c++ - Narrowing from literal doesn't cause warning -

int = 0; short b{a}; short c{0};

the compiler gives waring short b{a}. can understand this, because int narrowed short.

however, doesn't give warning short c{0}, weird me. remember literal integers, type of 0 should @ least int. narrowing int short happening here. why doesn't compiler give warning?

for short c{0};, narrowing conversion doesn't occur. because 0 constant expression , can stored in short.

(emphasis mine)

list-initialization limits allowed implicit conversions prohibiting following:

• ...

• conversion integer or unscoped enumeration type integer type cannot represent values of original, except source constant expression value can stored in target type

relevant explanations , examples standard, $8.6.4/7 list-initialization [dcl.init.list]:

(emphasis mine)

a narrowing conversion implicit conversion

• ...

• from integer type or unscoped enumeration type integer type cannot represent values of original type, except source constant expression value after integral promotions fit target type.

[ note: indicated above, such conversions not allowed @ top level in list-initializations. — end note ] [ example:

// ... const int z = 99; // ... char c4{z};               // ok: no narrowing needed unsigned char uc1 = {5};  // ok: no narrowing needed // ... float f2 { 7 };           // ok: 7 can represented float // ... 

— end example ]