Summary of the relationship between one-dimensional arrays and pointers in C ++

For the array int a [10];

A Indicates the address of the first element of the array, that is, & a [0];

If the pointer p points to the first element of the array, you can perform the following operations:

Int * p =;

Or

Int * p = & a [0];

Then p ++ points to the first element in the array, that is, a [1];

* P is the value in a [1.

In this case, a [I] = p [I] = * (a + I) = * (p + I)

The following is an example;

Output directly using a [I]

 

#include<iostream>  using namespace std; int main(){     int a[10]={1,2,3,4,5,6,7,8,9,10};     cout<<"Please input 10 intergers: "<<endl;     int i=0;     for(i=0;i<10;i++)     cout<<a[i]<<" ";     cout<<endl;     return 0; } #include<iostream>using namespace std;int main(){ int a[10]={1,2,3,4,5,6,7,8,9,10}; cout<<"Please input 10 intergers: "<<endl; int i=0; for(i=0;i<10;i++) cout<<a[i]<<" "; cout<<endl; return 0;}

Output with * (a + I)

 

#include<iostream>  using namespace std; int main(){     int a[10]={1,2,3,4,5,6,7,8,9,10};     cout<<"Please input 10 intergers: "<<endl;     int i=0;     for(i=0;i<10;i++)     cout<<*(a+i)<<" ";     cout<<endl;     return 0; } #include<iostream>using namespace std;int main(){ int a[10]={1,2,3,4,5,6,7,8,9,10}; cout<<"Please input 10 intergers: "<<endl; int i=0; for(i=0;i<10;i++) cout<<*(a+i)<<" "; cout<<endl; return 0;}

Output with * (p + I)

 

#include<iostream>  using namespace std; int main(){     int a[10]={1,2,3,4,5,6,7,8,9,10};     cout<<"Please input 10 intergers: "<<endl;     int i=0;     int * p=a;     for(i=0;i<10;i++)     cout<<*(p+i)<<" ";     cout<<endl;     return 0; } #include<iostream>using namespace std;int main(){ int a[10]={1,2,3,4,5,6,7,8,9,10}; cout<<"Please input 10 intergers: "<<endl; int i=0; int * p=a; for(i=0;i<10;i++) cout<<*(p+i)<<" "; cout<<endl; return 0;}

About * p ++

Because ++ and * have the same priority and the combination direction is from right to left, it is equivalent to * (p ++ ). The function is to first obtain the value of the variable pointed to by p (* p), and then add 1 to the value pointed to by p.

[

#include<iostream>  using namespace std; int main(){     int a[10]={1,2,3,4,5,6,7,8,9,10};     cout<<"Please input 10 intergers: "<<endl;     int i=0;     int * p=a;     while(p<a+10){         cout<<*p++<<"\t";     }     cout<<endl;     return 0; } #include<iostream>using namespace std;int main(){ int a[10]={1,2,3,4,5,6,7,8,9,10}; cout<<"Please input 10 intergers: "<<endl; int i=0; int * p=a; while(p<a+10){  cout<<*p++<<"\t"; } cout<<endl; return 0;}

 

Equivalent

#include<iostream>  using namespace std; int main(){     int a[10]={1,2,3,4,5,6,7,8,9,10};     cout<<"Please input 10 intergers: "<<endl;     int i=0;     int * p=a;     while(p<a+10){         cout<<*p<<"\t";         p++;     }     cout<<endl;     return 0; } #include<iostream>using namespace std;int main(){ int a[10]={1,2,3,4,5,6,7,8,9,10}; cout<<"Please input 10 intergers: "<<endl; int i=0; int * p=a; while(p<a+10){  cout<<*p<<"\t";  p++; } cout<<endl; return 0;}

* P ++ is equivalent to * (p ++). * (++ p) indicates that p + 1 is used first, and * p is used.

 

#include<iostream>  using namespace std; int main(){     int a[10]={1,2,3,4,5,6,7,8,9,10};     cout<<"Please input 10 intergers: "<<endl;     int i=0;     int * p=a;     while(p<a+10){         cout<<*(++p)<<"\t";     }     cout<<endl;     return 0; } #include<iostream>using namespace std;int main(){ int a[10]={1,2,3,4,5,6,7,8,9,10}; cout<<"Please input 10 intergers: "<<endl; int i=0; int * p=a; while(p<a+10){  cout<<*(++p)<<"\t"; } cout<<endl; return 0;}

Running the above program will output the values from a [2] to a [11], where a [11] is not defined.