What are the signs of crosses initialization?
C++InitializationC++ Problem Overview
Consider the 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:
- What is
crosses initialization
? - Why do the first initializer
x + y
pass the compilation, but the latter failed? - What are the problems of so-called
crosses initialization
?
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 a multi-case switch statement.
C++ Solutions
Solution 1 - C++
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).
Solution 2 - C++
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;
};
Solution 3 - C++
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.
Solution 4 - C++
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.