Homework: Help transalate pseudo code into C code

I don't see any "200 or more" or "more than 200" in the pseudo code. Maybe the right answer is if (i == 200) then exit.

Pretend you're the computer. Every time you write a character to the buffer, you increment i.

Assuming you start at 0, which is normal in C, after you've written a single character, i would be 1. After the second character, i is 2. Extrapolate.

Looks like "If (i >= 200) then exit" to me.

In C code, the angle bracket means exactly the same as in mathematics:
< = smaller than
> = greater than

And if you add an =sign, you get smaller/greater or equal.

If i is the number of characters written to the stack, then i >= 200 (i is greater or equal to 200) is your solution.

Well none of the answers is actually C. You are translating a statement into psudo code (sorry I know its symantics but there is still a distinction). In any case since it doesn't state more than 200 I would vote for 1. If (i >= 200) then exit myself as it states when you hit 200 so its inclusive.

"If we have written 200 characters to the buffer variable, the stack should stop because it cannot hold any more data."

The correct answer is 1: If (i >= 200) then exit.

Here's why.

Your pseudo code specifically says, if 200 characters are written, exit.

If the answer was 2: It would only exit after 201 characters were written (since 201 > 200).

This depends on how i is indexed and how your loop is setup. If it is indexed by 0 (likely) and you had a buffer (buf) you would do something like this.

data_type buf[200];
int i;
for (i = 0; i < 200; i++) {
  // write data to buffer
  buf[i] = data;
}

Therefore, when i becomes 200, the loop exits (and the condition is i < 200). data_type is what ever data you are using in the buffer. If i was indexed starting at one then your loop condition would be i <= 200.

i vote for 1.if(i>=200) since that is what your spec says. C doesnt have 'then' keyword, and it should have been 'exit(0);' instead of just 'exit'.

if(i >= 200){
    exit(0);
}