使用数组作为在C ++中有效的条件表达式？

Question

I have this code:

int main()
{
    char buffer[10];
    if( buffer ) {
       return 1;
    }
    return 0;
}

which Visual C++ 10 interprets like this: buffer decays to a pointer, then a pointer is compared against null. When this is compiled with /O2 the check gets eliminated and the code gets equivalent to just return 1; .

Is the code above valid? Does Visual C++ compile it right (I mean the decaying part, not the optimization)?

Answer 1

C++11, 6.4/4:

The value of a condition that is an expression is the value of the expression, contextually converted to bool for statements other than switch; if that conversion is ill-formed, the program is ill-formed.

So the standard says that the compiler has to perform any implicit conversions at its disposal to convert the array to a boolean. Decaying the array to pointer and converting the pointer to boolean with a test against against equality to null is one way to do that, so yes the program is well-defined and yes it does produce the correct result -- obviously, since the array is allocated on the stack, the pointer it decays to can never be equal to the null pointer.

Update: As to why this chain of two conversions is followed:

C++11, 4.2/1:

An lvalue or rvalue of type “array of NT” or “array of unknown bound of T” can be converted to a prvalue of type “pointer to T”. The result is a pointer to the first element of the array.

So, the only legal conversion from an array type is to a pointer to element type. There is no choice in the first step.

C++11, 4.12/1:

A prvalue of arithmetic, unscoped enumeration, pointer , or pointer to member type can be converted to a prvalue of type bool . A zero value, null pointer value, or null member pointer value is converted to false ; any other value is converted to true . A prvalue of type std::nullptr_t can be converted to a prvalue of type bool ; the resulting value is false .

There is an implicit conversion directly from bare pointer to boolean; so the compiler picks that as the second step because it allows the desired result (conversion to boolean) to be immediately reached.

Answer 2

Yes, the conversion from an array type to bool is well-defined by the standard conversions. Quoting C++11, 4/1 (with the relevant conversions highlighted):

A standard conversion sequence is a sequence of standard conversions in the following order:

— Zero or one conversion from the following set: lvalue-to-rvalue conversion, array-to-pointer conversion , and function-to-pointer conversion.

— Zero or one conversion from the following set: integral promotions, floating point promotion, integral conversions, floating point conversions, floating-integral conversions, pointer conversions, pointer to member conversions, and boolean conversions .

— Zero or one qualification conversion.

A standard conversion sequence will be applied to an expression if necessary to convert it to a required destination type.

Answer 3

Yes.

if( buffer ) means: check if buffer is not NULL . An array variable points to the start of the array (unless you move it) and is equivalent to a pointer.

The optimization just returns 1 because that buffer is allocated on the stack, so it definitely has a value (pointer to the location on the stack), so it's always true.

Answer 4

You said it yourself :

buffer decays to a pointer

Since the array is on the stack, it can not be NULL (unless something goes wrong, like stack smashing).