编程例子：通过交易码获取交易名称（待解决）

1.通过结构数组解决

A1:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MIX_SIZE 4
struct tag
{
	char code[8];
	char jymc[100];
}tags[MIX_SIZE]={
											{"1000","name1"},
											{"1001","name2"},
											{"1002","name3"},
											{"1003","name4"}
								};
									
char *getname(char *code)
{
	int i;
	for(i=0;i<MIX_SIZE;i++)
	{
		if(strcmp(tags[i].code,code)==0)
		{
				return tags[i].jymc;
		}
	}
	return NULL;
}
int main()
{ 
	char code[]="1000",name[100],*p;									
	p=getname(code);
	printf("name=%s\n",p);
	exit(1);
}									

　在上面的基础上改了一点点，用指针指向tag结构体，不知怎么就不行了，等下班后再查明原因

A2:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct tag
{
    char code[8];
    char jymc[100];
}tags[]={
                                            {"1000","name1"},
                                            {"1001","name2"},
                                            {"1002","name3"},
                                            {"1003","name4"},
                                            {NULL,NULL}
                 };
                                    
char *getname(char *code)
{
    int i;
    struct tag *p=tags;
    while(p->code!=NULL)
    {
        if(strcmp(p->code,code)==0)
        {
                return p->jymc;
        }
        printf("test %s\n",p->code);
        p++;
    }
    return NULL;
}
int main()
{ 
    char code[]="1010",name[100],*p;                                    
    p=getname(code);
    if(p!=NULL)
    {
        printf("name=%s\n",p);
    }    
    exit(1);
}    
运行结果：
[root@localhost program]# ./a.o
test 1000
test 1001
test 1002
test 1003
test 
test 
test 
test 
test 
test 
test 
test 
test 
test 
test 
test 
test 
test 
test 
test 
test 
Segmentation fault (core dumped)                                

 再稍微修改了一下，又可以了，与A2不同的是最后一个元素修改成一个指定字符串判定数组结束

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct tag
{
    char code[8];
    char jymc[100];
}tags[]={
                                            {"1000","name1"},
                                            {"1001","name2"},
                                            {"1002","name3"},
                                            {"1003","name4"},
                                            {"EOF",NULL}
                 };
                                    
char *getname(char *code)
{
    int i;
    struct tag *p=tags;
    while(strcmp(p->code,"EOF")!=0)
    {
        if(strcmp(p->code,code)==0)
        {
                return p->jymc;
        }
        p++;
    }
    return NULL;
}
int main()
{ 
    char code[]="1010",name[100],*p;                                    
    p=getname(code);
    
    if(p!=NULL)
        printf("name=%s\n",p);
    else 
        printf("not found \n");
        
    exit(1);
}                                    

 

 

2.通过三维数组解决，下班后再补上