C/CPP系類知識 What happens when a function is called before its declaration in C? 在C中當使用沒有聲明的函數時將發生什麼

標籤：

http://www.geeksforgeeks.org/g-fact-95/

 

1 在C語言中，如果函數在聲明之前被調用，那麼編譯器假設函數的傳回值的類型為INT型，

所以下面的code將無法通過編譯：

#include <stdio.h>int main(void){    // Note that fun() is not declared     printf("%d\n", fun());    return 0;} char fun(){   return ‘G‘;}

錯誤：其實就是fun函數定義了兩遍，衝突了

test1.c:9:6: error: conflicting types for ‘fun‘ char fun()      ^test1.c:5:20: note: previous implicit declaration of ‘fun‘ was here     printf("%d\n", fun());                    ^

 

將傳回值改成int行可以編譯並運行：

#include <stdio.h>int main(void){    printf("%d\n", fun());    return 0;} int fun(){   return 10;}

 

2 在C語言中，如果函數在聲明之前被調用，如果對函數的參數做檢測？

compiler assumes nothing about parameters. Therefore, the compiler will not be able to perform compile-time checking of argument types and arity when the function is applied to some arguments. This can cause problems.

編譯器對參數不做任何假設，所以無法做類型檢查。 下面code就會有問題，輸出是garbage

#include <stdio.h> int main (void){    printf("%d", sum(10, 5));    return 0;}int sum (int b, int c, int a){    return (a+b+c);}

 

輸出結果：

[email protected]:~/myProg/geeks4geeks/cpp$ ./a.out 1954607895[email protected]:~/myProg/geeks4geeks/cpp$ ./a.out 1943297623[email protected]:~/myProg/geeks4geeks/cpp$ ./a.out -16827881[email protected]:~/myProg/geeks4geeks/cpp$ ./a.out 67047927[email protected]:~/myProg/geeks4geeks/cpp$ ./a.out -354871129[email protected]:~/myProg/geeks4geeks/cpp$ ./a.out -562983177[email protected]:~/myProg/geeks4geeks/cpp$ ./a.out 33844135[email protected]:~/myProg/geeks4geeks/cpp$

 

 

所以It is always recommended to declare a function before its use so that we don’t see any surprises when the program is run (See this for more details).

 

C/CPP系類知識 What happens when a function is called before its declaration in C? 在C中當使用沒有聲明的函數時將發生什麼