有没有一种方法可以在C语言中指定函数的默认参数?
当前回答
是的,有了C99的功能,你就可以这样做了。这种工作方式不需要定义新的数据结构,也不需要函数在运行时决定如何调用 没有任何计算开销。
有关详细的解释,请参阅我的帖子
http://gustedt.wordpress.com/2010/06/03/default-arguments-for-c99/
Jens
其他回答
我改进了Jens Gustedt的回答:
不使用内联函数 默认值在预处理期间计算 模块化可重用宏 可以设置编译器错误,使其与允许的默认值参数不足的情况有意义地匹配 如果实参类型保持无二义性,则不需要默认值来形成形参列表的尾部 与C11 _Generic互操作 根据参数的数量来改变函数名!
variadic.h:
#ifndef VARIADIC
#define _NARG2(_0, _1, _2, ...) _2
#define NUMARG2(...) _NARG2(__VA_ARGS__, 2, 1, 0)
#define _NARG3(_0, _1, _2, _3, ...) _3
#define NUMARG3(...) _NARG3(__VA_ARGS__, 3, 2, 1, 0)
#define _NARG4(_0, _1, _2, _3, _4, ...) _4
#define NUMARG4(...) _NARG4(__VA_ARGS__, 4, 3, 2, 1, 0)
#define _NARG5(_0, _1, _2, _3, _4, _5, ...) _5
#define NUMARG5(...) _NARG5(__VA_ARGS__, 5, 4, 3, 2, 1, 0)
#define _NARG6(_0, _1, _2, _3, _4, _5, _6, ...) _6
#define NUMARG6(...) _NARG6(__VA_ARGS__, 6, 5, 4, 3, 2, 1, 0)
#define _NARG7(_0, _1, _2, _3, _4, _5, _6, _7, ...) _7
#define NUMARG7(...) _NARG7(__VA_ARGS__, 7, 6, 5, 4, 3, 2, 1, 0)
#define _NARG8(_0, _1, _2, _3, _4, _5, _6, _7, _8, ...) _8
#define NUMARG8(...) _NARG8(__VA_ARGS__, 8, 7, 6, 5, 4, 3, 2, 1, 0)
#define _NARG9(_0, _1, _2, _3, _4, _5, _6, _7, _8, _9, ...) _9
#define NUMARG9(...) _NARG9(__VA_ARGS__, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0)
#define __VARIADIC(name, num_args, ...) name ## _ ## num_args (__VA_ARGS__)
#define _VARIADIC(name, num_args, ...) name (__VARIADIC(name, num_args, __VA_ARGS__))
#define VARIADIC(name, num_args, ...) _VARIADIC(name, num_args, __VA_ARGS__)
#define VARIADIC2(name, num_args, ...) __VARIADIC(name, num_args, __VA_ARGS__)
// Vary function name by number of arguments supplied
#define VARIADIC_NAME(name, num_args) name ## _ ## num_args ## _name ()
#define NVARIADIC(name, num_args, ...) _VARIADIC(VARIADIC_NAME(name, num_args), num_args, __VA_ARGS__)
#endif
简化使用场景:
const uint32*
uint32_frombytes(uint32* out, const uint8* in, size_t bytes);
/*
The output buffer defaults to NULL if not provided.
*/
#include "variadic.h"
#define uint32_frombytes_2( b, c) NULL, b, c
#define uint32_frombytes_3(a, b, c) a, b, c
#define uint32_frombytes(...) VARIADIC(uint32_frombytes, NUMARG3(__VA_ARGS__), __VA_ARGS__)
使用_Generic:
const uint8*
uint16_tobytes(const uint16* in, uint8* out, size_t bytes);
const uint16*
uint16_frombytes(uint16* out, const uint8* in, size_t bytes);
const uint8*
uint32_tobytes(const uint32* in, uint8* out, size_t bytes);
const uint32*
uint32_frombytes(uint32* out, const uint8* in, size_t bytes);
/*
The output buffer defaults to NULL if not provided.
Generic function name supported on the non-uint8 type, except where said type
is unavailable because the argument for output buffer was not provided.
