簡單的使用計算器ç問題

Question

我新的C和我試圖做一個簡單的計算器

#include <stdio.h> 
#include <stdlib.h> 

int main() { 
    int firstNumber, secondNumber, answer; 
    char mathSign[20]; 

    printf("Enter first number \n"); 
    scanf("%d", &firstNumber); 

    printf("Enter sign \n a - Add \n s = Subtract \n m - Multiply \n d - Divide \n"); 
    scanf("%s", mathSign); 

    printf("Enter second number \n"); 
    scanf("%d", &secondNumber); 

    if (mathSign == 'a') { 
     answer = firstNumber + secondNumber; 
    } else if (mathSign == 's') { 
     answer = firstNumber - secondNumber; 
    } else if (mathSign == 'm') { 
     answer = firstNumber * secondNumber; 
    } else if (mathSign == 'd') { 
     answer = firstNumber/secondNumber; 
    } 

    printf("Your answer is %d", answer); 

    return 0; 
} 

答案最終總是作爲52出於某種原因。我應該讓整數加倍嗎？

Answer 1

您測試mathSign == 'a'但mathSign是一個數組，因此您應該編寫mathSign[0] == 'a'。編譯器應該已經產生了警告。你不應該忽視這些警告，它們表明問題通常是真正的錯誤。您可以通過提高警告級別獲得更多有用的診斷：gcc -Wall -W -Werror或clang -Weverything -Werror ...

注意scanf("%s", mathSign);會導致如果用戶鍵入超過19個字符長的單詞的緩衝區溢出。您應該保護與目標數組：

scanf("%19s", mathSign); 

此外，你應該檢查轉換成功，否則，你可能有不確定的行爲，如果輸入無法轉換，你使用的仍然未初始化的值。

您可以使用" %c"來讀取單個字符並使用經典操作符以提高可讀性。請注意0​​之前的空格：它允許scanf()跳過空白字符，例如前一個響應中的換行符。

下面是修改後的版本：

#include <stdio.h> 
#include <stdlib.h> 

int main(void) { 
    int firstNumber, secondNumber, answer; 
    char mathSign; 

    printf("Enter first number: "); 
    if (scanf("%d", &firstNumber) != 1) 
     return 1; 

    printf("Enter operation\n + - Add\n - = Subtract\n * - Multiply\n/- Divide\n"); 
    if (scanf(" %c", &mathSign) != 1) 
     return 1; 

    printf("Enter second number: "); 
    if (scanf("%d", &secondNumber) != 1) 
     return 1; 

    if (mathSign == '+') { 
     answer = firstNumber + secondNumber; 
    } else if (mathSign == '-') { 
     answer = firstNumber - secondNumber; 
    } else if (mathSign == '*') { 
     answer = firstNumber * secondNumber; 
    } else if (mathSign == '/') { 
     answer = firstNumber/secondNumber; 
    } else { 
     printf("unsupported operation: %c\n", mathSign); 
     return 1; 
    } 

    printf("Your answer is %d\n", answer); 

    return 0; 
} 

可以進一步與switch聲明簡化程序：

#include <stdio.h> 
#include <stdlib.h> 

int main(void) { 
    int firstNumber, secondNumber, answer; 
    char mathSign; 

    printf("Enter a simple expression: "); 
    if (scanf("%i %c%i", &firstNumber, &mathSign, &secondNumber) != 3) 
     return 1; 

    switch (mathSign) { 
     case '+': answer = firstNumber + secondNumber; break; 
     case '-': answer = firstNumber - secondNumber; break; 
     case '*': answer = firstNumber * secondNumber; break; 
     case '/': answer = firstNumber/secondNumber; break; 
     case '%': answer = firstNumber % secondNumber; break; 
     default: 
     printf("unsupported operation: %c\n", mathSign); 
     return 1; 
    } 
    printf("the result is %d\n", answer); 
    return 0; 
} 

Answer 2

當您僅使用mathSign時，它是指向數組的指針。 要訪問第一個元素，你必須使用mathSign[0]。

或者，因爲你只打算先簽後換一個角色，你可以聲明char mathSign而不是char mathSign[20]然後閱讀使用scanf(" %c", mathSign); 在這種情況下，比較mathSign == 'a是正確的符號字符。

Answer 3

試試這個代碼。這應該工作。此外，這將是最優化的一個。

#include <stdio.h> 
#include <stdlib.h> 

int main() { 
    int firstNumber, secondNumber, answer; 
    char mathSign; 

    printf("1.Enter first number \n"); 
    scanf("%d", &firstNumber); 

    printf("Enter sign \n a - Add \n s = Subtract \n m - Multiply \n d - Divide \n"); 
    scanf(" %c", &mathSign); 

    printf("Enter second number \n"); 
    scanf("%d", &secondNumber); 

    switch (tolower(mathSign)) { 
    case 'a': 
     answer = firstNumber + secondNumber; 
     break; 
    case 's': 
     answer = firstNumber - secondNumber; 
     break; 
    case 'm': 
     answer = firstNumber * secondNumber; 
     break; 
    case 'd': 
     answer = firstNumber/secondNumber; 
     break; 
    default: 
     printf("Invalid Choice"); 
     break; 
    } 

    printf("Your answer is %d", answer); 
    return 0; 
}