为什么函数inittriplet执行完后结构体中的数组a的值是乱的

Question

#include<stdio.h>
#include<stdlib.h>
#define OK 1
#define ERROR 0
typedef int status;
typedef struct{
   int a[3];
}triplet;
status inittriplet(triplet*e,int v1,int v2,int v3)
{
    e=(triplet*)malloc(sizeof(triplet));
    if(!e) return ERROR;
    e->a[0]=v1;
    e->a[1]=v2;
    e->a[2]=v3;
    return OK;
}
status delete(triplet*a){
    if (!a)
    {
        return ERROR;
    }
    free (a);
    return OK;
}
status get(triplet*e,int b,int c){
   if(!e&&1>b&&b>3) return ERROR;
   c=e->a[b-1];
   return OK;
}
status put(triplet*e,int b,int c){
   if (!e&&b<1&&b>3) return ERROR;
   e->a[b-1]=c;
   return OK;
}
status isascending(triplet*e){
    return(e->a[0]<e->a[1]&&e->a[1]<e->a[2]);
}
status max(triplet*e,int b){
    b=(e->a[0]>e->a[1])?((e->a[0]>e->a[2])?e->a[0]:e->a[2]):((e->a[1]>e->a[2])?e->a[1]:e->a[2]);
    return OK;
}

int main(void)
{
    triplet*e;
    int b,c,i;
    inittriplet(e,1,2,4);
    get(e,2,c);
      printf("%d",c);
    put(e,2,5);
    i=0;
    while(i<3)
    {
        printf("%d\t",e->a[i]);
        i++;
    }
     printf("\n");
     max(e,c);
     printf("%d\n",c);
     if(isascending(e))
     for(int k=0;k<3;k++)
     printf("%d\t",e->a[i]);
     system("pause");
     return 0;
}

Answer 1

修改如下，问题见注释，供参考：

#include<stdio.h>
#include<stdlib.h>
#define OK 1
#define ERROR 0
typedef int status;
typedef struct{
        int a[3];
}triplet;

status inittriplet(triplet *&e,int v1,int v2,int v3)//status inittriplet(triplet*e,int v1,int v2,int v3)
{
    e=(triplet*)malloc(sizeof(triplet));
    if(!e) return ERROR;
    e->a[0]=v1;
    e->a[1]=v2;
    e->a[2]=v3;
    return OK;
}
status del(triplet*a){ //status delete(triplet*a)
    if (!a)
    {
        return ERROR;
    }
    free (a);
    return OK;
}
status get(triplet*e,int b,int &c){//status get(triplet*e,int b,int c)
   if(!e&&1>b&&b>3) return ERROR;
   c=e->a[b-1];
   return OK;
}
status put(triplet*e,int b,int c){
   if (!e&&b<1&&b>3) return ERROR;
   e->a[b-1]=c;
   return OK;
}
status isascending(triplet*e){
    return(e->a[0]<e->a[1]&&e->a[1]<e->a[2]);
}
status max(triplet*e,int &b){ //status max(triplet*e,int b)
    b=(e->a[0]>e->a[1])?((e->a[0]>e->a[2])?e->a[0]:e->a[2]):((e->a[1]>e->a[2])?e->a[1]:e->a[2]);
    return OK;
}
int main(void)
{
    triplet*e;
    int b,c,i;
    inittriplet(e,1,2,4);
    get(e,2,c);
    printf("%d\n",c); //printf("%d",c);
    put(e,2,5);
    i=0;
    while(i<3)
    {
        printf("%d\t",e->a[i]);
        i++;
    }
    printf("\n");
    max(e,c);
    printf("%d\n",c);
    if(isascending(e))
      for(int k=0;k<3;k++)
         printf("%d\t",e->a[k]);//printf("%d\t",e->a[i]);
    system("pause");
    return 0;
}

//输出：
//2
//1       5       4
//5
//请按任意键继续. . .