Reformatting an entered string with spaces in it - c++

Question

Currently I am working on a problem to reformat the inputted string into the odd char then even char with no newline. ex. Input: Good Test. Ouput: Go etodTs. For some reason when I run the program it only outputs a "G".
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;
int main (int argc, char ** argv) {
char sWordOdd[100] = {0};
scanf("%s", sWordOdd);
int iNum = strlen(sWordOdd);
for (int i=0; i<=iNum && i%2==0; i++) {
printf("%c",sWordOdd[i]);
}
for (int a=0; a<=iNum && a%2!=0; a++) {
printf("%c",sWordOdd[a]);
}
printf("\n");
return 0;
}

Answer 1

Your i<=iNum && i%2==0 break condition terminates your loop early. To achieve the effect you want, put an if statement inside the loop, like so:
for (int i = 0; i <= iNum; i++){
if(i % 2 == 0)
printf("%c",sWordOdd[i]);
}
As for your second loop, I think you meant a++ instead of a--, because otherwise you'll try to access the array using a negative index. I think you meant for your loop to look like this:
for (int a = 0; a <= iNum; a++){
if(a % 2 != 0)
printf("%c",sWordOdd[a]);
}
Side note: Notice the spacing between the variables and the operators in my code. Using spaces like this makes your code easier to read.

Answer 2

int main()
{
string line;
getline(cin, line);
for (size_t start : {0,1})
for (size_t ii = start; ii < line.size(); ii += 2)
cout << line[ii];
cout << endl;
}
The above code handles arbitrarily long lines, is C++11 rather than C, and works.

Answer 3

As said before the loops terminate if i<=iNum && i%2==0 is true (for the first loop), which is the case for i=0. The second loop terminates with i=1.
Since you want to iterate all characters you have to move the i%2==0 part out of the for-statement:
for (int i=0; i<=iNum; i++) {
if (i%2==0)
{
printf("%c",sWordOdd[i]);
}
}
The second loop needs to be modified in the same way...

Answer 4

To fix this problem you have to use a function that can see stuff after whitespaces. This function is called getline. But with getline, you have to use string, so in this example I used string. I then found the size with the .size() function and then using just one constraint for the for loop instead of two in the question. I also took off the char array and replaced it with string as stated above. Everything else is pretty much the same. Using this answer lets me not having to use cstring and also simplifying my code into a short, easy way that is easy to follow through.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#define Set(a, s) memset(a, s, sizeof (a))
#define Rd(r) freopen(r, "r", stdin)
#define Wt(w) freopen(w, "w", stdout)
using namespace std;
int main (int argc, char ** argv)
{
string sWordOdd;
getline(cin, sWordOdd, '\n');
int iNum = (int)sWordOdd.size();
for (int i=0; i<iNum; i+=2){
cout << sWordOdd[i];
}
for (int a=1; a<iNum; a+=2){
cout << sWordOdd[a];
}
printf("\n");
return 0;
}