为什么编译器允许缩小转换

Question

Can anyone please explain to me, why the compiler allows initialize variables of built-in type if the initializer might lead to the loss of information?

For example C++ Primer, the 5th edition says, that The compiler will not let us list initialize variables of built-in type if the initializer might lead to the loss of information.

but my compiler gcc v 4.7.1 initialized variable a in the following code successfully:

long double ld = 3.1415926536; 
int a{ld};

there was just warning: narrowing conversion of 'ld' from 'long double' to 'int' inside { } [-Wnarrowing].

Answer 1

One of the features of initializer lists is that narrowing conversions are not allowed. But the language definition doesn't distinguish between warnings and errors; when code is ill-formed it requires "a diagnostic", which is defined as any message from a set of implementation-defined messages. Warnings satisfy this requirements. That's the mechanism for non-standard extensions: having issued a warning, the compiler is free to do anything it wants to, including compiling something according to implementation-specific rules.

Answer 2

You can set the compiler flag to flag all warnings as error. In that case only it will stop you from doing like that. Otherwise it will only be a warning.

Answer 3

This issue has been coming up lately. With gcc-4.7 a command line switch turns on the required behaviour:

g++ -Werror=narrowing ...