C, C ++ return string judgment (string pointer usage), return pointer

C, C ++ return string judgment (string pointer usage), return pointer

Function: enter a string to determine whether it is a return.

Mainly exercise pointer usage.

1. C Edition

#include<stdio.h>int main(){    char he[100];    char a;    int i=0,flag=1;    while((a=getchar())!='\n')    {        he[i]=a;        i++;    }    int n=i;    for(i=0;i<n/2;i++)    {        printf("%c\t%c\n",he[i],he[n-1-i]);        if(he[i]!=he[n-1-i])        {            printf("no");break;        }    }    if(flag==1)    {        printf("yes");    }    return 0;}

The getchar () function obtains an input in sequence, assigns it to the char variable a, and then assigns it to the char array he [] through a.

When "\ n" is input, that is, the Return key jumps out of the loop.

 

2. C ++

#include<iostream>#include<string.h>using namespace std;int main(void){    char *p="abcba";    int n=strlen(p);    bool flag=1;    int i;    for(i=0;i<n/2;i++)    {        cout<<p[i]<<"\t"<<p[n-1-i]<<endl;        if(p[i]!=p[n-1-i])        {            flag=0;            cout<<"no"<<endl;break;        }    }    if(flag==1)         cout<<"yes"<<endl;    return 0;}

Strlen () Get the length

3. C function call Edition

#include<stdio.h>#include<string.h>int pp(char *p){    int n=strlen(p),i;    //printf("%d",n);    if(p==NULL)return -1;    for(i=0;i<n/2;i++)    {       // printf("%c\t%c\n",p[i],p[n-1-i]);        if(p[i]!=p[n-1-i])        {            return 0;        }    }    return 1;}int main(){    char *p="abcba";    int a=pp(p);    printf("%d",a);}

Pp () is an int-type function, so an int-type value is returned. It is received by declaring an int a in the main function.