main arguments and functions in c [closed] - c

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#include <stdio.h>
int multiply(int n){
int doenstwork = n * 2;
return doesntwork;
}
int main(int argc, char *argv[]){
int n;
n = atoi(argv[1]);
return multiply(n);
}
I compile but when i run it with an argument I see nothing. Compiled it with gcc -std=c99 filename.c -g.

Answer 1

With the updated code in your question, you have several problems:
You have a typo: you declare a variable doenstwork, but then you try to refer to doesntwork.
Since you say you're able to compile and run your program, what you posted is obviously not the same as what you're actually running. Always copy-and-paste your exact code. If you re-type it, you'll make mistakes like this, and those of us trying to help you can't tell the difference between mistakes you made entering the code in your question and mistakes that are actually in the code you're compiling and running.
The value returned from main is the status of your program, not its output. On Unix-like systems, all but the low-order 8 bits of this value are ignored, so this is not a good way for your program to return detailed information.
Once I corrected the typo, I was able to use your program:
$ gcc c.c -o c
$ ./c 7 ; echo $?
14
but it works only for a small range of values:
$ ./c 1000 ; echo $?
208
The status of a program is used to indicate whether it succeeded or failed, and possibly to return a small amount of information about why it failed. Returning a value of 0 denotes successful execution; returning EXIT_FAILURE (defined in <stdlib.h>) denotes failure. On Unix-like systems, usually EXIT_FAILURE == 1; sometimes a value like 2 is used to denote a different kind of failure. For example, grep returns a status of 0 if a match was found, 1 if no match was found, and 2 if something else went wrong (such as a malformed regular expression or a missing file).
To get information out of a program, the simplest way is to print that information in text form to standard output. You can replace
return multiply(n);
by
printf("%d\n", multiply(n));
You can add a return 0; after that; as of C99 it's not required, but it's not a bad idea to be explicit.

Answer 2

You don't define the variable "doesntwork". This shouldn't compile.
You don't end your function "multiply" with a closing bracket. This shouldn't compile.
You don't include a function prototype for the function "multiply". This shouldn't compile.
If your IDE is so lazy that it will compile this then how do you know that it "didn't work"? Maybe it returned some int to the OS.

Answer 3

Actually Billy Bob is correct. Different IDEs and compilers require different levels of strictness. Full ANSI/ISO compliance, requiring prototypes, etc. can save you a lot of trouble. Your compiler SHOULD require a prototype.
The line:
n = atoi(argv[1]);
is suspicious as well because in C arrays are indexed from zero so the first item in your array should be argv[0].