筆者介紹：姜雪偉，IT公司技術合夥人，IT高階講師，CSDN社群專家，特邀編輯，暢銷書作者，已出版書籍：《手把手教你架構3D遊戲引擎》電子工業出版社和《Unity3D實戰核心技術詳解》電子工業出版社等。

專案開發中，經常會遇到C與C++之間互相呼叫問題，但是有時會遇到在C語言沒啥問題，但是將其放到C++中就會出現問題，本篇部落格在此給讀者總結一下，在遇到下面這些情況時就要注意了。

一、在C＋＋使用函式時，事先要先宣告否則就會報錯，但是在C語言就不存在這種問題，如下所示：

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

int fun()
{
   return 10;
}
 

#include <stdio.h> 
int main(void)
{
    int i = 10;   
    int j = 20;
    const int *ptr = &i;    /* ptr is pointer to constant */
  
    printf("ptr: %d\n", *ptr); 
    *ptr = 100;        /* error: object pointed cannot be modified
                     using the pointer ptr */
  
    ptr = &j;          /* valid */
    printf("ptr: %d\n", *ptr);
  
    return 0;
}
 

 error: assignment of read-only location ‘*ptr’

#include <stdio.h> 
 
int main(void)
{  
    int const i = 10;    /* i is stored in read only area*/
    int j = 20;
 
    int const *ptr = &i;        /* pointer to integer constant. Here i 
                                 is of type "const int", and &i is of 
                                 type "const int *".  And p is of type                            
                                "const int", types are matching no issue */
 
    printf("ptr: %d\n", *ptr); 
 
    *ptr = 100;
 
        /* error */
 
    ptr = &j;          /* valid. We call it as up qualification. In 
                         C/C++, the type of "int *" is allowed to up 
                         qualify to the type "const int *". The type of 
                         &j is "int *" and is implicitly up qualified by 
                         the compiler to "cons tint *" */
 
    printf("ptr: %d\n", *ptr);
 
    return 0;
}

error: assignment of read-only location ‘*ptr’

#include <stdio.h>
  
int main(void)
{
   int i = 10;
   int j = 20;
   int *const ptr = &i;    /* constant pointer to integer */
  
   printf("ptr: %d\n", *ptr);
  
   *ptr = 100;    /* valid */
   printf("ptr: %d\n", *ptr);
  
   ptr = &j;        /* error */
   return 0;
}

error: assignment of read-only variable ‘ptr’

#include <stdio.h>
  
int main(void)
{
    int i = 10;
    int j = 20;
    const int *const ptr = &i;        /* constant pointer to constant integer */
  
    printf("ptr: %d\n", *ptr);
  
    ptr = &j;            /* error */
    *ptr = 100;        /* error */
  
    return 0;
}

  error: assignment of read-only variable ‘ptr’
     error: assignment of read-only location ‘*ptr’

三、在C中，一個void指標可以直接分配給其他一些指標，如int *，char *。 但是在C ++中，一個void指標必須被明確地指定型別。

int main()
{
   void *vptr;
   int *iptr = vptr; //In C++, it must be replaced with int *iptr=(int *)vptr; 
   return 0;
}

int main()
{
    const int a;   // LINE 4
    return 0;
}

int main(void)
{
    int new = 5;  // new is a keyword in C++, but not in C
    printf("%d", new);
}

#include <stdio.h>
int main()
{
    char *c = 333;
    printf("c = %u", c);
    return 0;
}

以上是在程式設計時經常遇到的，在此給讀者列出來，供參考。。。。。。。。。