*/
#include "variadic.h"
#define uint16_tobytes_2(a, c) a, NULL, c
#define uint16_tobytes_3(a, b, c) a, b, c
#define uint16_tobytes(...) VARIADIC( uint16_tobytes, NUMARG3(__VA_ARGS__), __VA_ARGS__)
#define uint16_frombytes_2( b, c) NULL, b, c
#define uint16_frombytes_3(a, b, c) a, b, c
#define uint16_frombytes(...) VARIADIC(uint16_frombytes, NUMARG3(__VA_ARGS__), __VA_ARGS__)
#define uint32_tobytes_2(a, c) a, NULL, c
#define uint32_tobytes_3(a, b, c) a, b, c
#define uint32_tobytes(...) VARIADIC( uint32_tobytes, NUMARG3(__VA_ARGS__), __VA_ARGS__)
#define uint32_frombytes_2( b, c) NULL, b, c
#define uint32_frombytes_3(a, b, c) a, b, c
#define uint32_frombytes(...) VARIADIC(uint32_frombytes, NUMARG3(__VA_ARGS__), __VA_ARGS__)
#define tobytes(a, ...) _Generic((a), \
const uint16*: uint16_tobytes, \
const uint32*: uint32_tobytes) (VARIADIC2( uint32_tobytes, NUMARG3(a, __VA_ARGS__), a, __VA_ARGS__))
#define frombytes(a, ...) _Generic((a), \
uint16*: uint16_frombytes, \
uint32*: uint32_frombytes)(VARIADIC2(uint32_frombytes, NUMARG3(a, __VA_ARGS__), a, __VA_ARGS__))
和可变函数名选择,不能与_Generic组合:
// winternitz() with 5 arguments is replaced with merkle_lamport() on those 5 arguments.
#define merkle_lamport_5(a, b, c, d, e) a, b, c, d, e
#define winternitz_7(a, b, c, d, e, f, g) a, b, c, d, e, f, g
#define winternitz_5_name() merkle_lamport
#define winternitz_7_name() winternitz
#define winternitz(...) NVARIADIC(winternitz, NUMARG7(__VA_ARGS__), __VA_ARGS__)
还有一个选项使用结构体:
struct func_opts {
int arg1;
char * arg2;
int arg3;
};
void func(int arg, struct func_opts *opts)
{
int arg1 = 0, arg3 = 0;
char *arg2 = "Default";
if(opts)
{
if(opts->arg1)
arg1 = opts->arg1;
if(opts->arg2)
arg2 = opts->arg2;
if(opts->arg3)
arg3 = opts->arg3;
}
// do stuff
}
// call with defaults
func(3, NULL);
// also call with defaults
struct func_opts opts = {0};
func(3, &opts);
// set some arguments
opts.arg3 = 3;
opts.arg2 = "Yes";
func(3, &opts);
使用宏的另一个技巧:
#include <stdio.h>
#define func(...) FUNC(__VA_ARGS__, 15, 0)
#define FUNC(a, b, ...) func(a, b)
int (func)(int a, int b)
{
return a + b;
}
int main(void)
{
printf("%d\n", func(1));
printf("%d\n", func(1, 2));
return 0;
}
如果只传递一个参数,则b接收默认值(在本例中为15)
我知道如何更好地做到这一点。 您只需将NULL赋值给一个参数,因此,您将没有值。然后检查参数值是否为NULL,将其更改为默认值。
void func(int x){
if(x == NULL)
x = 2;
....
}
尽管如此,它还是会引起警告。一个更好的选择是,如果参数值是:
void func(int x){
if(x == 1)
x = 2;
....
}
在上面的例子中,如果x是1,函数将其更改为2;
感谢@user904963,编辑: 如果必须覆盖所有的数字范围,那么添加另一个参数来告诉函数是否将形参设置为默认并不难
void func(int x, bool useDefault){
if(useDefault) //useDefault == true
x = 2;
....
}
但是,请记住包含stdboolh
不是真的。唯一的方法是编写一个varargs函数,并手动填充调用者没有传递的参数的默认值。