用C语言模仿Python函数

首先得说明一点，C 语言不是函数式编程语言，要想进行完全的函数式编程，还得先写个虚拟机，然后再写个解释器才行（相当于 CPython ）。

下面我们提供一个例子，说明即便不写虚拟机， C 语言函数也是可以“适度地模仿” Python 函数。

我们有如下的 Python 程序：

def line_conf(a, b):
    def line(x):
        return a*x + b
    return line

line1 = line_conf(1, 1)
line2 = line_conf(4, 5)
print(line1(5), line2(5))

我们在C程序中适度地模拟其中的line_conf函数：

/* MIT License

Copyright (c) 2017 Yuandong-Chen

Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:

The above copyright notice and this permission notice shall be included in all
copies or substantial portions of the Software.

THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
SOFTWARE. */

///////////////////////////////////////////////////////////////////////////////

// Note: The C program is almost equivalent to the Python program as follows:
// def line_conf(a, b):
//     def line(x):
//         return a*x + b
//     return line
//
// line1 = line_conf(1, 1)
// line2 = line_conf(4, 5)
// print(line1(5), line2(5))

#include <stdio.h> 
#include <stdlib.h> 
#include <unistd.h> 
#include <stdarg.h>

typedef int Func();

Func *line_conf(int x, int y,...)
{ 
    va_list ap; 
    va_start(ap, y);

    asm volatile(
        "push %%eax
	"
        "subl $40, %%esp
	"
        "movl 8(%%ebp), %%eax
	"
        "movl %%eax, -36(%%ebp)
	"
        "movl 12(%%ebp), %%eax
	"
        "movl %%eax, -40(%%ebp)
	"
        "addl $40, %%esp
	"
        "pop %%eax
	"
        :::"memory"
        );

if(va_arg(ap,int) == 1){

LINE:

    asm volatile(
        "push %%ebp
	"
        "movl %%esp, %%ebp
	"
        "movl 8(%%ebp), %%eax
	"
        "imul -36(%%ebp), %%eax
	"
        "addl -40(%%ebp), %%eax
	"
        "movl %%ebp, %%esp
	"
        "pop %%ebp
	"
        "ret
	"
        :::"memory","%eax"
        );
}   
__END:  
    va_end(ap);
    return (Func *)(&&LINE);
}

int main(int argc, const char *argv[]){  
    printf("====TEST START====
");
    printf("34*234+6 ?= %d
",line_conf(34,6)(234));
    printf("1*3+2 ?= %d; 324*65+3 ?= %d; 13*66+2 ?= %d
",line_conf(1,2)(3),line_conf(324,3)(65),line_conf(13,2)(66));

    int fd = line_conf(1,6)(4);
    Func *fun = line_conf(3,3);
    int a = 1;  // Limited point
    printf("3*3+3 ?= %d; 1*4+6 ?= %d
",fun(3),fd);
    printf("====TEST END====
");
    return 0;  
}

// Compile it by the following command:
// gcc -m32 -O0 -fno-stack-protector CFunctional.c; ./a.out
// The terminal output should looks like:
// ====TEST START====
// 34*234+6 ?= 7962
// 1*3+2 ?= 5; 324*65+3 ?= 21063; 13*66+2 ?= 860
// 3*3+3 ?= 12; 1*4+6 ?= 10
// ====TEST END====
//Note: The limitation happens between line 86 and line 88, we cannot insert any function here
// whose stack is larger than 40 bytes.(Why is 40? check the inline assembler language)

结果在MacOSX和Ubuntu上（i386）都能通过简单的测试。但是可以看到，仅仅是简单的模拟，我们也得用到大量（按比例）的汇编，可读性很差，而且模拟程度非常有限，代码长度也更长。

注意：这只是个玩具，别这么写代码，除非想搞破坏（充斥着各种漏洞，极易被侵入）。