When I call a function within a function it skips code and terminates program?

Question

It is a big program. I stripped off unnecessary code. I left only one of the key functions

When I call ss(); in any function the function gives control back to main() without accepting a string. The code works if I don't use a function to accept the string. I can't find anything wrong with it.

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<ctype.h>
void ss();
void casechange();
using namespace std;
char str[100];
int main (){
int choice;

cout<<"Make a choice"<<endl; 
cout<<"Press 1 to change the case of alphabets"<<endl;
cout<<"Press 2 to count number of vowels"<<endl;
cout<<"Press 3 to check if entered string is a palindrome or not"<<endl;
cout<<"Press 4 to reverse a string"<<endl;
cout<<"Press 5 to count number of words"<<endl;
cin>>choice;
switch(choice){
case 1: casechange();
    break; 
case 2: vowelcount();
        break;
case 3:pal();
    break;
case 4: rev();
    break;
case 5: wordcount();
    break;
default: cout<<"Wrong choice"<<endl;
 }
 return 0;
}
void casechange(){
ss();
for(int i=0;str[i]!='\0';i++)
 {
 if(isupper(str[i]))
 str[i]=tolower(str[i]);
 else str[i]=toupper(str[i]);
 }
 puts(str);
}
void ss()
{
cout<<"Enter a string"<<endl;
 gets(str);
}

p.s. I am using code blocks. gcc compiler I guess.

Answer 1

You asked user to make a choice. User typed a number and enter. You then read a single character. enter still sitting there in the buffer. When it comes to gets, it reads it as an empty string.

Also please pay attention to all the comments about IO, gets etc.