I've been working through some examples in The C Programming Language by K&R, but I'm kind of confused on a test that I did on exercise 1-12 where I switched the if and else statements.
The question goes as: Write a program that prints its input one word per line.
So basically to test out the \n while ignoring multiple spaces in between words, I first wrote this.
#include <stdio.h>
int main()
{
int c;
while ((c = getchar()) != EOF) {
if (c == ' ' || c == '\t') {
printf("\n");
}
else {
putchar(c);
}
}
}
In this one, it would print out a new paragraph upon seeing blank spaces or tabs, and if it doesn't, then it copies exactly the input.
This code produced the results that I wanted (well there's still extra paragraphs from extra spaces in between words on input, but I'm ignoring that for now)
Then I decided to see what happened if I switched my if and else statements while changing the conditions, but it turns out that the result is not the same.
#include <stdio.h>
int main()
{
int c;
while ((c = getchar()) != EOF) {
if (c != ' ' || c != '\t') {
putchar(c);
}
else {
printf("\n");
}
}
}
In this one, I wanted it to print out whatever wasn't a blank space or tab, and if it doesn't, it would create a new paragraph.
I expected that the new conditions would produce the same results. Instead it just copied what I put inputted. Does anyone know why this doesn't work?
if (c != ' ' || c != '\t') {
This condition checks to see whether either c is not a space, or c is not a tab. This is always true, since c can't be both a space and a tab at the same time. Try:
if (c != ' ' && c != '\t') {
You can read about this principle in more detail at De Morgan's Laws.
An alternative approach would let the compiler do the logical negation for you:
if (!(c == ' ' || c == '\t')) {
however, some people (me included) would consider that harder to read because of the extra parentheses and the tiny ! on the left.
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