A question regarding the syntax of loop in C - c

Question

I saw this little problem in the practice book, which is used to print whatever is typed in backward. And here, I just can't understand this loop for (; i >= 0; i--), why doesn't it assign the i to a variable in the first, is this a normal syntax as well? Thanks!
#include <stdio.h>
#include <string.h>
main() {
int i;
char msg[25];
printf("Type up to 25 characters and then press Enter...\n");
for (i = 0; i < 25; i++) {
msg[i] = getchar(); //Outputs a single character
if (msg[i] == '\n') {
i--;
break;
}
}
putchar('\n'); // One line break after the loop is done.
for (; i >= 0; i--) { // <--- this is the line
putchar(msg[i]);
}
putchar('\n');
return (0);
}

Answer 1

why doesn't it assign the i to a variable in the first
Because i already has the value we want to start at, from prior logic.
is this a normal syntax as well?
Yes. Any of the three parts of the for intro may be empty.

Answer 2

The three expressions which appear between the parentheses in a for() loop are all optional, and theoretically arbitrary.
Theoretically, you could write
for(a = b; c < d; e++)
printf("%d\n", f);
This is probably meaningless and useless, but as far as the C language is concerned, there's nothing wrong with it, and I wouldn't expect a compiler to emit any error messages or even warnings about it.
Now, conventionally, the first expression initializes something, the second expression tests whether we should make another trip through the loop, and the third expression increments. But that's all by convention, and again, all three expressions are optional.
For example, you could write
int i = 5;
for(; i < 10; i++)
printf("%d\n", i);
In this case, i gets its initial value when it's declared. This looks weird, to be sure, and it's arguably bad style, but it's not illegal.
The other expressions are optional, too. For example, I could write it like this:
int i = 5;
for(; i < 10; )
printf("%d\n", i++);
Now i gets incremented within the printf call, so there's no need to increment it in the for loop header. But by now, we've basically got a while loop:
int i = 5;
while(i < 10)
printf("%d\n", i++);
Anyway, this explains the loop you saw. The variable i already had the value it needed, from code up above, so there was no reason to set it in the for loop header.

Answer 3

The initial clause of the for loop can be empty: in the case you mention, i already contains the number of characters stored into the array, 25 or less if the user typed fewer characters before the newline. Yet the program has a bug because i should be decremented to avoid accessing msg beyond the last character stored in the array. Decrementing i if the character typed was '\n' is inconsistent and starting the output at msg[i] is incorrect too.
Your practice book is obsolete and confusing. Consider discarding it and using a more up to date book or online tutorial.
Any of the clauses of the for loop can be empty:
the initial clause may not be needed
the test clause can be empty, causing an infinite loop that ends either via a break or a return statement, or via a direct or indirect call to exit().
the update clause can contain any update expression, or nothing if the code already updates the required variables in the body of the loop, if any update is needed at all.
The first for loop is a very idiomatic example: for (i = 0; i < 25; i++) { ... }
Here is another classic for loop: for (;;) { /* repeat some action for ever */ }
Note the following problems in the posted code:
main must be defined with a return type int. The code uses an obsolete syntax where the return type was implicitly int. This syntax is invalid in C99.
the first for loop stores up to 25 bytes to the array, if the user typed more than 24 characters before the newline or signalled an end of file before the newline.
the program has undefined behavior in this latter case because msg[i] points beyond the end of msg if the user typed 25 characters before the newline.
the contents of the array is printed backwards, with potentially some funny characters such as ÿ at the beginning if indeed the input was ended prematurely.
there is no need to include <string.h>
the parentheses around the return value (0) are useless.
Here is a modified version:
#include <stdio.h>
int main() {
int i, c;
char msg[25];
printf("Type up to 25 characters and then press Enter...\n");
for (i = 0; i < 25; i++) {
c = getchar();
if (c == EOF || c == '\n')
break;
msg[i] = c;
}
putchar('\n'); // One line break after the loop is done.
while (--i >= 0) { // equivalent to for (; --i >= 0;)
putchar(msg[i]);
}
putchar('\n');
return 0;
}