1
當我在一個數字我打字我看到 輸入一個數字1 鍵入操作錯誤:未知的操作! 累加器= 0.000000 鍵入一個數字一個簡單的「印刷」計算器
爲什麼步驟 - 輸出( 「在操作員鍵入 」)被跳過,被替換爲 - 默認: 的printf(「 ERROR:未知操作者\ N!」); 休息;
感謝您的幫助!
// Program to produce a simple printing calculator
#include <stdio.h>
#include <stdbool.h>
int main (void)
{
double accumulator = 0.0, number; // The accumulator shall be 0 at startup
char operator;
bool isCalculating = true; // Set flag indicating that calculations are ongoing
printf("You can use 4 operator for arithmetic + -/*\n");
printf("To set accumulator to some number use operator S or s\n");
printf("To exit from this program use operator E or e\n");
printf ("Begin Calculations\n");
while (isCalculating) // The loop ends when operator is = 'E'
{
printf("Type in a digit ");
scanf ("%lf", &number); // Get input from the user.
printf("Type in an operator ");
scanf ("%c", &operator);
// The conditions and their associated calculations
switch (operator)
{
case '+':
accumulator += number;
break;
case '-':
accumulator -= number;
break;
case '*':
accumulator *= number;
break;
case '/':
if (number == 0)
printf ("ERROR: Division by 0 is not allowed!");
else
accumulator /= number;
break;
case 'S':
case 's':
accumulator = number;
break;
case 'E':
case 'e':
isCalculating = false;
break;
default:
printf ("ERROR: Unknown operator!\n");
break;
}
printf ("accumulator = %f\n", accumulator);
}
printf ("End of Calculations");
return 0;
}
謝謝你讓·弗朗索瓦·法布爾!它現在工作! – Yellowfun
我敢打賭:我測試過了,那些輸入函數很棘手! –
順便說一句,你可以接受答案,如果它伎倆。 –