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by bergesenha
2786 days ago
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And in C++, the type of a string literal expression is not const char* , but in fact const char (&)[], ie a reference to an array of chars. I was surprised to learn this as I always assigned a literal to a const char* , relying on pointer decay without knowing it. |
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In C++, it's "array of n const char", or
It's not of a reference type. See the ISO C++11 standard, 2.4.15 [lex.string].(In both C and C++, array expressions "decay" to pointer expressions in most contexts. But for example
is 6 (the size of the string), not the size of a char* pointer.