有用的函数atoi()

字符串转整型

#include <stdlib.h>
#include <stdio.h>
 
int main(void)
{
    int n;
    char *str = "12345.67";
    n = atoi(str);
    printf("n=%d\n",n);
    return 0;
}

result:

字符串转浮点型

#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
    char *str = "12345.67";
    printf("%f\n", strtof(str, NULL));
    printf("%.2f\n", strtof(str, NULL));
    return 0;
}

result:

整型转字符串：

#include<stdio.h>
int main()
{
    int n=9999;
    char str[20];
    printf("%s\n",itoa(n,str,10));//10表示基数，是什么进制的数字
    return 0;
}

result：

https://blog.csdn.net/qq_37640597/article/details/79772948 

再加个异或运算：

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int x=12;
    int y=0xf;;
    printf("%d\n",x^y);
    return 0;
}

result:

1100^1111=0011=3

#include<stdio.h>
int main()
{
    char x='1';
    if(isdigit(x))
        cout<<"yes"<<endl;
    else
        cout<<"no"<<endl;
    if(isalpha(x))
        cout<<"yes"<<endl;
    else
        cout<<"no"<<endl;
    return 0;
}

result: