初级问题送分

[faintboy]

main()
{
char *t="oooooop";
doit(t);//为什么t的返回值不变还是"oooooop";而不是ttttttttttttttttttttttttttt
delete [] t;
}

void doit(char * u)
{
u="ttttttttttttttttttttttttttttt";
}

[youarestupid]

main()
{
char *t="oooooop";
doit(t）
}

void doit(char* &u){ //注意这里
u="ttttttttttttttttttttttttttttt";
}

使用引用，这样就可以实现你要的功能了

[begar]

该为返回指针最好！！！

[sywj115447]

还有，程序中没有出现new,就不要使用delete.

[cz_happy]

main()
{
char *t="oooooop";
doit(&t);
delete [] t;
}

void doit(char **u)
{
*u="ttttttttttttttttttttttttttttt";
}

[pcman1990]

main()
{
char *t = new char(100);
lstrcpy(t, "oooooop");
doit(t);
delete[] t;
}

void doit(char * u)
{
lstrcpy(u, "ttttttttttttttttttttttttttttt");
}

[atdawn]

void doit(char * u)
{
strcpy(u,"ttttttttttttttttttttttttttttt");
}

[atdawn]

在C里面参数传递是按值传递，调用doit时，实参t和虚参u的值是一样的，为字符串"oooooop"的地址，当使用u="ttttttttttttttttttttttttttttt";是把
"ttttttttttttttttttttttttttttt"的地址给了u，而t的值还没有变。
所以要改变u指向的字符串区域的内容，而不是改变u的值。

[finalvictory]

up

[boredom_vc]

可改为用引用：
void doit(char &u)
{
}

[atdawn]

引用是C++的技术.