What is the problem with this structure?

Question

#include <iostream>
#include <string>
using namespace std;

void Sepchr(char*k);
void CntWorlds(char*s);
void yearlysalary(float i);

struct employee{

char name[20];
char family[20];
int person;
int age;
float salary;
void Sepchr (char *k){
for(int i=0;*(k+i)!='\0';i++)
{
cout << "*(k+i)";
}}

void CntWorlds(char *s){
int j=0;
for(int i=0;i<strlen(s);i++)
{
if(isalpha(s[i]))
j++;
}
cout << "\nj";
}

void yearlysalary(float i){
float y;
y=i*12;
cout << ("\n%f",y);
}

};

int main(){
struct employee p1;

cout << "name:";
cin >> p1.name ;
cout << "family:";
cin >> p1.family ;
cout << "person:";
cin >> p1.person ;
cout <<"age:";
cin >> p1.age ;
cout << "salary:";
cin >> p1.salary ;
Sepchr(p1.name);
CntWorlds(p1.family);
yearlysalary(p1.salary);
return 0;
}

Answer 1

There are lots of bug in your code strlen () function is a part of string.h header file which belong to c library and you included string header file. Then in line no 36
Where u have used loop in CntWorlds(chars *s ) function here in loop your int variable i=0 is not compatible you need to change it with long unsigned or you can write size_t after that wherr u have written
if(isalpha....)
inside body cout<<"\n "; here j is unnecessary if u want to print value then u need to write like this
cout<<"\n"<<j;
Then in void yearlysalary(...) Function u used cout but you printing value of y in c language syntax format which is incorrect u dont need to write%f or any format specifiers to print values.

you have two more error but i didn't understood those