Any idea how to get this warning to go away. The code runs fine, I just don't like warnings in my project. I have never come across this warning before so I am making a mountain out of a mole hill I think. Boxing syntax? Is that referring to the square brackets? This warning shows up when trying to modernize an old project in Objective-C using Xcode.
任何想法如何让这个警告消失。代码运行正常，我只是不喜欢我的项目中的警告。我之前从来没有遇到过这个警告，所以我想在一座鼹鼠山上建造一座山。拳击语法？这是指方括号吗？尝试使用Xcode在Objective-C中对旧项目进行现代化时会出现此警告。
for (int i = 0; i <= 6; i++) {
[sequence addObject:[NSNumber numberWithInt:random()% 6]];
}
It throws an error stating:
它抛出一个错误说明：
Converting to boxing syntax requires casting 'long' to 'int'
转换为拳击语法需要将'long'转换为'int'
1 个解决方案
#1
1
"Boxing" refers to the new syntax for boxing C expressions, e.g.
“拳击”指拳击C表达式的新语法，例如
NSNumber *n = @(2*3+4)
instead of
代替
NSNumber *n = [NSNumber numberWithInt:(2*3+4)];
(see http://clang.llvm.org/docs/ObjectiveCLiterals.html for details).
（有关详细信息，请参阅http://clang.llvm.org/docs/ObjectiveCLiterals.html）。
In your case,
在你的情况下，
[NSNumber numberWithInt:random()% 6]
creates a number object containing an int, but
创建一个包含int的数字对象，但是
@(random()% 6)
would create a number object containing a long, because random() is declared as
会创建一个包含long的数字对象，因为random（）被声明为
long random(void);
So to get exactly the same behavior as before the conversion, you would have to write
因此，为了获得与转换之前完全相同的行为，您必须编写
[NSNumber numberWithInt:(int)(random()% 6)]
which is then converted to
然后转换为
@((int)(random()% 6))
If you don't care which "flavor" of number object you get, then just convert that line manually to
如果你不关心你得到的数字对象的“味道”，那么只需手动将该行转换为
[sequence addObject:@(random()% 6)];
but Xcode cannot decide that for you.
但Xcode无法为您决定。
#1
1
"Boxing" refers to the new syntax for boxing C expressions, e.g.
“拳击”指拳击C表达式的新语法，例如
NSNumber *n = @(2*3+4)
instead of
代替
NSNumber *n = [NSNumber numberWithInt:(2*3+4)];
(see http://clang.llvm.org/docs/ObjectiveCLiterals.html for details).
（有关详细信息，请参阅http://clang.llvm.org/docs/ObjectiveCLiterals.html）。
In your case,
在你的情况下，
[NSNumber numberWithInt:random()% 6]
creates a number object containing an int, but
创建一个包含int的数字对象，但是
@(random()% 6)
would create a number object containing a long, because random() is declared as
会创建一个包含long的数字对象，因为random（）被声明为
long random(void);
So to get exactly the same behavior as before the conversion, you would have to write
因此，为了获得与转换之前完全相同的行为，您必须编写
[NSNumber numberWithInt:(int)(random()% 6)]
which is then converted to
然后转换为
@((int)(random()% 6))
If you don't care which "flavor" of number object you get, then just convert that line manually to
如果你不关心你得到的数字对象的“味道”，那么只需手动将该行转换为
[sequence addObject:@(random()% 6)];
but Xcode cannot decide that for you.
但Xcode无法为您决定。