Possible Duplicates: main(int argc, char *argv[]) Main's Signature in C++ If i write: int main(int argc, char** argv) i get proper commandline input. What would happen if i wrote say, int main(int foo, double fooDouble, fooType FooVar) Is this OS-dependent or does it depend on the compiler? Given that it does compile, it will still only be called with the argc and argv arguments.
可能重复： main（int argc，char * argv []） Main's Signature in C ++ 如果我写： int main(int argc, char** argv) 我得到正确的命令行输入。 如果我写的话会发生什么， int main(int foo, double fooDouble, fooType FooVar) 这是依赖操作系统还是依赖于编译器？ 鉴于它编译，它仍然只能用argc和argv参数调用。 所以你的fooDouble将得到argv的指针值，并且FooVar将获得用于该参数位置的寄存器/堆栈空间
I have a bunch of code where objects of type std::string are compared for equality against string literals. Something like this: //const std:string someString = //blahblahblah; if( someString == "(" ) { //do something } else if( someString == ")" ) { //do something else } else if// this chain can be very long The comparison time accumulates to a serious amount (yes, I profiled) and so it
我有一堆代码，其中std::string类型的对象与字符串文字进行比较。 像这样的东西： //const std:string someString = //blahblahblah; if( someString == "(" ) { //do something } else if( someString == ")" ) { //do something else } else if// this chain can be very long 比较时间累积到一个很大的数量（是的，我简介），所以它会很快加快。 该代码将字符串与许多短字符串进行比较，这种比较难以避免。 留下
So, I'm wanting to get a better grasp on how string literals in C++ work. I'm mostly concerned with situations where you're assigning the address of a string literal to a pointer, and passing it around. For example: char* advice = "Don't stick your hands in the toaster."; Now lets say I just pass this string around by copying pointers for the duration of the program. Sure, it'
所以，我想更好地掌握C ++中的字符串文字是如何工作的。 我主要关心的是你将一个字符串文字的地址赋给一个指针，然后传递它。 例如： char* advice = "Don't stick your hands in the toaster."; 现在我们可以说我只是通过在程序期间复制指针来传递这个字符串。 当然，这可能不是一个好主意，但我很好奇幕后会发生什么。 再举一个例子，假设我们创建一个返回字符串文字的函数： char* foo() { // function does does
This question already has an answer here: “#include” a text file in a C program as a char[] 15 answers The C/C++ preprocessor acts in units of tokens, and a string literal is a single token. As such, you can't intervene in the middle of a string literal like that. You could preprocess script.py into something like: "some coden" "some more code that will be appendedn" and #include tha
这个问题在这里已经有了答案： “#include”一个C程序中的文本文件为char [] 15个答案 C / C ++预处理器以令牌为单位进行操作，字符串文字是单个令牌。 因此，您不能像这样干预字符串的中间。 您可以将script.py预处理为如下所示的内容： "some coden" "some more code that will be appendedn" 然而#include那。 或者，您可以使用xxd -i生成一个C静态数组，以供包含。 这不会让你一直在那里，但它会让你相当接近。
Is there any way to have multi-line plain-text, constant literals in C++, à la Perl? Maybe some parsing trick with #include ing a file? I can't think of one, but boy, that would be nice. I know it'll be in C++0x. Well ... Sort of. The easiest is to just use the fact that adjacent string literals are concatenated by the compiler: const char *text = "This text is pretty long, but w
有没有办法在C ++中使用多行纯文本，常量字面值，在Perl中？ 也许一些#include文件的解析技巧？ 我想不出一个，但男孩，那很好。 我知道它将在C ++ 0x。 那么...排序。 最简单的方法就是使用相邻字符串文字由编译器连接的事实： const char *text = "This text is pretty long, but will be " "concatenated into just a single string. " "The disadvantage is that you have to quote " "each part, and newline
Suppose I have a base class which cloning of derived classes: class Base { public: virtual Base * clone() { return new Base(); } // ... }; I have a set of derived classes which are implemented using a curiously recurring template pattern: template <class T> class CRTP : public Base { public: virtual T * clone() {
假设我有一个克隆派生类的基类： class Base { public: virtual Base * clone() { return new Base(); } // ... }; 我有一组派生类，它们使用奇怪的循环模板模式实现： template <class T> class CRTP : public Base { public: virtual T * clone() { return new T(); } // ... }; 我试图从这样的进一步得出： cl
I've got, probably trivial question but i can't understand it. I've wrote the simple code: fstream file; file.open("data", ios::in); if(!file.good()){ file.close(); file.open("data", ios::out); if(!file.good()) cout<<"not good"<<endl; file<<"test"<<endl; file.close(); } in fresh VS8 C++ Express project. When I run it and "data" does
我有，可能微不足道的问题，但我不明白。 我写了简单的代码： fstream file; file.open("data", ios::in); if(!file.good()){ file.close(); file.open("data", ios::out); if(!file.good()) cout<<"not good"<<endl; file<<"test"<<endl; file.close(); } 在新鲜的VS8 C ++ Express项目中。 当我运行它并且“数据”不存在时，它创建文件，但也返回“不好”（第二个），所以输出不会写入
I am trying to simulate a mouse click on a QMenu item from a QMenuBar, for example clicking on "Save As" QAction using the QTestLib framework. I am triyng this under Windows XP 32 bit and Qt 5.0.2. Any Ideas? Probably this question is not relevant for question owner, but I suppose it could be helpful for others. Unlike QToolBar , QMenu doesn't have method widgetForAction .
我试图模拟鼠标点击QMenuBar上的QMenu项目，例如使用QTestLib框架点击“另存为”QAction。 我在Windows XP 32位和Qt 5.0.2下运行。 有任何想法吗？ 可能这个问题与问题所有者无关，但我想这可能对其他人有帮助。 不像QToolBar ， QMenu没有办法widgetForAction 。 我发现了这种情况的一个简单的解决方法。 尝试使用QTest::mouseClick弹出菜单，如果没有任何反应尝试使用QTest::keyClick(..., first_char_in_menu_tite, Q
I am getting this error every time, while compiling programs, configuring and installing some things like binutils, textinfo, etc.. /usr/local/bin/ld: this linker was not configured to use sysroots collect2: error: ld returned 1 exit status I want to know clearly about this. When will will come and what is the actual problem, and also how to solve it? Try to use gcc with the following op
每次在编译程序，配置和安装时都会收到此错误 有些东西比如binutils，textinfo等。 /usr/local/bin/ld: this linker was not configured to use sysroots collect2: error: ld returned 1 exit status 我想清楚地知道这一点。 什么时候会来，什么是实际问题，以及如何解决它？ 尝试使用以下选项的gcc --sysroot=/usr/local ： gcc --sysroot=/usr/local 但是，正如其他人在评论中告诉你的，除非你知道你在做什么，否
什么是最可靠的方法来检测架构在C ++中使用一个或两个补码表示法？ You shouldn't have to worry - there aren't too many ones complement machines out there :) But the easiest thing might be to compare "-1" with ~0.
什么是最可靠的方法来检测架构在C ++中使用一个或两个补码表示法？ 你不必担心 - 有没有太多补充机器:) 但最简单的事情可能是将“-1”与〜0进行比较。