C语言练习

#include <stdio.h>
int main(void )
{
    
    double number1=0.0;
    double number2=0.0;
    char operation =0;
    
    printf ("
Enter the calculation 
");
    scanf("%lf %c %lf ",&number1,operation,&number2 );
    


    switch (operation)
    {
        
        case '+':
        printf("=%lf",number1+number2);
        break;
        
        case'-':
        printf("=%lf",number1-number2);
        break;
        
        case'*':
        printf("=%lf",number1*number2);
        break;
        
        case '/':
            if (number2!=0)
            printf("=%lf",number1/number2);
            else 
                printf("wrong number2");
                break ;
                
        case '%':
            if ((long )number2==0)
                printf("Error");
            else ;
            printf("=%lf",(long)number1%(long)number2);
            break;
            
        default:
            printf("aiiiiiiiii");
            break;
        
    }
    
    return 0;
}
 
机房的系统执行不了，也是很无语。。。。   

#include <stdio.h>
int main()
{
    int i=1;
    printf("********************");
    for (;i<12;i++)
    printf("
*                   *");
    
    printf("
********************");
    return 0;

}
打印小方格

#include <stdio.h>
int main()
{
    
    
    for(int k=1,j=2;k<=5;++k,j=j+2)
        printf("
%-5d",k*j);
    return 0;


}打印几个数相乘

#include <stdio.h>
int main()
{
    long sum =0l;
    int count = 0;
    
    printf("
Enter the number of integers you want to summed");
    scanf("%d",&count );
    
    for (int i=1;i<=count;i++)
    sum+=i;
    
上述代码可以替换为

 for (int i=1;i<=count;sum+=i++);//循环语句为空，其次可以发现for循环的范围是管到第一个分号就结束了

上述代码可以替换为

 for (int i=count;i>=1;sum+=i--);//逆向运算

    printf("%ld",sum);
    return 0;
    
}
>>6
返回6+5+4+3+2+1

一个无限循环只有输入n 才跳出
#include <stdio.h>
#include <ctype.h>//切记，tolower()函数必须要有
int main()
{
    char a=0;
    for (;;)
    {
    printf("      y/n ?");
    scanf("%c",&a);
    if(tolower(a)=='n')
    break;
    }
    return 0;
}

#include <stdio.h>
#include <ctype.h>
int main()
{
    char a=0;
    double total=0.0;
    double value=0.0;
    int count =0;
    
    printf("
this program ....");
    
    for(;;)
    {
        
        printf("
Enter a value :");
        scanf ("%lf",&value);
        total+=value;
        ++count;
        
        printf("Do you want another value?");
        scanf("%c",&a);
        
        if (tolower(a)=='q')
        break;
        
    
        
        
            
    }
    printf("%-2lf",total/count);
    return 0;
}//字母和那个数字他如何识的前后？

#include <stdio.h>
#include <ctype.h>
int main()
{
    int chosen =12;
    int guess=0;
    int count=3;
    
    for (;count>=1;count--)
    {
        printf("
Enter the number you guess");
        scanf("%d",&guess);
            if (guess==chosen)
    {printf("Congratulation");
    goto m;}
    if((guess>15)||(guess<1))
    printf("Error");
        

    
    
    }
        printf("

shibai");
        m :
        printf("
yes");
    
}//goto m;多此一举。直接可以用return 0；
跳出所有直接结束，更不必用break
// 或  ||
且  &&
莫搞错