为const char **和char ** [英] const char ** and char **

问题描述

code 1：没有警告，没有错误。工作完美。

 的#include＆LT;＆stdio.h中GT;无效printP（为const char * P）
{
    的printf（为const char * p是：％P \\ N，P）;
    如果（p）的
        的printf（％S \\ n，p）的;
}无效printP2P（为const char ** P）
{
    的printf（为const char ** P指针：％P \\ N，* P）;
    如果（P＆放大器;及（* P））
        的printf（％S \\ n，* P）;
}诠释的main（）
{
    所以char a [] =Hello World的;
    为const char * p = A;
    为const char ** P2P =安培; P;
    printP（P）;
    printP2P（P2P）;
    返回0;
}
 

code 2：可以正常工作

。

  

警告：从兼容的指针类型[默认启用]初始化

 为const char * p = A;
==＆GT;
的char * p = A;
 

code 3：Segementation故障

  

警告：从兼容的指针类型[默认启用]初始化

 为const char ** P2P =安培; P;
==＆GT;
为const char ** P2P =安培; A;
 

问题：

1. 为什么code 2可以工作，但code 3有一个节段的错吗？

2，当了一通聊天* 到为const char * ，没有警告或错误发生了，但是当我分配的char ** 到为const char ** ，我得到了一个警告，为什么？

解决方案

假设

  A [] = 0x100处 -  ^ hèL LØW 0 R L D = 0x48 0x65 0x6c 0x6c ...
P =为0x200
P2P = 0x500处
 

有关案件3

  P2P =放大器;一
P2P = 0x100的
 

和当您尝试使用 * P2P 您正试图 *为0x100 这是读取内存地址打印保存在的0x100 是 0x4865 （假设2个字节的地址和做客场字节序）。地址 0x4865 最有可能是无效的，读它会导致分段错误。

有关第二种情况，其直线前进......

  P2P =安培，P
P2P =为0x200
* P2P = * =为0x200 0x100处
 

---

混淆的部分是，阵列地址＆放大器;一个是一样的 A 这是一样的＆安培; A [0]

Code 1: no warning , no error . work perfect.

#include <stdio.h>

void printP(const char *p)
{
    printf("const char *p is :  %p\n",p);
    if( p )
        printf("%s\n",p);
}

void printP2P(const char **p)
{
    printf("const char **p pointer to :     %p\n",*p);
    if( p &&(*p) )
        printf("%s\n",*p);
}

int main()
{
    char a[] = "Hello World";
    const char *p = a;
    const char **p2p = &p;
    printP(p);
    printP2P(p2p);
    return 0;
}

Code 2: can work.

warning: initialization from incompatible pointer type [enabled by default]

const char *p = a;
==>
char *p = a;

Code 3 : Segementation fault.

warning: initialization from incompatible pointer type [enabled by default]

const char **p2p = &p;
==>
const char **p2p = &a;

Problem:

1. Why code 2 could work but code 3 got a segement fault ?

2.When a pass chat * to const char * , no warning or error happened ,but when I assignment char ** to const char **, I got a warning, why?

解决方案

Assume

a[] = 0x100 - H e l l o  w o r l d = 0x48 0x65 0x6c 0x6c ...
p = 0x200 
p2p = 0x500

For 3rd case

p2p = &a
p2p = 0x100

and when you try to print it using *p2p you are trying *0x100 which is reading memory at address stored at 0x100 which is 0x4865 (assuming 2 bytes address and doing away to endianness). The address 0x4865 most likely is not valid and reading it causes segmentation fault.

For 2nd case, its straight forward ...

p2p = &p
p2p = 0x200
*p2p = *0x200 = 0x100

---

Confusing part is that for array address &a is same as a which is same as &a[0].

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