为什么C允许我调用未声明的函数? [英] Why does C allow me to call an undeclared function?

问题描述

我有两个文件: test1.c 和 test2.c ，其中包含 main()函数.

I have two files: test1.c, and test2.c, which contains the main() function.

test1.c:

#include <stdio.h>  // printf() function declaration/prototype

// function definition
void say_hello() {
  printf("\tHello, world!\n");
}

test2.c:

#include <stdio.h>  // printf() function declaration/prototype
int main() {
  printf("Inside main()\n");
  say_hello();

  return 0;
}

这是我的makefile:

And this is my makefile:

a.out: test1.o test2.o
    $(CXX) -o a.out test1.o test2.o

test1.o: test1.c
    $(CXX) -c test1.c

test2.o: test2.c
    $(CXX) -c test2.c

现在应该清楚问题出在哪里:test2.c中的main()函数无需声明就可以调用say_hello()！我运行以下命令，以使用gcc编译器: make CXX = gcc 我在屏幕上收到此警告:

Now it should be clear where the problem lies: The main() function in test2.c calls say_hello() without declaring it! I run the following command, to use the gcc compiler: make CXX=gcc I get this warning to the screen:

gcc -c test1.c
gcc -c test2.c
test2.c: In function ‘main’:
test2.c:16:3: warning: implicit declaration of function ‘say_hello’ [-Wimplicit-function-declaration]
   say_hello();
   ^
gcc -o a.out test1.o test2.o

尽管* .o文件已编译并链接到可执行文件中.那真是怪了.想象一下，当我运行a.out文件时感到惊讶，并且看到 main()成功调用了 say_hello()，该函数未在main的翻译单元中声明，好像根本没有问题！我的理由是，由于 say_hello()先前未在 test2.c 中声明，因此完全不应该由 main()调用它..请注意，我已经在 #include< stdio.h> 中添加了注释.这些预处理器指令包括函数声明/原型，它们将其范围扩展到相应的* .c文件中.这就是为什么我们能够使用它们.在该翻译单元中没有函数声明/原型==没有作用域，所以直到现在我还是这么想.

Although the *.o files got compiled and linked into the executable. That's weird. Imagine my surprise when I run the a.out file, and I see that main() successfully called say_hello(), a function which is not declared inside of main's translation unit, as if there were no issue at all! I reason that since say_hello() was not previously declared in test2.c, it should not allowed to be called by main() at all. Notice that I've added comments to the #include <stdio.h>. These preprocessor directives include the function declarations/prototypes, which extend their scope into the corresponding *.c files. That is why we are able to use them. No function declaration/prototype == no scope in that translation unit, or so I thought until now.

然后我将上面的代码编译为C ++代码: make CXX = g ++

Then I compiled the above code as C++ code: make CXX=g++

我在屏幕上看到此错误:

I get this error to the screen:

test2.c: In function ‘int main()’:
test2.c:16:13: error: ‘say_hello’ was not declared in this scope
   say_hello();
             ^
makefile:18: recipe for target 'test2.o' failed
make: *** [test2.o] Error 1

g ++完成了应做的工作，并停止了编译过程.但是gcc没有做到这一点！这是怎么回事?它是C编程语言的风范吗?编译器有问题吗?

g++ does what it's supposed to do, and stops the compilation process. But gcc did not do this! What's going on? Is it a perk of the C programming language? Is it an issue with the compiler?

推荐答案

简单，因为C允许调用未声明的函数，而C ++不允许.无论哪种方式， gcc 都会警告您，您可能要认真对待警告.

Simple, because C allows undeclared functions to be called and C++ does not. Either way, gcc warns you and you may want to take warnings seriously.

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