can not execute a C script call "if"
 

Hi,

I am using Ubuntu 7.04.

I am starting to test C on Linux and something I do not really understand has happened. I created this simple script called "if.c":

#include <stdio.h>
int main(void) {
int test = 1;
if (test) {
printf ("*** Value of the test is %d.\n", test);
}
return 0;
}

So, the thing is...

1) If I compile it and name it "if":

$ gcc -g -o if if.c

If compiles OK, but when I execute it:

$ if
>

It doesn't run. I need to stop it with ctrl+c.


2) However, if I compile it with another name:

$ gcc -g -o maybe if.c

It does run!!!

$ maybe
*** Value of the test is 1.

Somebody knows what's going on? Is there any conflict with using "if" as the name?

Yes. Specifically the bash builtin command "if" that you see used in scripts. More specifically, you will have a conflict with any other executable found in your PATH. That's why you should always preface your newly compiled executable with "./" as in:
Also, a C program is not commonly referred to as a "script"; it's more commonly referred to as a program. Calling it a "script" will likely lead to confusion.

No, bash builtin's "if" runs and enters a kind of multiline mode after you hit enter, showing the prompt ">" waiting for you to enter the rest of the IF condition. It will terminate normally (not gracefully) by end-if keyword: "fi". try it:
-Mustafa

What everyone is telling you:
1. You created a compiled program that is named the same thing as a word in bash. Bash is your shell language.

2. will tell you the type of command or file you get by typing the word "if".

3. putting ./ in front of a filename means "use the current working directory" to find the file.