Compiling process in terms of a simple c++ program

This question already has an answer here: How does the compilation/linking process work? 5 answers This is a very broad question in general but i'll try to answer as briefly as possible. A typical language processing system has the following phases : 1. Preprocessing Phase - In this phase all preprocessors and macros are handled and code is generated which is free from these. This in

用一个简单的c ++程序编译过程

这个问题在这里已经有了答案: 编译/链接过程如何工作? 5个答案 这是一个非常广泛的问题,但我会尽可能简短地回答。 典型的语言处理系统有以下几个阶段: 1.预处理阶段 - 在这个阶段中,所有的预处理器和宏都被处理,并且代码被生成,而这些代码是没有这些的。 这涉及用宏体替换宏调用并用实际参数替换形式参数。 2.编译阶段 - 这有几个较小的阶段,例如:词法分析,语法分析,语义分析,中间代码生成,代码优化,目

How is the header file connected to the corresponding .cpp file?

This question already has an answer here: How does the compilation/linking process work? 5 answers C++: Compiler and Linker functionality 8 answers Short answer: there is no relationship between the header and its implementation. One can exist without the other, or the two could be placed in files with unrelated names. while compiling the main.cpp how will the compiler know to look for

头文件如何连接到相应的.cpp文件?

这个问题在这里已经有了答案: 编译/链接过程如何工作? 5个答案 C ++:编译器和链接器功能8个答案 简而言之:头部与其实现之间没有关系。 一个可以没有另一个存在,或者两个可以放在与名称无关的文件中。 同时编译main.cpp编译器如何知道yum.h中yum.cpp提到的任何函数的定义? 编译器不知道。 每次看到对yum.h中或其他头文件中声明的yum.h东西的引用时,它都会停留在寻找相应的定义上。 如果定义在编译器到达翻

How do libraries work at compile & link times

This question already has an answer here: How does the compilation/linking process work? 5 answers Short answer: At compile time you compile your program using the library's headers. At link time the linker basically looks up the symbols that the compiler found in the library's headers in the library's binaries so that your program knows what code in the library's binarie

图书馆如何在编译和链接时间工作

这个问题在这里已经有了答案: 编译/链接过程如何工作? 5个答案 简短的回答: 在编译时,你使用库的头文件来编译你的程序。 在链接时,链接器基本上查找编译器在库的二进制文件的库头中找到的符号,以便当程序使用该库中的符号(运行时)时,程序知道库的二进制文件中要执行哪些代码。

Why #include<.hpp> in .cpp, not <.cpp> in .hpp?

This question already has an answer here: How does the compilation/linking process work? 5 answers Why have header files and .cpp files? [closed] 9 answers You can #include arbitrary files in a C++ translation unit (the *.cpp you are compiling) provided the preprocessed form (after pre-processing) is valid C++. Read some documentation on the preprocessor and the C preprocessor wikipage.

为什么在.cpp中包含<.hpp>,而不是在.hpp中包含<.cpp>?

这个问题在这里已经有了答案: 编译/链接过程如何工作? 5个答案 为什么有头文件和.cpp文件? [已完成] 9个答案 如果预处理表单(在预处理之后)是有效的C ++,您可以在C ++翻译单元中#include任意文件(您正在编译的*.cpp )。 阅读有关预处理器和C预处理器wikipage的一些文档。 不要忘记,预处理是C或C ++编译器的第一阶段。 (从历史上看,它甚至是一个不同的进程/lib/cpp ;现在它出于性能原因在编译器内部)。

Translating source code into a foreign language

I'm running an educational website which is teaching programming to kids (12-15 years old). As they don't all speak English in the code source of the solutions we are using French variables and functions names. However we are planing to translate the content into other languages (German, Spanish, English). To do so I would like to translate the source code as fast as possible. We mos

将源代码翻译成外语

我正在运行一个教育网站,为孩子们(12-15岁)教授节目。 由于他们并不都在解决方案的代码源中使用英语,所以我们使用法语变量和函数名称。 不过,我们计划将内容翻译成其他语言(德语,西班牙语,英语)。 为此,我想尽可能快地翻译源代码。 我们主要有C / C ++代码。 我打算使用的解决方案: 从源代码中提取所有变量/函数名称,并将它们在文件中的位置(声明,使用,称为...的位置) 删除所有语言关键字和库函数

What do linkers do?

