在Objective-C中循环对象

我有许多

ui按钮

s，我正在以编程方式为它们设置自定义字体，如下所示：

Button1.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button2.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button3.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button4.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button5.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button6.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];

for (int i = 1; i <= 5; i++) {
    Button(i).titleLabel.font = etc.. // How would I get the value of 'i'?
}

NSMutableArray *buttons = [NSMutableArray array];
[buttons addObject: Button1];
[buttons addObject: Button2];
[buttons addObject: Button3];
[buttons addObject: Button4];
[buttons addObject: Button5];
[buttons addObject: Button6];

for (UIButton *button in buttons) {
    button.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
}

这段代码是正确的，它可以工作，但它似乎不是最简洁的编写方式。有没有办法在这些元素之间循环

我曾经想象过这样的事情：

Button1.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button2.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button3.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button4.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button5.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button6.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];

for (int i = 1; i <= 5; i++) {
    Button(i).titleLabel.font = etc.. // How would I get the value of 'i'?
}

NSMutableArray *buttons = [NSMutableArray array];
[buttons addObject: Button1];
[buttons addObject: Button2];
[buttons addObject: Button3];
[buttons addObject: Button4];
[buttons addObject: Button5];
[buttons addObject: Button6];

for (UIButton *button in buttons) {
    button.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
}

---

对于（int i=1；i您可以使用NSArray来迭代：

UIFont *font = [UIFont fontWithName:@"myCustomFont" size:16];
NSArray *buttons = [NSArray arrayWithObjects: Button1, Button2, Button3, 
                                              Button4, Button5, Button6, 
                                              nil];
for (UIButton *button in buttons) {
  button.titleLabel.font = font;
}

嗯，我想你可以把所有的按钮放在一个数组中，然后快速枚举


NSArray *buttons; 
//Put all of your buttons inhere
for (UIButton *button in buttons)
{
    button.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
}

您可以这样做：

Button1.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button2.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button3.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button4.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button5.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
Button6.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];

for (int i = 1; i <= 5; i++) {
    Button(i).titleLabel.font = etc.. // How would I get the value of 'i'?
}

NSMutableArray *buttons = [NSMutableArray array];
[buttons addObject: Button1];
[buttons addObject: Button2];
[buttons addObject: Button3];
[buttons addObject: Button4];
[buttons addObject: Button5];
[buttons addObject: Button6];

for (UIButton *button in buttons) {
    button.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
}

或

for（inti=1；i这个怎么样？currentView是包含所有按钮的UIView

NSArray* buttonsArray = currentView.subviews;

for((UIButton*) button in buttonsArray)
{
     button.titleLabel.font = [UIFont fontWithName:@"myCustomFont" size:16];
}

按钮是在一个数组中还是在单个实例中？您需要多久执行一次（在多少个不同的位置）？您可以将它们添加到NSdictionary中，并循环通过以将它们拉出并更改属性，但如果替换6行代码，则几乎不值得。只有当按钮在某种集合中的顺序已知时，循环通过才有意义，这种集合非常整洁。遗憾的是，我只能选择一个正确的答案！我选择不要忘记
[NSArray arrayWithObjects:]的最后一个参数
nil
以避免未定义的行为。这是一个所谓的可变参数列表，需要
nil
来确定没有更多参数。您不能将对象添加到
NSArray
，因为它是不可变的（在init之后不能修改）。但您可以使用
NSMutableArray
来实现这一点。