c - 返回值或传递指针作为参数？

Question

我了解到，为了通过调用函数访问或修改变量的值，我们需要将指针作为参数传递，如下所示:

#include <stdio.h>

//create a function 
void Increment(int* x) {    
   *x = *x+1;
   printf("Address of variable x in increment = %p\n",x); 

}

int main() {
   int a;
   a = 10;
   Increment(&a);   
   printf("Address of variable 'a' in main = %p\n",&a);
   printf("Value of a in main function = %d\n",a); 


}

但是我又做了一次测试，发现通过调用函数并返回值，我也可以达到同样的结果。

#include <stdio.h>

//create a function 
int Increment(int x) {        // do not use VOID
   x = x+1;
   printf("Address of variable x in increment = %p\n",x); 
   return x;                  

}

int main() {
   int a;
   a = 10;
       int hasil;
       hasil = Increment(a);   
   printf("Address of variable 'a' in main = %p\n",&a);
   printf("Value of a in main function = %d\n",hasil); 


}

我的问题:

1) 如果我只能使用返回值来获得相同的结果，是否必须将指针作为参数传递？

2) 当我从返回值的函数打印变量“x”的内存地址时，我观察到内存地址非常短 0xb ，知道为什么吗？通常地址很长。

Answer 1

1) Do I have to pass pointers as argument if I can just use return value to achieve the same result?

不，您不必总是这样做。但是观察它释放了函数的返回值用于错误报告。一个非常简单的例子:

enum error_code {
  E_SUCCESS,
  E_GENERAL_FAILURE,
  E_MEMORY_ALLOCATION_FAILED,
  E_INVALID_ARGUMENT,
  E_FILE_NOT_EXIST
};

struct my_important_data {
  // stuff
};

enum error_code fill_important_data_from_file(char const *file_name,
                                              struct my_important_data **data)
{
  if(!data || !file_name)
    return E_INVALID_ARGUMENT;

  *data = malloc(sizeof(**data));

  if(!*data)
    return E_MEMORY_ALLOCATION_FAILED;

  // Return different error codes based on type of failure
  // so the caller can know what exactly went wrong

  return E_SUCCESS;
}

2) I observe when I print the address of the memory of variable 'x' from the function that returns value, the memory address is very short 0xb , any idea why? normally the address is very long.

那是因为您正在使用 %p 转换说明符来打印常规 int，而不是实际地址。严格来说，这会导致 printf 函数具有未定义的行为。