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

不,那是c++语言的特性。

其他回答

我们可以为默认值创建(仅)使用命名形参的函数。这是bk.答案的延续。

#include <stdio.h>                                                               

struct range { int from; int to; int step; };
#define range(...) range((struct range){.from=1,.to=10,.step=1, __VA_ARGS__})   

/* use parentheses to avoid macro subst */             
void (range)(struct range r) {                                                     
    for (int i = r.from; i <= r.to; i += r.step)                                 
        printf("%d ", i);                                                        
    puts("");                                                                    
}                                                                                

int main() {                                                                     
    range();                                                                    
    range(.from=2, .to=4);                                                      
    range(.step=2);                                                             
}    

C99标准定义了初始化中后面的名称覆盖前面的项。我们也可以有一些标准的位置参数,只是相应地改变宏和函数的签名。默认值参数只能在命名参数样式中使用。

项目输出:

1 2 3 4 5 6 7 8 9 10 
2 3 4 
1 3 5 7 9

不,但你可以考虑使用一组函数(或宏)来近似使用默认参数:

// No default args
int foo3(int a, int b, int c)
{
    return ...;
}

// Default 3rd arg
int foo2(int a, int b)
{
    return foo3(a, b, 0);  // default c
}

// Default 2nd and 3rd args
int foo1(int a)
{
    return foo3(a, 1, 0);  // default b and c
}

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__)

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)

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。