char *和int *的區別是什麼？C++ char * vs int *

Question

char *和int *有什麼區別？

例如，當我跑這樣的代碼：

#include<iostream> 
using namespace std; 


int main(){ 
    int a[3] = {10, 20, 30}; 
    int *ptr2a = &a[0]; 
    int i; 
    char line[] = "Hello"; 
    char *ptr2line = &line[0]; 

    for(i=0; i<3; i++){ 
    cout<<"Value of a["<<i<<"] is: "<<a[i]<<" same as "<<*(ptr2a+i); 
    cout<<" Address is: "<<ptr2a+i<<" , same as "<<&a[i]<<endl; 
    } 

    for(i=0; i<strlen(line); i++){ 
    cout<<"Value of line["<<i<<"] is: "<<line[i]<<" same as "<<*(ptr2line+i); 
    cout<<" Address is: "<<ptr2line+i<<" , same as "<<&line[i]<<endl; 
    } 

return 0; 
} 

我得到以下輸出：

Value of a[0] is: 10 same as 10 Address is: 0x7fff5bf29b1c , same as 0x7fff5bf29b1c 
Value of a[1] is: 20 same as 20 Address is: 0x7fff5bf29b20 , same as 0x7fff5bf29b20 
Value of a[2] is: 30 same as 30 Address is: 0x7fff5bf29b24 , same as 0x7fff5bf29b24 
Value of line[0] is: H same as H Address is: Hello , same as Hello 
Value of line[1] is: e same as e Address is: ello , same as ello 
Value of line[2] is: l same as l Address is: llo , same as llo 
Value of line[3] is: l same as l Address is: lo , same as lo 
Value of line[4] is: o same as o Address is: o , same as o 

當我改變這一行：

cout<<" Address is: "<<ptr2line+i<<" , same as "<<&line[i]<<endl; 

到：

cout<<" Address is: "<<&ptr2line+i<<" , same as "<<&line[i]<<endl; 

我得到這樣的輸出：

Value of line[0] is: H same as H Address is: 0x7fff5054bad0 , same as Hello 
Value of line[1] is: e same as e Address is: 0x7fff5054bad8 , same as ello 
Value of line[2] is: l same as l Address is: 0x7fff5054bae0 , same as llo 
Value of line[3] is: l same as l Address is: 0x7fff5054bae8 , same as lo 
Value of line[4] is: o same as o Address is: 0x7fff5054baf0 , same as o 

在這種情況下 地址0x7fff5054bad0相當於「你好」 0x7fff5054bad8對應值「ELLO」等。

創建存儲字符串每個字符地址的指針的正確方法是什麼？

解決方案

一種方法以可視化的指針是對printf使用代替

printf("Value of line[%d] is: %c same as %c, Address is: %p, same as %p\n", i, line[i], *(ptr2line+i),ptr2line+i,&line[i]); 

，得到所需的輸出

Value of line[0] is: H same as H, Address is: 0x7fff54ce6ade, same as 0x7fff54ce6ade 
Value of line[1] is: e same as e, Address is: 0x7fff54ce6adf, same as 0x7fff54ce6adf 
Value of line[2] is: l same as l, Address is: 0x7fff54ce6ae0, same as 0x7fff54ce6ae0 
Value of line[3] is: l same as l, Address is: 0x7fff54ce6ae1, same as 0x7fff54ce6ae1 
Value of line[4] is: o same as o, Address is: 0x7fff54ce6ae2, same as 0x7fff54ce6ae2 

另一種解決方案已經概述通過下述@Daniel怨婦。

Answer 1

正如已經指出的那樣，char *和int *之間沒有任何區別（當然指出類型除外）。你看到的不同的輸出是因爲the non member overloads的operator<<這導致特殊處理char和char *。

「看到」 的指針，就可以轉換指針到void const *或直接調用成員函數（live on ideone）：

#include <iostream> 

int main() { 
    char const * string = "Hello"; 
    std::cout << "string: " << string << std::endl 
     << "casted: " << static_cast<void const*>(string) << std::endl 
     << "member: "; 
    std::cout.operator<<(string); 
    std::cout << std::endl; 
} 

輸出示例：

string: Hello 
casted: 0x2b1c038fbb7d 
member: 0x2b1c038fbb7d 

Answer 2

您創建的指針已將每個字符的地址存儲在c字符串中。輸出問題的方式是std::coutoperator<<處理c字符串。它將繼續讀取內存中的下一個條目，直到達到空字符\0。這就是爲什麼每行你都得到整個C字符串中剩餘的字符。

編輯：爲了更好地理解發生了什麼，請參閱Daniel Jour的答案。