2014练习1
2014-01-06
memcpy函数
功能简介：该函数是标准的C库函数。提供了一般内存的复制，函数原型如下
void memcpy(void *dest,const void *src,size_t count)
从原型中可以看到第二个函数是一个只读（const）的指针常量，即该指正指向的内容在函数体中是不会被改变的。第三个参数是复制内存的个数（类型为size_t，这是一个和机器位数相关的类型，32位系统被定义成unsigned int,64位系统是long unsigned int,C语言中位于头文件stddef.h中）。Linux kernle实现
void *memcpy(void *dest, const void *src, size_t count)
{
assert(dest != NULL && src != NULL);
char *tmp = dest;
const char *s = src;
while (count--)
{
*tmp++ = *s++ ;
}
return dest;
}
若断言（assert）被证明非真，程序将在标准错误流输出一条适当的提示信息，并且执行异常终止。
写个小程序验证下功能。
#include<stdio.h>
#include<string.h>
void *memcpy(void *dest, const void *src, size_t count)
{
if( dest == NULL && src == NULL){
printf("Null pointer in src or dest,pls check!\n");
return NULL;
}
char *tmp = dest;
const char *s = src;
while (count--)
{
*tmp++ = *s++ ;
}
return dest;
}
int main(int argc,char* argv[])
{
char *s="Hello,yunzhi!";
char d[20]={ [0 ... 18] = 'a','\0'};
memcpy(d,s,(strlen(s)+1) );
printf("%s",d);
return 0;
}
编译及结果
gcc -Wall test.c
a.exe
Hello，yunzhi！
一个小小的编程练习
假设某电影院卖票给男人，女人，孩子时票价不同，结算时，知道收入，票数，和各个单价。求可能的男人，女人，孩子组合数。
这是一个高中时代的3元一次方程，列两个方程，求解即可。
编程如下：
#include<stdio.h>
#include<malloc.h>
#include<string.h>
#include<stddef.h>
#include <assert.h>
//The most least number should be defined in CHILDCOST, or will be error
//the three cost should not equal
#define MANCOST 3
#define WOMANCOST 2
#define CHILDCOST 1
#define MANCOST_MINUS_CHILDCOST ((MANCOST)-(CHILDCOST))
#define WOMANCOST_MINUS_MANCOST ((WOMANCOST)-(MANCOST))
#define WOMANCOST_MINUS_CHILDCOST ((WOMANCOST)-(CHILDCOST))
#define size_t int
/************************************************************************
* Algorithm:
* money = man * MANCOST + woman *WOMANCOST + children * CHILDCOST
* people = man + woman + children
*
*************************************************************************/
void flush();
size_t find_woman_men_children(size_t w_offset,size_t c_offset,size_t threshold);
int main(int argc,char *argv[])
{
size_t money,people;
size_t a,b; //temp variable
size_t method = 0; // to get the method numbers
size_t i;
size_t *order[3];
size_t t_women,t_children,t_women_left,t_children_left;
size_t final_m; // temp max number
assert(MANCOST_MINUS_CHILDCOST);
assert(WOMANCOST_MINUS_CHILDCOST);
printf("pls input money:");
while( ( scanf("%d",&money) != 1 ) || ( money < 6 ) ){
printf("\nMoney should be a number and not less than 6,pls input again:");
flush();
}
printf("pls input people:");
while( ( scanf("%d",&people) !=1 ) || ( people < 3 ) ){
printf("\npeople should be a number and not less than 3,pls input again:");
flush();
}
a = money - CHILDCOST*people;
b = people*WOMANCOST - money;
final_m = (a-WOMANCOST_MINUS_CHILDCOST)/MANCOST_MINUS_CHILDCOST;
method = find_woman_men_children(a,b,final_m);
printf("we have %d method to handle this problem!\n",method);
for(i=0;i<3;i++){
order[i] = (size_t *)malloc(sizeof(size_t)*method); // to save the order
if(order[i] == NULL){
printf("failed to alloc space\n");
return -1;
}
memset(order[i],0,method);
}
for(i=0;i<method;i++){
t_women = (a - MANCOST_MINUS_CHILDCOST*(i+1))/WOMANCOST_MINUS_CHILDCOST;
t_women_left = (a -MANCOST_MINUS_CHILDCOST*(i+1))%WOMANCOST_MINUS_CHILDCOST;
t_children = (b - WOMANCOST_MINUS_MANCOST*(i+1))/WOMANCOST_MINUS_CHILDCOST;
t_children_left = (b - WOMANCOST_MINUS_MANCOST*(i+1))%WOMANCOST_MINUS_CHILDCOST;
if( (t_women_left == 0) && (t_children_left== 0) \
&& (t_women > 0) && (t_children > 0) ){
*(order[0]+i) = i + 1;
*(order[1]+i) = t_women;
*(order[2]+i) = t_children;
}
}
printf(" MAN WOMEN CHILDREN\n");
for(i=0;i<method;i++){
printf("%5d%5d%5d\n",*(order[0]+i),*(order[1]+i),*(order[2]+i));
}
return 0;
}
void flush()
{
char c;
while((c=getchar()) != '\n' && c != EOF);
}
size_t find_woman_men_children(size_t w_offset,size_t c_offset,size_t threshold)
{
size_t t_women,t_children,t_women_left,t_children_left;
size_t i;
size_t length=0;
printf(" MAN WOMEN CHILDREN\n");
for(i=0;i<threshold;i++){
t_women = (w_offset - MANCOST_MINUS_CHILDCOST*(i+1))/WOMANCOST_MINUS_CHILDCOST;
t_women_left = (w_offset -MANCOST_MINUS_CHILDCOST*(i+1))%WOMANCOST_MINUS_CHILDCOST;
t_children = (c_offset - WOMANCOST_MINUS_MANCOST*(i+1))/WOMANCOST_MINUS_CHILDCOST;
t_children_left = (c_offset - WOMANCOST_MINUS_MANCOST*(i+1))%WOMANCOST_MINUS_CHILDCOST;
if( (t_women_left == 0) && (t_children_left== 0) \
&& (t_women > 0) && (t_children > 0) ){
//printf("%5d%5d%5d\n",i+1,t_women,t_children);
length++;
}
}
return length;
}
说明： 1.出于简单输入的缘故，程序中宏定义了男人，女人，孩子的票价。导致如果要变更价格需要重新编译。另外，某两个人的票价相同也可以。还需要假设孩子票价最低，如果不是可以将最低的票价放到孩子处，修改打印标题栏即可。 2.由于想将总结果存在一个动态数组中，先计算可能性，然后再进行存储。从代码上看有些重复，因为没有想到什么更好的办法。 3.用到的几个C语言知识：scanf,assert,动态数组，指向数组的指针。当然，还可以继续加ifdef这种宏来加入输入价格这样子的代码。顺便还看了看局部未初始化变量的存储。
