运行的时候要把表达式输两遍才能出结果
不知道哪里出了问题 /(ㄒoㄒ)/~~
刚开始学 多多指点
#include <stdio.h>
#include <stdlib.h>
int main()
{
float operand,sum;
char ch;
printf(“Enter an expression:”);
scanf(“%f”,&operand);
sum=operand;
while (ch!=”\n”)
{
scanf(“%c”,&ch);
scanf(“%f”,&operand);
switch (ch)
{
case “+”:
sum += operand;
break;
case “-“:
sum -= operand;
break;
case “*”:
sum *= operand;
break;
case “/”:
sum /= operand;
break;
}
}
printf(“Value of expression:%.1f\n”,sum);
return 0;
}
不知道哪里出了问题 /(ㄒoㄒ)/~~
刚开始学 多多指点
#include <stdio.h>
#include <stdlib.h>
int main()
{
float operand,sum;
char ch;
printf(“Enter an expression:”);
scanf(“%f”,&operand);
sum=operand;
while (ch!=”\n”)
{
scanf(“%c”,&ch);
scanf(“%f”,&operand);
switch (ch)
{
case “+”:
sum += operand;
break;
case “-“:
sum -= operand;
break;
case “*”:
sum *= operand;
break;
case “/”:
sum /= operand;
break;
}
}
printf(“Value of expression:%.1f\n”,sum);
return 0;
}
解决方案
60
原因是你一次while循环后，你的ch并不等于”\n”,感觉你scanf(“%c”, &ch);应该换个位置
#include <stdio.h>
#include <stdlib.h>
int main()
{
float operand, sum;
char ch = " ";
printf("Enter an expression:");
scanf("%f", &operand);
scanf("%c", &ch);
sum = operand;
while (ch != "\n")
{
scanf("%f", &operand);
switch (ch)
{
case "+":
sum += operand;
break;
case "-":
sum -= operand;
break;
case "*":
sum *= operand;
break;
case "/":
sum /= operand;
break;
}
scanf("%c", &ch);
}
printf("Value of expression:%.1f\n", sum);
return 0;
}
20
scanf(“%c”,&ch); 会将之前输入留下的换行符 \n接收了
改成ch = getchar();
改成ch = getchar();
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