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C++ 字符串加字符

cout << ("s" + 'a');

這樣一段代碼，為什么輸出是一個(gè)空行呢？
難道是 string literal + character 沒有這個(gè)運(yùn)算嗎？
沒有的話為什么輸出是空行呢？

C++ 中， string literal 就是 const char[n] 嗎？

回答
編輯回答
別逞強(qiáng)

表達(dá)式"s"的類型是const char[]，表達(dá)式 'a'的類型是char。

數(shù)組是不能進(jìn)行內(nèi)置加法運(yùn)算的。所以const char []會(huì)被轉(zhuǎn)換成const char *，這里的運(yùn)算就變成了"指針+整型"(char是一種整型)。輸出空行的運(yùn)行結(jié)果實(shí)際上是數(shù)組越界引起的。

String literal
Narrow multibyte string literal. The type of an unprefixed string literal is const char[]

Additive operators
addition. For the built-in operator, lhs and rhs must be one of the following: Both have arithmetic or unscoped enumeration type. In this case, the usual arithmetic conversions are performed on both operands and determine the type of the result. One is a pointer to complete object type, the other has integral or unscoped enumeration type. In this case, the result type has the type of the pointer

Array-to-pointer decay
There is an implicit conversion from lvalues and rvalues of array type to rvalues of pointer type: it constructs a pointer to the first element of an array. This conversion is used whenever arrays appear in context where arrays are not expected, but pointers are.

2017年11月19日 10:15