如何在同一时间打印两个输入的输出？

Question

代码：

#include <stdio.h> 

int main() 
{ 
    int w,x,y,z; 
    float v; 
    printf("Enter the driver salary\n"); 
    scanf("%d",&x); 
    printf("Enter the car mileage in km per litre\n"); 
    scanf("%d",&y); 
    printf("Enter the cost of petrol per litre\n"); 
    scanf("%d",&z); 
    printf("Enter the taxi fare for a km\n"); 
    scanf("%d",&w); 
    printf("Enter the distance of travel\n"); 
    scanf("%f",&v); 
    if(w==200 && y==10 && z==60 && x== 20 && v==10.5) 
    printf("Minimal cost travel is by taxi\n"); 
    else 
    printf("Minimal cost travel is by audi\n"); 
    return 0; 
} 

对于两种不同的w组输入值，y，z，x，v，我需要打印在同一时间两个输出语句。我获得第一个输出，但是如何同时获得两个输出？

Answer 1

您需要将核心功能包装在for/while循环中。我将建议将核心功能也放在一个功能中。

void processInput() 
{ 
    int w,x,y,z; 
    float v; 

    printf("Enter the driver salary\n"); 
    scanf("%d",&x); 
    printf("Enter the car mileage in km per litre\n"); 
    scanf("%d",&y); 
    printf("Enter the cost of petrol per litre\n"); 
    scanf("%d",&z); 
    printf("Enter the taxi fare for a km\n"); 
    scanf("%d",&w); 
    printf("Enter the distance of travel\n"); 
    scanf("%f",&v); 
    if(w==200 && y==10 && z==60 && x== 20 && v==10.5) 
     printf("Minimal cost travel is by taxi\n"); 
    else 
     printf("Minimal cost travel is by audi\n"); 
} 

int main() 
{ 
    int i; 
    for (i = 0; i < 2; ++i) 
    { 
     processInput(); 
    } 
    return 0; 
} 

Answer 2

如果你想同时输出一个标志数组来存储results.Also你可能希望存储在一个浮点数10.5的值。 Read more。我加在你的代码检查的几行，如果它的工作原理：

#include<stdio.h> 
int main() 
{      
    float l=10.5;    //to be safe about float rounding up 
    int i,fl[2];    //stores results for output 

    for(i=0;i<2;i++)   //add this 
    { 
    int w,x,y,z; 
    float v; 
    printf("Enter the driver salary\n"); 
    scanf("%d",&x); 
    printf("Enter the car mileage in km per litre\n"); 
    scanf("%d",&y); 
    printf("Enter the cost of petrol per litre\n"); 
    scanf("%d",&z); 
    printf("Enter the taxi fare for a km\n"); 
    scanf("%d",&w); 
    printf("Enter the distance of travel\n"); 
    scanf("%f",&v); 

    if(w==200 && y==10 && z==60 && x== 20 && v==l) 
     fl[i]=1; 
    else 
     fl[i]=0; 
    } 
    for(i=0;i<2;i++) 
    { 
    if(fl[i]==1) 
     printf("Case %d : Minimal cost travel is by taxi\n",i+1); 
    if(fl[i]==0) 
     printf("Case %d : Minimal cost travel is by audi\n",i+1); 
    } //close braces 

    return 0; 
}