您的位置:首页 > 编程语言 > C语言/C++

C++可变参数的另一种实现

2013-05-19 21:23 736 查看
大家熟知的C库函数printf函数就是一个可变参数函数,它是怎么实现的呢?不过他实现是有条件的,必须函数参数的入栈顺序为从右向左的顺序,也即函数的形参,在函数调用之前,必须是最右边的参数先入栈,并且参数都必须通过栈传递,以1个例子说明,如函数func(arg1, arg2,arg3),那么函数的堆栈应是:



ebp是帧指针寄存器,一般用来存取堆栈,有了堆栈结构,下面我们看看C可变参数的具体实现原理:

#include <stdio.h>
enum {
ptChar,
ptInt,
ptFloat,
ptDouble,
};
void printSum(unsigned long paramFormat, ...)
{
/*高16位为可变参数类型,低16位为可变参数个数*/
int paramType = (paramFormat >> 16);
int paramNum = paramFormat & 0xffff;

/*¶mFormat = ebp + 8,第一个参数的地址*/
unsigned long *pArg = ¶mFormat;
/*ebp + 0x0c, 第二个参数地址*/
pArg++;
switch(paramType)
{
case ptChar:
{
int sum = 0;
for (int i = 0; i < paramNum; i++)
{
char *pValue = (char *)pArg;
sum += *pValue;
pArg++;
}
printf("%d\n", sum);
}
break;
case ptInt:
{
int sum = 0;
for (int i = 0; i < paramNum; i++)
{
int *pValue = (int *)pArg;
sum += *pValue;
pArg++;
}
printf("%d\n", sum);
}
break;
case ptFloat:
{
float sum = 0;
/**/
pArg++;

/*浮点参数,堆栈占8个字节,所以指针偏移为8*/
for (int i = 0; i < paramNum; i++)
{
float *pValue = (float *)pArg;
sum += *pValue;
pArg++;
pArg++;
}
printf("%f\n", sum);
}
break;
case ptDouble:
{
double sum = 0;
/*双精度浮点参数,堆栈占8个字节,所以指针偏移为8*/
for (int i = 0; i < paramNum; i++)
{
double *pValue = (double *)pArg;
sum += *pValue;
pArg++;
pArg++;
}
printf("%f\n", sum);
}
break;
default:
printf("unknowned type!\n");
break;
}
}

void main()
{
unsigned long paramFormat = 3;
char a = 1, b = 2, c = 3;
printSum(paramFormat, a, b, c);

paramFormat = ptInt << 16;
paramFormat += 3;
int ia = 1, ib = 2, ic = 3;
printSum(paramFormat, ia, ib, ic);

paramFormat = ptFloat << 16;
paramFormat += 3;
float fa = 1, fb = 2, fc = 3;
printSum(paramFormat, fa, fb, fc);

paramFormat = ptDouble << 16;
paramFormat += 3;
double da = 1, db = 2, dc = 3;
printSum(paramFormat, da, db, dc);
}

上面这种方法对函数参数的入栈顺序有限制,必须从右向左入栈,这就是为什么pascal调用方式不能实现printf的原因,并且函数形参都要通过栈来传递,这对有些编译器为了优化处理,函数参数通过寄存器来传递,从而不满足要求。鉴于次,本文采用C++的默认形参实现可变参数的方法,没有上面的这些限制,下面是实现代码:

#include <stdio.h>
enum {
ptChar,
ptInt,
ptFloat,
ptDouble,
};
void printSum(unsigned long paramType,
void *arg1 = NULL,
void *arg2 = NULL,
void *arg3 = NULL,
void *arg4 = NULL,
void *arg5 = NULL,
void *arg6 = NULL,
void *arg7 = NULL,
void *arg8 = NULL,
void *arg9 = NULL,
void *arg10 = NULL)
{
void *arg[10] = {
arg1,
arg2,
arg3,
arg4,
arg5,
arg6,
arg7,
arg8,
arg9,
arg10,
};
switch(paramType)
{
case ptChar:
{
int sum = 0;
for (int i = 0; i < 10; i++)
{
if (arg[i] != NULL)
{
char *pValue = (char *)arg[i];
sum += *pValue;
}
else
break;
}
printf("%d\n", sum);
}
break;
case ptInt:
{
int sum = 0;
for (int i = 0; i < 10; i++)
{
if (arg[i] != NULL)
{
int *pValue = (int *)arg[i];
sum += *pValue;
}
else
break;
}
printf("%d\n", sum);
}
break;
case ptFloat:
{
float sum = 0;
for (int i = 0; i < 10; i++)
{
if (arg[i] != NULL)
{
float *pValue = (float *)arg[i];
sum += *pValue;
}
else
break;
}
printf("%f\n", sum);
}
break;
case ptDouble:
{
double sum = 0;
for (int i = 0; i < 10; i++)
{
if (arg[i] != NULL)
{
double *pValue = (double *)arg[i];
sum += *pValue;
}
else
break;
}
printf("%f\n", sum);
}
break;
default:
printf("unknowned type!\n");
break;
}
}

void main()
{
unsigned long paramType = ptChar;
char a = 1, b = 2, c = 3;
printSum(paramType, &a, &b, &c);

paramType = ptInt;
int ia = 1, ib = 2, ic = 3;
printSum(paramType, &ia, &ib, &ic);

paramType = ptFloat;
float fa = 1, fb = 2, fc = 3;
printSum(paramType, &fa, &fb, &fc);

paramType = ptDouble;
double da = 1, db = 2, dc = 3;
printSum(paramType, &da, &db, &dc);
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: