做...邏輯運算符while循環

Question

我從我的教科書中複製了一個簡單的「計算器」示例，它接受2個值和一個運算符。但是，它不檢查輸入的表達式是否有效。代碼應該提示用戶，直到輸入正確的表達式。

我該如何解決這個問題？什麼是更好的方式來做到這一點？

/* 
Input: A simple expression with 2 values and 1 operator 
Format: value1 (operator) value2 
Ex: 5*5 
Output: Answer 
*/ 

#include <stdio.h> 

int main (void) 
{ 
float value1, value2; 
char operator = ' '; 

//Issue Area, the while section does not work 
do 
{ 
    printf ("Type in your expression:"); 
    scanf ("%f %c %f", &value1, &operator, &value2); 
} 
while ((operator != '+' || operator != '-' || operator != '*' || operator != '/')); 

//This is fine, code is for an else...if example in textbook 
if (operator == '+') 
    printf ("%.2f\n",value1 + value2); 
else if (operator == '-') 
    printf ("%.2f\n",value1 - value2); 
else if (operator == '*') 
    printf ("%.2f\n",value1 * value2); 
else if (operator == '/') 
    printf ("%.2f\n",value1/value2); 

return 0; 
} 

Answer 1

您有：

do 
{ 
    ... 
} 
while ((operator != '+' || operator != '-' || operator != '*' || operator != '/')); 

在有條件的while的使用||是錯誤的。

比方說operator的值是'+'。然後，while計算結果爲：

while ((false || true || true || true)); 

計算結果爲

while (true); 

無論什麼operator值，其中至少有三個子表達式將評估對true。因此，條件將始終評估爲true。您需要使用&&而不是||。

do 
{ 
    ... 
} 
while ((operator != '+' && operator != '-' && operator != '*' && operator != '/')); 

一種可能的方式，使代碼更清晰是：

do 
{ 
    ... 
} 
while (!isValidOperator(operator)); 

其中

int isValidOperator(char operator) 
{ 
    return (operator == '+' || 
      operator == '-' || 
      operator == '*' || 
      operator == '/'); 
} 

可以使代碼在isValidOperator較短的使用：

int isValidOperator(char operator) 
{ 
    return (operator != '\0' && strchr("+-*/", operator) != NULL); 
}