C++ std::max - 需要一个标识符
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std::max - expected an identifier
提问by Jay
I'm having a problem with std::max. I can't figure it out.
我有问题std::max。我想不通。
int border = 35;
int myInt = 2;
int myOtherInt = 3;
int z = std::max(myInt + 2 * border, myOtherInt + 2 * border);
I've included the algorithm standard header. When I mouse over max, I am getting:
我已经包含了算法标准标题。当我将鼠标悬停在 max 上时，我得到：
Error: expected an identifier
错误：需要一个标识符
And a compile errors of:
和一个编译错误：
error C2589:
'(': illegal token on right side of'::'
error C2059: syntax error :'::'
错误 C2589：错误 C2059
'('右侧的非法标记'::'
：语法错误：'::'
What is wrong?
怎么了？
回答by ildjarn
Hazarding a guess, since you're using VC++ – put this before any #includes:
危险的猜测，因为您使用的是 VC++ - 将其放在任何#includes之前：
#define NOMINMAX
windows.hdefines macros named minand maxlike so:
windows.h定义命名的宏min，max如下所示：
#define min(a,b) (((a) < (b)) ? (a) : (b))
#define max(a,b) (((a) > (b)) ? (a) : (b))
The Windows SDK has contained these macros since before C++ was standardized, but because they obviously play havoc with the C++ standard library, one can define the NOMINMAXmacro to prevent them from being defined.
Windows SDK 在 C++ 标准化之前就包含了这些宏，但由于它们显然对 C++ 标准库造成了严重破坏，因此可以定义NOMINMAX宏以防止它们被定义。
As a rule, if you're using C++ (as opposed to C) and including windows.h, alwaysdefine NOMINMAXfirst.
通常，如果您使用 C++（而不是 C）并包含windows.h，请始终NOMINMAX先定义。
回答by Dave S
If you're on VC++, you can either use #define NOMINMAXprior to including any headers, or do (std::max)(myInt + 2 * border, myOtherInt + 2 * border)
如果您使用的是 VC++，则可以#define NOMINMAX在包含任何标头之前使用，或者执行(std::max)(myInt + 2 * border, myOtherInt + 2 * border)
回答by Jim Buck
I would say that either maxis #define's to something else or you need to explicitly invoke the template via std::max<int>.
我会说要么max是#define对别的东西，要么你需要通过std::max<int>.
回答by PeterN
The "using" declaration (see using Declaration) is yet another way to work around the issue:
“using”声明（参见using Declaration）是另一种解决问题的方法：
int border = 35;
int myInt = 2;
int myOtherInt = 3;
using std::max;
int z = max(myInt + 2 * border, myOtherInt + 2 * border);
It allows using std::max without explicit qualification.
它允许在没有明确限定的情况下使用 std::max。
回答by jweyrich
Have you tried using ::std::maxinstead? If this doesn't work, something is messing with your std namespace.
您是否尝试过使用::std::max？如果这不起作用，则说明您的 std 命名空间有问题。
