[C/CPP系列知识] 在C中使用没有声明的函数时将发生什么 What happens when a function is called before its declaration in C

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

1 在C语言中，如果函数在声明之前被调用，那么编译器假设函数的返回值的类型为INT型，

所以下面的code将无法通过编译：

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

char fun()
{
   return 'G';
}

错误：其实就是fun函数定义了两遍，冲突了

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

将返回值改成int行可以编译并运行：

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

int fun()
{
   return ;
}

2 在C语言中，如果函数在声明之前被调用，如果对函数的参数做检测？

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.

编译器对参数不做任何假设，所以无法做类型检查。 下面code就会有问题，输出是garbage

#include <stdio.h>

int main (void)
{
    printf("%d", sum(, ));
    return ;
}
int sum (int b, int c, int a)
{
    return (a+b+c);
}

输出结果：

diego@ubuntu:~/myProg/geeks4geeks/cpp$ ./a.out 

diego@ubuntu:~/myProg/geeks4geeks/cpp$ ./a.out 

diego@ubuntu:~/myProg/geeks4geeks/cpp$ ./a.out
-
diego@ubuntu:~/myProg/geeks4geeks/cpp$ ./a.out 

diego@ubuntu:~/myProg/geeks4geeks/cpp$ ./a.out
-
diego@ubuntu:~/myProg/geeks4geeks/cpp$ ./a.out
-
diego@ubuntu:~/myProg/geeks4geeks/cpp$ ./a.out 

diego@ubuntu:~/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 thisfor more details).