why does case 2 yields error(error: 'ch' undeclared (first use in this function)). when its perfectly okey to declare 'i' inside

Question

why does case 2 yields error(error: 'ch' undeclared (first use in this function)). when its perfectly okey to declare 'i' inside the loop (used in both) so why can't we declare 'ch' inside loop????

CASE 1 :

char ch;

for (int i = 0; ch != '\n'; i++)
{
scanf("%c", &ch);
name[i] = ch;
}

CASE 2 :

for (int i = 0; char ch != '\n'; i++)
{
scanf("%c", &ch);
name[i] = ch;
}

Answer 1

in the expression: char ch != '\n'

its not correct, at first it seems like it is stating a declaration/initialization of the variable ch but suddenly it changes to a comparison != (not equal to).

I believe it is a nice and interesting syntax error.👍

Answer 2

A for loop is basicaly :

declaration;
while(condition) {
//do this
//update
}

short hand:
for(declaration;condition;update){
// Do this
}

So you need to decalare ch in the declaration part in order to work in all parts

So if you had decalred int i in the condition it also wouldnt work

Answer 3

Arturop gottcha...makes sense

Answer 4

Hiii