在Objective C中定义斯威夫特通话variadics C函数 [英] Swift call variadics C function defined in Objective C
问题描述
我总结这个问题的步骤,我已在目标C定义的C函数:
I have summarized the steps of the problem, I have a C function defined in Objective C:
ObjC.h
#import <Foundation/Foundation.h>
void cuslog(NSString *format, ...);
@interface ObjC : NSObject
@end
ObjC.m
ObjC.m
#import "ObjC.h"
@implementation ObjC
@end
void cuslog(NSString *format, ...)
{
// Implementation
}
我接触它桥接-Header.h:
I exposed it in Bridging-Header.h:
#import "ObjC.h"
// Also tried to put this line in bridging header
void cuslog(NSString *format, ...);
在我迅速的意图这样调用该函数:
In swift I intent to call the function like this:
cuslog("Some log")
得到的错误说:
"Use of unresolved identifier cuslog"
什么是调用快捷功能的正确方法是什么?
What is the correct way to call the function in swift?
推荐答案
据雨燕的开发者, ç参数可变型函数不能与斯威夫特variadics 兼容,所以你将不能够直接调用你的函数。
According to the Swift devs, C variadic functions are not compatible with Swift variadics, so you won't be able to call your function directly.
在这个时候,唯一的解决方法是编写C或对象 - 一个非可变参数的包装,并调用从斯威夫特。
The only workaround at this time is to write a non-variadic wrapper in C or Obj-C, and call that from Swift.
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