[check code] : Why does Code 1 prints the characters as it is sent through "scanf" and code 2 doesn't. Please Explain

Question

[check code] : Why does Code 1 prints the characters as it is sent through "scanf" and code 2 doesn't. Please Explain
CODE 1 :
#include <stdio.h>
int main()
{
char name[100];
int i;
char ch;
for (i = 0; ch != '\n'; i++)
{
scanf("%c", &ch);
name[i]=ch;
}
name[i] = '\0';
puts(name);
return 0;
}
input : Anirban Mitra
output : Anirban MItra
CODE 2 :
#include <stdio.h>
int main()
{
char name[100];
int i;
for (i = 0; name[i] != '\n'; i++)
{
scanf("%c", &name[i]);
}
name[i] = '\0';
puts(name);
return 0;
}
input : Anirban Mitra
output : prints nothing.. :/ idk what happens here

Answer 1

hmmm
In first case, you must enter \n character explicitly to stop loop..
It works fine then but you are using
ch != '\n' , in first time you are comparing uninitialized ch so it may behave undefined way some times.. Better to use do-while loop here in these cases..
In 2nd program , also it reading undefined times, so there because of infinite loop, you don't getting no output..
You are reading character into name[i] , after this i++ executes so when you comparing name[i] != '\n' , here name[i] is still undefined, not assigned character variable , it is next location you are going to read, not already read character.
So name[i] != '\n' is always true, never false..

Answer 2

Its works.... Just compile again...

Answer 3

Please don't tag languages like "java" when this has nothing to do with it.