请教一个弱智问题，关于指针转换的

[Developerljm]

如何把一个整数转换为指针？
怎样处理函数setInfo中的代码？下面代码：

#include <stdio.h>
#include <malloc.h>
#include <string.h>

typedef struct _tagInfo{
char* name ;
int age;
}sINFO;

void setInfo(long l){
sINFO inf;
//inf = (sINFO)(*(void*)l);
//to do
printf("The name: %s",inf.name);//得不到预想的结果啊！
}

main(){
sINFO* s;
s = (sINFO*)malloc(sizeof(sINFO));
s->age=26;
s->name = malloc((sizeof(char)*3));
strcpy(s->name,"jim");
setInfo((long)&s);
free(s->name);
free(&s);
}

[Afo1st]

#include <stdio.h>
#include <malloc.h>
#include <string.h>

typedef struct _tagInfo{
char* name ;
int age;
}sINFO;

void setInfo(long l){
sINFO* inf;
inf = (sINFO*)l;
//to do
printf("The name: %s",inf->name);}

main(){
/*
sINFO* s;
s = (sINFO*)malloc(sizeof(sINFO));
s->age=26;
s->name = (char*)malloc((sizeof(char)*3));
strcpy(s->name,"jim");
setInfo((long)&s);
free(s->name);
free(&s);
*/
sINFO s;
s.age = 26;
s.name = "abc";
setInfo( ( long )&s );
return 0;
}

[doer_ljy]

#include <stdio.h>
#include <malloc.h>
#include <string.h>

typedef struct _tagInfo{
char* name ;
int age;
}sINFO;

void setInfo(long l){
sINFO * inf;
inf = (sINFO *)l;
printf("The name: %s",inf->name);//得不到?想的?果?！
}

main(){
sINFO* s;
s = (sINFO*)malloc(sizeof(sINFO));
s->age=26;
s->name = (char *)malloc((sizeof(char)*3));
strcpy(s->name,"jim");
setInfo((long)s);
free(s->name);
free(&s);
}

[Afo1st]

If the 's' is a pointer, you doesn't need reference '&' anymore.

[zteliubin]

问题不少哦，刚入门吧，兄弟：
1。s->name = malloc((sizeof(char)*3)); //3应修改为4哦，越界！
2。free(&s); //free(s).
3. sInfo inf;// 应为sINFO * inf;
4. inf = (sINFO *)l;

[zteliubin]

If the 's' is a pointer, you doesn't need reference '&' anymore

但是：借人家的东西还是要还的吧(free),:)

[taozhangzhi9]

s->name = malloc((sizeof(char)*3)); //3应修改为4哦，越界！
。free(&s); //free(s).
3. sInfo inf;// 应为sINFO * inf;
4. inf = (sINFO *)l;

[Developerljm]

如果使用这个语句setInfo(long(&s)),那么函数该怎么写啊？