Question

C++ syntax error : ')'

#include"iostream.h"
struct stu
{
int num;
char *name;
char sex;
float score;
} student[5]={{101,"Li ping",'M',45},
{102,"Zhang ping",'M',62.5},
{103,"He fang",'F',92.5),
{104,"Cheng ling",'F',87},
{105,"Wang ming",'M',58}
};
float ave(struct stu *ps)
{
int i;
float s=0;
for(i=0;i<5;i++,ps++)
s+=ps->score;
return s/5;
}
int nopass(struct stu *ps)
{
int c=0,i;
for(i=0;i<5;i++,ps++)
if(ps->score<60)
c++;
return c;
}
void main()
{
struct stu *ps;
ps=student;
cout<<"aver="<<ave(ps)<<endl;
cout<<"nopass="<<nopass(ps)<<endl;
}

--------------------Configuration: 71 - Win32 Debug--------------------
Compiling...
71.cpp
F:\1\71\71.cpp(10) : error C2059: syntax error : ')'
F:\1\71\71.cpp(11) : error C2078: too many initializers
F:\1\71\71.cpp(13) : error C2059: syntax error : ';'
执行 cl.exe 时出错.

71.obj - 1 error(s), 0 warning(s)

Answer 1

简单看了下是语法错误，{103,"He fang",'F',92.5)中的最后一个）改为｝，就行了

Answer 2

struct stu
{
int num;
char name[256]; //结构体里面尽量不要用指针。这是大神的领域，我等小民老实点吧，虽然浪费，但稳定。
char sex;
float score;
}
.....
void main()
{
stu student[5]={....}; //建议用时再定义 应该能解决 “too many initializers”

stu *ps=&student[0]; //这样写虽然麻烦，但看的清楚。
cout<<"aver="<<ave(ps)<<endl;
cout<<"nopass="<<nopass(ps)<<endl;
}
为啥非要用指针呀，既然操作的是数组，直接用下标更清晰呀。指针应该常用在链表吧。
上面的仅做参考哈，没时间验证。

Answer 3

{103,"He fang",'F',92.5)改为{103,"He fang",'F',92.5}

Answer 4

用构造函数赋值试一下