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Home/ Questions/Q 8668095
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T18:07:52+00:00 2026-06-12T18:07:52+00:00

Can anyone please explain to me, why the compiler allows initialize variables of built-in

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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].

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  1. Editorial Team
    Editorial Team
    2026-06-12T18:07:53+00:00Added an answer on June 12, 2026 at 6:07 pm

    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.

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