「Puts（）」函數如何在沒有參數的情況下工作？

Question

我在網站上看到這段代碼。

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

此代碼編譯並運行良好！

它得到一個字符串作爲來自用戶並打印輸入!!!!

但是，我的問題是，怎麼樣？

（注意：puts()函數不包含任何參數！）

Answer 1

的C舊版本有隱含類型變量和函數，這個代碼利用了這一點，一些其他的東西。實際返回價值也很鬆懈。

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

那是因爲你剛剛叫gets()用正確的參數調用puts()發現堆棧不變。在具有多個寄存器的CPU上，除非gets()不使用包含第一個參數的寄存器，否則這可能會中斷。編譯啓用優化，這可能就足夠了。

如果把兩者之間的任何函數調用，它會破壞了。

用相同數量代碼的清潔方法是：

puts(gets(&i)); 

Answer 3

你gets(&i)功能實際上得到的字符串。 puts()在您聲明兩個語句的順序中沒有影響。

Answer 4

通過它不一定每臺機器和執行工作組的魔法。 puts();僅僅意味着你沒有參數傳遞，但有些東西在堆棧上，它是指向字符串的指針（確實是這樣）。 puts需要它（它不知道你沒有在堆棧上推動任何東西，它只是「相信」你做到了）並且它工作。由於調用者需要清理堆棧，所以一切都很順利（如果它是一個被調用者任務，就會出現問題）。它運作的事實是一個「機會」（可能會發生，但你不能太信任的東西）;它編譯的事實是由標準或編譯器確定的，警告但不停止編譯（可能會添加選項以嚴格遵守特定標準，然後代碼可能無法編譯）

Answer 5

當您說「編譯和運行正常「，你的意思是（a）你忽略了編譯器警告，（b）代碼出現」運行正常「。您的編譯器應該生成多個警告，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，如果你嘗試這種在一個以上的平臺，你會發現它並不總是「運行良好」 - 它很可能打印垃圾和/或崩潰。