错误:左值要求为一元&操作数 [英] error: lvalue required as unary & operand

问题描述

在编译时(使用Linux Server 6.1版)

While compiling (with Linux Server release 6.1)

strftime(AppTime, sizeof(AppTime),"%Y/%m/%d %T", localtime(&((long)u32_Time)));

获取错误错误:必须为一元'&'的左值操作数"

getting error "error: lvalue required as unary '&' operand"

，但相同的代码已在Red Hat Enterprise Linux AS发行版3中成功编译.

but the same code compiled successfully with Red Hat Enterprise Linux AS release 3.

为什么呢?该如何纠正?

Why so? How to correct this?

推荐答案

地址操作符& 需要一个变量来获取地址.强制转换(long)u32_Time 的结果是临时的，不一定驻留在内存中，因此没有可以使用的地址.因此，如果那段代码曾经在某个地方编译过，那么它就是一个非标准的编译器扩展.

The address-operator & requires a variable to take the address from. The result of your cast (long)u32_Time is a temporary that does not necessarily reside in memory and therefore has no address that could be taken. So if that piece of code ever compiled somewhere it was a nonstandard compiler extension.

该标准§5.3.1,3要求:

The standard, §5.3.1,3 demands:

一元&的结果运算符是指向其操作数的指针.操作数应为左值[...]

The result of the unary & operator is a pointer to its operand. The operand shall be an lvalue [...]

如何解决此问题: std :: localtime 需要一个指向 std :: time_t 的指针，因此您最好提供它.您没有提供任何解释或进一步的代码，因此我只能猜测 u32_Time 是某种4字节无符号算术类型，应该以某种方式表示时间.如何将其正确转换为 std :: time_t 取决于您的编译器如何实现后者以及如何获得进一步的价值.简单地应用C-cast是 not 可移植的，而强制转换为 long 则更不易移植.
当且仅当当前平台上的 std :: time_t 也是无符号的32位类型，并且使用与相同的表示形式> u32_Time ，使用

How to fix this: std::localtime expects a pointer to a std::time_t so you best provide that. You did not provide any explanation or further code, so I can only guess that u32_Time is some 4 byte unsigned arithmetic type that is supposed to represent a time in some manner. How that is properly converted into a std::time_t depends on how your compiler implements the latter and how you got the value of the further. Simply applying a C-cast is not portable, and casting to long is even less portable.
If, and only if the std::time_t on your current platform is also a unsigned 32 bit type using the same representation as your u32_Time, it might suffice to use

 localtime(reinterpret_cast<std::time_t*>(&u32_Time));

更多可移植的设备会首先将值存储在正确的数据类型中:

More portable would be storing the value in the correct data type first:

 std::time_t time = u32_Time;
 localtime(&time);

这样，如果 time_t 和 u32_Time 的类型不兼容，您将得到必要的警告和/或错误.

That way you will get the necessary warnings and/or errors if time_t and the type of u32_Time are not compatible.

我强烈建议您不要使用C-cast，因为一旦您不得不将这段代码移植到另一个平台上，您将无法轻松找到这种讨厌的演员.

I would strongly advice against using C-casts, because once you have to port this piece of code to another platform you will have no means to find that nasty cast easily.

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