Simple program not working

Question

Following an example from "C: The Programming Language" (Kernighan & Ritchie), I typed this program into XCode, using the Standard Tool Project under Command Line Utility:
#include <stdio.h>
/* count characters in input; 1st version */
main()
long nc;
nc = 0;
while (getchar() != EOF)
++nc;
printf("%ld\n", nc);
When I clicked Build & Go, I didn't get any errors, and the program ran fine. However, no matter what my input, I don't get an output. What's the problem with this program, and/or my computer?

Answer 1

musicwind95 wrote:
If I don't get a reply within a day or two here, I'm making a new thread with this.
Once the A-team sees you've learned the cleverly hidden secret of how to say thanks around here, a new thread is like wearing a short skirt five minutes before closing.
That said, your question is too interesting to pass up:
When I click Build & Go, it tells me that the line "if (c == ' ' || c == '
' || c = '     ')" has an invalid lvalue in assignment.
Your followers know c='t' was just a typo, but the Lval error it produced is very instructive.
Typos that swap '=' for '==' often cause very tenacious bugs. In many cases the operators can be swapped without making the syntax invalid or causing a runtime error. Not too long ago I found code like this in a game program that had been online for over 12 years (ok, not a game played by half of all teenagers in North America, but literally 12+ yrs just the same):
if (c = 2)
activate = TRUE;
else
continue;
The expression (c=2) not only assigns the value 2 to c, the value of the entire expression takes the value of the Lvar, which has become 2. When evaluating a boolean expression, everything but zero is TRUE, so the else branch is never reached. Most modern compilers will warn about a simple assignment (in the absence of any boolean operators) where a boolean is expected. But some of the compilers in use 12 yrs ago weren't that helpful.
But (c==' ' || c=='
' || c='     ') is much more interesting! Boolean operators take precedence over assignments, so the boolean expressions are evaluated first, as if the larger expression were this:
( (c == ' ' || c == '
' || c) = '     ')
When the inner expression is evaluated, the result is going to be either TRUE or FALSE; i.e. non-zero or zero (almost always TRUE in this case since c has just eaten a non-zero char). So now the expression becomes:
1 = '     ' // <-- attempting to assign a constant (actually the integer 9) to a constant
Whether the left side would evaluate to 1 or 0 is beside the point. Only a variable may be an Lval, hence the error.
- Ray