I've always wondered. I know that compilers convert the code you write into binaries but what do linkers do? They've always been a mystery to me. I roughly understand what 'linking' is. It is when references to libraries and frameworks are added to the binary. I don't understand anything beyond that. For me it "just works". I also understand the basics of dyn

连接器有什么作用?

我一直在想。 我知道编译器会将您编写的代码转换为二进制文件,但连接器有什么作用? 他们对我来说一直是个谜。 我大致了解'连接'是什么。 这是对库和框架的引用被添加到二进制文件。 除此之外我什么都不明白。 对我来说,它“只是工作”。 我也了解动态链接的基础知识,但没有太深入的内容。 有人可以解释这些条款吗? 要理解链接器,首先了解当将源文件(如C或C ++文件)转换为可执行文件(可执行文件是可以

What is the rationale for this undefined behavior?

Warning[...]: undefined behavior: the order of volatile accesses is undefined in this statement x.cpp xxx Why this line is undefined behavior? case 2: Vdda = 3.3 * (*VREFINT_CAL) / ADC_DR->DATA; Where the declarations/initializations are: volatile short const *VREFINT_CAL = (short *) 0x1FFFF7BA; and volatile STRUCT_ADC_DR *ADC_DR = (STRUCT_ADC_DR*) 0x40012440; defined by: typede

这种未定义的行为的基本原理是什么?

警告[...]:未定义的行为:在此语句x.cpp xxx中,未定义易失性访问的顺序 为什么这条线是未定义的行为? case 2: Vdda = 3.3 * (*VREFINT_CAL) / ADC_DR->DATA; 声明/初始化为: volatile short const *VREFINT_CAL = (short *) 0x1FFFF7BA; 和 volatile STRUCT_ADC_DR *ADC_DR = (STRUCT_ADC_DR*) 0x40012440; 被定义为: typedef struct { unsigned DATA : 16; unsigned : 16; } ST

c++ default assignment operator

int a[10]; int b[10]; a = b; // struct test { int a[10]; }; test a,b; a = b; First code doesn't compile, since we cant assign array, but the second does. Isn't the default assignment operator of class simply call assignment for each data members? Why does the the second code compile? From the C++11 draft, section 12.8: The implicitly-defined copy/move assignment operator for

c ++默认赋值运算符

int a[10]; int b[10]; a = b; // struct test { int a[10]; }; test a,b; a = b; 第一个代码不能编译,因为我们不能分配数组,但第二个代码不能。 不是类的默认赋值运算符只是为每个数据成员调用赋值? 为什么要编译第二个代码? 从C ++ 11草案,第12.8节: 非联合类X的隐式定义的复制/移动赋值运算符会执行其子对象的成员复制/移动赋值。 首先分配X的直接基类,按照它们在base-specifier-list中的声明顺序,然后

C/C++ post

Possible Duplicate: Undefined Behavior and Sequence Points I am using microsoft visual c++. Look at the following example: int n = 5; char *str = new char[32]; strcpy(str, "hello world"); memcpy(&str[n], &str[n+1], 6+n--); printf(str); // output is "hell world" So unexpectadly my compiler produces code that first decrements n and then executes memcpy. The following source will do

C / C ++后

可能重复: 未定义的行为和序列点 我正在使用Microsoft Visual C ++。 看看下面的例子: int n = 5; char *str = new char[32]; strcpy(str, "hello world"); memcpy(&str[n], &str[n+1], 6+n--); printf(str); // output is "hell world" 因此,我的编译器生成的代码首先递减n,然后执行memcpy。 以下来源将做我预计会发生的事情: int n = 5; char *str = new char[32]; strcpy(str, "hello world"); memcpy(&

how does increment work?

Possible Duplicate: Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…) Undefined Behavior and Sequence Points Ok we all know that i++ increments value by 1 on next line and ++i increments on same line (please correct me if i am wrong there ) so for a sample statement of c as follows: int a=0; printf("%d , %d",++a,a); the expected output should be 1 , 1

增量如何工作?

可能重复: 任何人都可以解释这些未定义的行为(i = i ++ + ++ i,i = i ++等) 未定义的行为和序列点 好的,我们都知道i ++在下一行增加1,而++ i在同一行增加 (如果我在那里错了,请纠正我) 所以对于c的一个样本陈述如下: int a=0; printf("%d , %d",++a,a); 预期的输出应该是1 , 1但是它会给出1 , 0所以人们可能会猜测我在这里问的是为什么当值已经增加时, i的第二个链接打印0而不是1 。 所以如果后增