c编程语言：应用程序不会停止后输入x

我做了这个程序，它的工作只是我想如何，但它应该停止时，我键入x但它不任何人可以告诉我为什么？，如果有任何快捷方式或更小的方式键入此代码？在此先感谢。c编程语言：应用程序不会停止后输入x

#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 
int main() 
{ 
    int meat[6]; 
    int i; 
    int total =0; 
    int avg; 
    char what[20]; 
    char end; 
    int days = 0; 
    char food[20]; 
    char drink[20]; 
    printf("what thing do u want to calculate the average of?(food/drink)\n"); 
    scanf(" %s", &what); 
    if(strcmp(what, "food")==0) 
    { 
     printf("what sort of food did u ate ?\n"); 
     scanf(" %s", food); 
    } 
    if(strcmp(what, "drink")==0) 
    { 
     printf("what sort of drink did u drank ?\n"); 
     scanf(" %s", drink); 
    } 
    for(i=0; i<6; i++)   
    { 
     days++; 
     if(strcmp(what, "food")==0) 
     { 
      printf("how many %s did u ate in day %d\n", food, i+1); 
     } 
     else if(strcmp(what, "drink")==0) 
     { 
      printf("how many liters of %s did u drank in day %d\n", drink, i+1); 
     } 
     scanf(" %d", &meat[i]); 
     total += meat[i]; 
     printf("if u want to continue type y\n"); 

     printf("type x if u want to finish\n"); 
     scanf(" %c", &end); 
     if((end = 'y')&&(i<5)) 
     { 
      continue; 
     } 
     else if(end = 'x' && strcmp(what, "drink")==0) 
     { 
      avg = total/days; 
      printf("\nyour average amount of liters of %s you had %d\t the total is %d\n", drink, avg, total); 
     } 
     else if(end = 'x' && strcmp(what, "food")==0) 
     { 
      avg = total/days; 
      printf("\nyour average amount of %s you had %d\t the total is %d\n", food, avg, total); 
     } 
     break; 
    } 
    if(strcmp(what, food)==0) 
    {   
     printf("\nyour average amount of %s you had is %d\t the total is %d\n", food, avg, total); 
    } 
    else if(strcmp(what, drink)==0) 
    {   
     printf("\nyour average amount of liters of %s you had is %d\t the total is %d\n", drink, avg, total); 
    } 

    return 0; 
}

来源

2015-03-08 talal alsharaa

else if(end = 'x' ...

应该是：

else if(end == 'x' ...

您使用==在if语句来测试平等。你已经在代码中的几个地方得到了这个，这是无意中执行的任务，而不是通过比较用户输入是否与特定字符相等来实现的。

更换=与==这里：

else if(end = 'x' && strcmp(what, "drink")==0)

这里：

else if(end = 'x' && strcmp(what, "food")==0)

和这里：

if((end = 'y')&&(i<5))

来源

2015-03-08 19:35:23

那些中的两个，也'如果（（端= 'y' 的）...）' – 2015-03-08 19:38:05

x工作，但现在进入6天时它不会损害平均或总数 – 2015-03-08 19:48:29

我怀疑你在这里最后的比较还有一个逻辑错误，如果（strcmp（what，food）== 0）'和here' else if（strcmp（what，drink）== 0）'，因为这期望用户为他们吃的东西输入了“食物”或“饮料”以及“食物”或“饮料”，这似乎并不是你要求用户输入的内容。我的猜测是，你想把这些比较中的“食物”和“饮料”分别转换为“食物”和“饮料”。 – 2015-03-08 19:58:14

c编程语言：应用程序不会停止后输入x

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