char* 和 int* 有什么区别?
例如,当我运行此代码时:
#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;
}
我得到以下输出:
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
当我改变这一行时:
cout<<" Address is: "<<ptr2line+i<<" , same as "<<&line[i]<<endl;
至:
cout<<" Address is: "<<&ptr2line+i<<" , same as "<<&line[i]<<endl;
我得到了这个输出:
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
在这种情况下,地址 0x7fff5054bad0 对应于 'Hello' 的值,0x7fff5054bad8 对应于 'ello' 等等。
创建存储字符串每个字符地址的指针的正确方法是什么?
解析度
可视化指针的一种方法是使用 printf 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]);
给出所需的输出
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
下面的@Daniel Jour 概述了另一种解决方案。
正如已经指出的那样, achar *
和 an之间没有区别(当然,指向类型除外)int *
。不同的输出,你所看到的是,由于非成员重载的operator<<
造成了特殊的处理char
和char *
。
要“查看”指针,您可以将指针转换为void const *
或直接调用成员函数(位于 ideone 上):
#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;
}
示例输出:
string: Hello
casted: 0x2b1c038fbb7d
member: 0x2b1c038fbb7d
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