Convert simple javascript program into C

Question

Hello I have some experience with javascript but I would really like to learn how to program in C and one of the ways I am trying to learn is by converting some simple javascript code into C. My current attempt at converting a simple program compiles without any errors however doesn`t produce the answer I want it to. The javscript code produces the correct answer and I wrote it to solve project euler problem number 5 which can be found here: https://projecteuler.net/problem=5

Here is the working js code:

var number = 2520;
var count = 1;
var solved = false;

while (!solved) {
    if (number % count === 0) {
        if (count === 20) {
            solved = true;
            console.log(number);
            } else {
                count++;
            }
    } else {
      number++;
      count = 1;
    }
}

Here is the C conversion which does not work:

#include <stdio.h>

int main () {
    unsigned int number = 2520;
    unsigned int count = 1;
    unsigned int solved = 0;
    while ((solved = 0)) {
        if (number % count == 0) {
            if (count == 20) {
                solved = 1;
                printf("%number");
            } else {
                count++;
            }
        } else {
            number++;
            count = 1;
        }
    }
    return 0;
}

"will not work" isn't super descriptive, but it should be printf("%i\n",number); — IdeaHat
– IdeaHat, Commented Jul 11, 2014 at 19:19

wholerabbit · Accepted Answer · 2014-07-11 21:26:42Z

1

while ((solved = 0)) {

You can use the same syntax you would use in js here, namely, while (!solved), or ==, but just = is an assignment.

printf("%number");

Doesn't mean what you think it means, which is why it's not an actual error (%n is a distinct specifier, and with no corresponding input, you'd get umber as the output). To reproduce console.log() you'd want:

printf("%d\n", (int)number);

Or

printf("%u\n", number);

Notice the explicit \n, since printf() does not add a newline otherwise.

edited Jul 11, 2014 at 21:26

answered Jul 11, 2014 at 19:19

wholerabbit

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2 Comments

mafso Over a year ago

%n means: fetch a pointer to int via va_arg and write the number of bytes written so far to the pointed-to value. It, by no means, means “nothing” and it expects an argument. I'd suggest only writing “(%n is a specifier)”. The rest, to me, seems a little off-topic.

wholerabbit Over a year ago

@mafso Thanks for the correction (edited). I've never used %n for anything, so that was a quick (mis)reading of some docs. My basic point was that it meant something other than what was intended.

Emanuele Paolini · Accepted Answer · 2014-07-11 19:18:42Z

0

Replace

while (solved = 0)

with

while (!solved)

and

print("%number")

with

print("%u\n",number)

answered Jul 11, 2014 at 19:18

Emanuele Paolini

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Comments

IdeaHat · Accepted Answer · 2014-07-11 19:25:52Z

0

I know its already answered, but you should know why.

while ((solved = 0))

Will actually set solved to 0 AND return 0 (which is interpreted as false). So the while loop is exited right away.

printf also takes a pretty strictly formatted string for the first one (just typeing what makes sense is guarenteed to be wrong). The compiler has know way to know what is inside the string is anything other than a string (C++ has (almost) NO reflection, unlike javascript: your written code dissapears into ones and zeros). printf needs to take number in as the second argument. Try printf("%i\n",number);. That says "Print an integer followed by a newline. The integer's value is number."

Welcome to C! Your biggest problem going into is is going to be my biggest problem with Java Script: C is strictly typed with no reflection, while javascript has no types with almost everything relying on some sort of reflection.

answered Jul 11, 2014 at 19:25

IdeaHat

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1 Comment

wholerabbit Over a year ago

Javascript has types, but uses only dynamic type checking, which maybe implies reflection.

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