Question

我是C ++的初学者，我遇到了这段代码的问题，应该在Superbowl决赛中显示分数：

#include <iostream>

enum POINTS { EXTRA_POINT = 1, SAFETY = 2, FIELD_GOAL = 3, TOUCHDOWN =6 };

unsigned short giantsScore = 0, patriotsScore = 0;

int main()
{
std::cout << " Giants: " << giantsScore << "\n";
std::cout << " Patriots: " << patriotsScore << "\n\n";

std::cout << " Giants: " << giantsScore = giantsScore + SAFETY << "\n";
std::cout << " Patriots: " << patriotsScore << "\n\n"; 

std::cout << " Giants: " << giantsScore = giantsScore + TOUCHDOWN + EXTRA_POINT << "\n";
std::cout << " Patriots: " << patriotsScore << "\n\n";

std::cout << " Giants: " << giantsScore << "\n";
std::cout << " Patriots: " << patriotsScore = patriotsScore + FIELD_GOAL << "\n\n";

std::cout << " Giants: " << giantsScore << "\n";
std::cout << " Patriots: " << patriotsScore = patriotsScore + TOUCHDOWN + EXTRA_POINT << "\n\n";

std::cout << " Giants: " << giantsScore << "\n";
std::cout << " Patriots: " << patriotsScore = patriotsScore + TOUCHDOWN + EXTRA_POINT  << "\n\n";

std::cout << " Giants: " << giantsScore = giantsScore + FIELD_GOAL << "\n";
std::cout << " Patriots: " << patriotsScore << "\n\n";

std::cout << " Giants: " <<  giantsScore = giantsScore + FIELD_GOAL << "\n";
std::cout << " Patriots: " << patriotsScore << "\n\n";

std::cout << " Giants: " << giantsScore << "\n";
std::cout << " Patriots: " << patriotsScore = patriotsScore + FIELD_GOAL + EXTRA_POINT  << "\n\n";

return 0;
}

忽略这是非常不优雅的，当我通过编译器，G ++运行时，我收到错误消息

错误：类型'int'和'const char [2]'到二进制'运算符＆lt;＆lt;'
的操作数无效

如果我删除常量并在每个std::cout之前添加它们，那么它运行正常。我只是想知道为什么我不能在每个输出行中添加常量？

Answer 1

您的错误消息指出：int << char，这当然是一个奇怪的操作。

这是因为运营商的优先事项。

每个运算符都具有优先级，这意味着它将在评估其他运算符之前或之后进行评估。

+在=

之前进行评估

和<<应在= cout<<"stuff"成为其原始目的后进行评估。

<<最初是位移运算符（仍然是），因此这就是您遇到这种奇怪行为的原因。添加括号，你会很好。

Answer 2

检查http://cs.smu.ca/~porter/csc/ref/cpp_operators.html以获取运营商优先级规则的概述。当你这样写：

std::cout << " Patriots: " << patriotsScore = patriotsScore + FIELD_GOAL + EXTRA_POINT  << "\n\n";

然后根据优先级规则，首先执行+运算符，为您提供：

std::cout << " Patriots: " << patriotsScore = result  << "\n\n";

然后＆lt;＆lt;运算符被执行，这也意味着`result＆lt;＆lt; “\ n \ n”。但是这个运算符没有在int和char [2]之间定义。

要解决您的问题，请在赋值操作周围加上括号，如下所示：

std::cout << " Patriots: " << (patriotsScore = patriotsScore + FIELD_GOAL + EXTRA_POINT)  << "\n\n";

C ++二进制文件的操作数无效

2 个答案: