没有头文件的 C 程序 [英] C program without header

问题描述

我用 C 编写了hello world"程序.

I write "hello world" program in C.

void main()
{ printf("Hello World"); }
// note that I haven't included any header file

程序编译时出现警告

vikram@vikram-Studio-XPS-1645:~$ gcc hello.c 
hello.c: In function ‘main’:
hello.c:2:2: warning: incompatible implicit declaration of built-in function ‘printf’
vikram@vikram-Studio-XPS-1645:~$ ./a.out 
Hello Worldvikram@vikram-Studio-XPS-1645:~$

这怎么可能?操作系统如何在不包含任何标头的情况下链接库?

How is this possible? How does the OS link a library without including any header?

推荐答案

编译器通过引用一个名为 printf() 的函数来构建你的源文件，它实际采用什么参数或它的返回类型是什么.生成的程序集包含程序静态数据区中字符串"Hello World"的地址的push，后跟callprintf.

The compiler builds your source file with a reference to a function called printf(), without knowing what arguments it actually takes or what its return type is. The generated assembly contains a push of the address of the string "Hello World" in the static data area of your program, followed by a call to printf.

将目标文件链接到可执行文件时，链接器会看到对printf 的引用并提供C 标准库函数printf().巧合，您传递的参数 (const char*) 与真正的 printf() 的声明兼容，因此它可以正常运行.但是，请注意您的程序隐式声明的 printf() 具有返回类型 int(我认为)，标准 printf() 也拥有;但是如果它们不同，并且您要将调用 printf() 的结果分配给一个变量，那么您将处于未定义行为的领域，并且您可能会得到不正确的值.

When linking your object file into an executable, the linker sees a reference to printf and supplies the C standard library function printf(). By coincidence, the argument you have passed (const char*) is compatible with the declaration of the real printf(), so it functions correctly. However, note that the printf() that your program implicitly declares has return type int (I think), which the standard printf() also has; but if they differed, and you were to assign the result of calling printf() to a variable, you would be in the land of undefined behaviour and you would likely get an incorrect value.

长话短说:#include 为您使用的函数获取正确声明的正确标头，因为这种隐式声明已被弃用，因为它容易出错.

Long story short: #include the correct headers to get the correct declarations for functions you use, because this kind of implicit declaration is deprecated, because it is error-prone.

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