I accidentally put the opening brace of my function definition after the return statement
我不小心把函数定义的左括号放在return语句之后
int id(int k) return k; { }
But GCC answered with a weird error message
但是GCC用一个奇怪的错误消息来回答
error: named return values are no longer supported
错误:不支持已命名的返回值
Can anyone please explain what that weird feature might be? I've never heard about it.
谁能解释一下这个奇怪的功能是什么吗?我从来没听说过。
3 个解决方案
#1
38
See here - early NRVO implementation by explicit definition of the named return value in the function header.
请参见这里——通过函数头中命名的返回值的显式定义实现的早期NRVO实现。
Native support for NRVO without this extension was added here - GCC 3.1 Release Series.
这里添加了对NRVO的本地支持，但没有扩展——GCC 3.1发布系列。
Brief cut and paste for context:
简单剪切粘贴上下文:
G++ now supports the "named return value optimization": for code like
g++现在支持“命名返回值优化”:对于类似的代码
A f () { A a; ... return a; }G++ will allocate a in the return value slot, so that the return becomes a no-op. For this to work, all return statements in the function must return the same variable.
G+将在返回值槽中分配a，这样返回就变成了no-op。要使其工作，函数中的所有返回语句必须返回相同的变量。
#1
38
See here - early NRVO implementation by explicit definition of the named return value in the function header.
请参见这里——通过函数头中命名的返回值的显式定义实现的早期NRVO实现。
Native support for NRVO without this extension was added here - GCC 3.1 Release Series.
这里添加了对NRVO的本地支持，但没有扩展——GCC 3.1发布系列。
Brief cut and paste for context:
简单剪切粘贴上下文:
G++ now supports the "named return value optimization": for code like
g++现在支持“命名返回值优化”:对于类似的代码
A f () { A a; ... return a; }G++ will allocate a in the return value slot, so that the return becomes a no-op. For this to work, all return statements in the function must return the same variable.
G+将在返回值槽中分配a，这样返回就变成了no-op。要使其工作，函数中的所有返回语句必须返回相同的变量。