How to count characters that appear twice consecutively - c++

Question

I'm using C++. So far, my code goes like this:
#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <windows.h>
int main() {
char word[100]; int ctr, count = 0;
printf("Enter string: "); gets(word);
ctr = 1;
while (word[ctr] != '\0') {
if (word[ctr-1] == word[ctr]) count++;
ctr++;
}
printf("%d", count);
return 0;
}
Sample Run
Enter string: mississippi
3
Enter string: mmmmrrnzzz
6
I've got the first sample run correctly (mississippi) with only 3 characters appearing twice consecutively but not on the second sample run (mmmmrrnzzz) with output 6.
My problem is that, it should not be 6 but 4 instead. 1 for the first two consecutive m, another separate 1 for the next two consecutive m, 1 for r, and 1 for z. I want a separate count for the first "mm" and the second "mm" and also for the "zz" but I don't know how.
I'm a freshman and very new to programming. I wish I could explain better. I'm hoping you could help me. Thank you.

Answer 1

In case of multiple couples like mmmm you need to make a double incrementation of your counter:
#include <stdio.h>
#include <string.h>
int main()
{
char word[100];
int ctr;
int count = 0;
printf("Enter string: ");
gets(word);
int len = strlen(word);
ctr = 1;
while (ctr<len) {
if (word[ctr-1] == word[ctr])
{
count++;
ctr++;
}
ctr++;
}
printf("%d", count);
return 0;
}

Answer 2

First of all the program looks like a C program. In fact you are not using C++. You are using C.:) At least for example in C++ you should use header
#include <cstdio>
instead of
#include <stdio.h>
and so on.
And moreover it has a bug because in general the string can be empty. In this case the condition of the loop skips the first zero-terminating character and the program has undefined behaviour.
Here is a correct approach
#include <stdio.h>
int main( void )
{
const char *s = "mmmmrrnzzz";
size_t count = 0;
while ( *s++ )
{
if ( *s == *( s - 1) )
{
++count;
++s;
}
}
printf( "count = %zu\n", count );
}
The output is
count = 4
Take into account that function gets is unsafe and is not supported by the C (or C++) Standard any more.
You should use function fgets instead of gets.

Answer 3

This will work
#include <stdio.h>
#include <string.h>
int main() {
char word[100]; int ctr, count = 0;
printf("Enter string: "); gets(word);
int len=strlen(word);
ctr = 1;
while (ctr<len) {
if (word[ctr-1] == word[ctr])
{
count++;
ctr++;
}
ctr++;
}
printf("%d", count);
return 0;
}

Answer 4

A standard library version:
#include <algorithm>
#include <iostream>
#include <string>
int main()
{
int count{};
std::string s;
std::cin >> s;
for (auto it = s.begin(); (it = std::adjacent_find(it, s.end())) != s.end(); it += 2)
++count;
std::cout << count << '\n';
}