- 使用变量编译器说ERROR [英] -Wunused-variable compiler says ERROR

问题描述

最近我开始在C ++中编程（我来自Java，我花了一点点改变哈哈）。在Windows下一切权利。问题是，我切换到Linux，这是我有编译器的问题。它通常是当你声明一个变量，不使用，编译器显示一个警告说，该变量不使用，但我（在linuxmint 15）抛出我为错误，没有编译：C.我做不知道有没有发生这种情况，但我在生产大型程序时（多个类）。

一个例子：

解决方案

未使用的变量警告是使用

$ b调用GCC的结果$ b

  g ++ -Wunused-variable ... 
  

如果是这种情况，请不要指定该参数。或者是因为-Wall：

  g ++ -Wall ... 
  pre> 
 
 
在这种情况下，指定-Wno-unused-variable 
 
 
 
它被抛出为错误，也有'-Werror'参数。
 
 
 
这个警告有几个原因：在C ++的范围之间可以阴影变量，因此是一个常见的原因未使用的变量是当有两个同名的变量时。
  int i = 5; 
 for（int i = 0; i <10; ++ i）{//<这是SECOND变量，称为i隐藏以前的
} 
 if（day ==Monday）{
 int i; //<<你不能看到第二个我在这里，这是第三也隐藏第一。 
 ... 
} 
 // std :: cout< i =< i<< std :: endl; //我们可以看到原来的我再次在这里
  
有两个 我这里。如果你取消注释最后一行，它将打印5，与两个额外的变量[i]无关。
 
 
 
没有std :: cout，虽然，原，外，我从来没有使用。也许最后一个int i是一个错误。
 
 
 
另一个常见的问题是在C ++中拥有全局变量的能力
  int Whoops; // GLOBAL：NEVER EVER TOUCH THIS。 
 
 int func（）{
 int whoops; //本地：总是触摸这个。 
 Whoops = 42; //> W< hoops！ 
} 
  
您会收到一条警告：whoops检测您修改了错误的变量。
 
Recently I'm starting to program in C + + (I come from Java, and it costs me a little change haha). Under Windows everything right. The problem is that I switched to Linux and this is where I have problems with the compiler. It is usually when you declare a variable and is not used, the compiler displays a "warning" saying that the variable is not used, but I (under linuxmint 15) throws me as "error" and leaves no compile: C. I do not know if anyone has this happened, but I'm sick at the time of making large programs (more than one class). 

a little example:
解决方案
The unusued variable warning is the result of invoking GCC with either
g++ -Wunused-variable ...
If this is the case, don't specify that argument. Or it's because of -Wall:
g++ -Wall ...
In which case, specify -Wno-unused-variable

It's being thrown as an error because you have the '-Werror' argument too.

There are a few reasons for this warning: It is possible to "shadow" variables between scopes in C++ and so a common cause of unused variables is when you have two variables of the same name.
int i = 5;
for (int i = 0; i < 10; ++i) { // << this is SECOND variable called i that hides the previous
}
if (day == "Monday") {
    int i; // << you can't see the second i here, this is a third that also hides the first.
    ...
}
// std::cout << "i = " << i << std::endl; // we can see original i again here
There are two variables called "i" here. If you uncommented the last line, it would print 5, unrelated to the two additional variables called 'i'.

Without the std::cout, though, the original, outer, i is never used. Perhaps that last "int i" is an error.

Another common problem relates to the ability to have global variables in C++
int Whoops; // GLOBAL: NEVER EVER TOUCH THIS.

int func() {
    int whoops; // LOCAL: ALWAYS TOUCH THIS.
    Whoops = 42; // >W<hoops!
}
You would receive a warning that "whoops" was an unused variable to aide in detecting you had modified the wrong variable.

                        
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