“Puts（）”函数如何在没有参数的情况下工作？

Question

我在网站上看到这段代码。

main(i) 
{ 
    gets(&i); 
    puts(); 
} 

此代码编译并运行良好！

它得到一个字符串作为来自用户并打印输入!!!!

但是，我的问题是，怎么样？

（注意：puts()函数不包含任何参数！）

Answer 1

的C旧版本有隐含类型变量和函数，这个代码利用了这一点，一些其他的东西。实际返回价值也很松懈。

main(i) // i is implicitly an integer (the default type for old C), and normally named argc 
// int main(int i) or void main(int i) 
{ // The stack (which lives in high memory but grows downward) has any arguments and 
    // probably the environmental variables and maybe even other (possibly blank/filler) 
    // stuff on it in addition to the return address for whatever called main and possibly 
    // the argument i, but at this point that could either be on the stack just under the 
    // return address or in a register, depending on the ABI (application binary interface) 


// extern int gets(int) or extern void gets(int) 
// and sizeof(int) is probably sizeof(char *) 
gets(&i); // By taking the address of i even if it wasn't on the stack it will be pushed to 
      // it so that it will have an address (some processors have addressable registers 
      // but they are rarely used by C for many reasons that I won't go into). 


      // The address of i is either also pushed onto the stack or put into a register 
      // that the ABI says should be used for the first argument of a function, and 
      // and then a call is made to gets (push next address to stack; jump to gets) 

      // The function gets does what it does, but according to the ABI there are 
      // some registers that it can do whatever it wants to and some that it must 
      // make sure are the same as they were before it was called and possibly one 
      // or more where it is supposed to store a return value. 
      // If the address of i was passed to it on the stack then it probably would be 
      // restricted from changing that, but if it was passed in a register it may 
      // have just been luckily left unchanged. 
      // Another possiblity is that since gets returns the string address it was 
      // passed is that it returns that in the same location as the first argument 
      // to functions is passed. 

puts(); // Since, like gets, puts takes one pointer argument it will be passed this 
      // this argument in the same way as gets was passed it's argument. Since we 
      // were somehow lucky enough for gets to not overwrite the argument that we 
      // passed to it and since the C compiler doesn't think it has anything new to 
      // pass to puts it doesn't change any registers' values or do too much to the 
      // stack. This leaves us in the situation where puts is called with the stack 
      // and registers set up in the same way as they would be if it were passed the 
      // address of i, just the same as gets. 

    // The gets call with the stack variable's address (so an address high on the stack) 
    // could have left main's return address intact, but also could have overwritten it 
    // with garbage. Garbage as main's return address would likely result in a jump to 
    // a random location (probably not part of your program) and cause the OS to kill the 
    // program (possibly with an unhandled SIGSEGV) which may have looked to you like a 
    // normal exit. Since puts appended a '\n' to the string it wrote and stdout is 
    // line buffered by default it would have been flushed before returning from puts 
    // even if the program did not terminate properly. 
} 

Answer 2

那是因为你刚刚叫gets()用正确的参数调用puts()发现堆栈不变。在具有多个寄存器的CPU上，除非gets()不使用包含第一个参数的寄存器，否则这可能会中断。编译启用优化，这可能就足够了。

如果把两者之间的任何函数调用，它会破坏了。

用相同数量代码的清洁方法是：

puts(gets(&i)); 

Answer 3

你gets(&i)功能实际上得到的字符串。 puts()在您声明两个语句的顺序中没有影响。

Answer 4

通过它不一定每台机器和执行工作组的魔法。 puts();仅仅意味着你没有参数传递，但有些东西在堆栈上，它是指向字符串的指针（确实是这样）。 puts需要它（它不知道你没有在堆栈上推动任何东西，它只是“相信”你做到了）并且它工作。由于调用者需要清理堆栈，所以一切都很顺利（如果它是一个被调用者任务，就会出现问题）。它运作的事实是一个“机会”（可能会发生，但你不能太信任的东西）;它编译的事实是由标准或编译器确定的，警告但不停止编译（可能会添加选项以严格遵守特定标准，然后代码可能无法编译）

Answer 5

当您说“编译和运行正常“，你的意思是（a）你忽略了编译器警告，（b）代码出现”运行正常“。您的编译器应该生成多个警告，例如

ub.c:2: warning: return type defaults to ‘int’ 
ub.c: In function ‘main’: 
ub.c:3: warning: implicit declaration of function ‘gets’ 
ub.c:4: warning: implicit declaration of function ‘puts’ 
ub.c:5: warning: control reaches end of non-void function 

另外，如果你尝试这种在一个以上的平台，你会发现它并不总是“运行良好” - 它很可能打印垃圾和/或崩溃。