.

[Example: Code:

void f()
{
  // ...
  goto lx;    // ill-formed: jump into scope of `a'
  // ...
 ly:
    X a = 1;
  // ...
 lx:
   goto ly;    // ok, jump implies destructor
 // call for `a' followed by construction
 // again immediately following label ly
}

--end example]

The transfer from the condition of a switch statement to a case label is considered a jump in this respect.

回答4:

I suggest you promote your r variable before the switch statement. If you want to use a variable across the case blocks, (or the same variable name but different usages), define it before the switch statement:

#include <iostream>
using namespace std;

int main()
{
    int x, y, i;
    cin >> x >> y >> i;
// Define the variable before the switch.
    int r;
    switch(i) {
        case 1:
            r = x + y
            cout << r;
            break;
        case 2:
            r = x - y;
            cout << r;
            break;
    };
}

One of the benefits is that the compiler does not have to perform local allocation (a.k.a. pushing onto the stack) in each case block.

A drawback to this approach is when cases "fall" into other cases (i.e. without using break), as the variable will have a previous value.