Question

如果无法比较多个textField.keyboardType;

BOOL b = textField.keyboardType == (UIKeyboardTypeDefault || UIKeyboardTypeASCIICapable || UIKeyboardTypeEmailAddress);

我在UIKeyboardTypeEmailAddress上获得了“使用逻辑与常量操作数”？

Answer 1

正确地说，它应该是

BOOL b = (textField.keyboardType == UIKeyboardTypeDefault)
        || (textField.keyboardType == UIKeyboardTypeASCIICapable)
        || (textField.keyboardType == UIKeyboardTypeEmailAddress);

检查keyboardType是否等于3个值中的一个。

Answer 2

如果要测试keyboardType是否为列出的值之一，请使用

BOOL b = textField.keyboardType == UIKeyboardTypeDefault || textField.keyboardType == UIKeyboardTypeASCIICapable || textField.keyboardType == UIKeyboardTypeEmailAddress;

Answer 3

最接近你的尝试，理论上可以完成你想要做的事情（但实际上不会起作用）：

BOOL b = textField.keyboardType & (UIKeyboardTypeDefault | UIKeyboardTypeASCIICapable | UIKeyboardTypeEmailAddress);

您只需要使用一个|来执行按位，或者因为||执行逻辑运算，或者无法实现您想要的功能。然后你需要对该值进行按位和（&），只有当两者有共同的位时才能产生非零值。

然而，问题是UIKeyboardType enum的值不是2的幂，允许你这样做：

typedef NS_ENUM(NSInteger, UIKeyboardType) {
    UIKeyboardTypeDefault,                // Default type for the current input method.
    UIKeyboardTypeASCIICapable,           // Displays a keyboard which can enter ASCII characters, non-ASCII keyboards remain active
    UIKeyboardTypeNumbersAndPunctuation,  // Numbers and assorted punctuation.
    UIKeyboardTypeURL,                    // A type optimized for URL entry (shows . / .com prominently).
    UIKeyboardTypeNumberPad,              // A number pad (0-9). Suitable for PIN entry.
    UIKeyboardTypePhonePad,               // A phone pad (1-9, *, 0, #, with letters under the numbers).
    UIKeyboardTypeNamePhonePad,           // A type optimized for entering a person's name or phone number.
    UIKeyboardTypeEmailAddress,           // A type optimized for multiple email address entry (shows space @ . prominently).
    UIKeyboardTypeDecimalPad NS_ENUM_AVAILABLE_IOS(4_1),   // A number pad with a decimal point.
    UIKeyboardTypeTwitter NS_ENUM_AVAILABLE_IOS(5_0),      // A type optimized for twitter text entry (easy access to @ #)
    UIKeyboardTypeWebSearch NS_ENUM_AVAILABLE_IOS(7_0),    // A default keyboard type with URL-oriented addition (shows space . prominently).

    UIKeyboardTypeAlphabet = UIKeyboardTypeASCIICapable, // Deprecated

};

由于它们不是2的幂，每个都没有唯一的位，这意味着按位 - 或然后按位 - 并且不能保证工作，并且大多数时间不会。

因此，您必须使用相等性单独比较每个：

BOOL b = (textField.keyboardType == UIKeyboardTypeDefault) || (textField.keyboardType == UIKeyboardTypeASCIICapable) || (textField.keyboardType == UIKeyboardTypeEmailAddress);

iOS Objective-C逻辑或在keyboardType上

3 个答案: