Help: Really strange output for simple calculation.

This is a discussion on Help: Really strange output for simple calculation. within the C Programming forums, part of the General Programming Boards category; Hey everyone, I have been reading up on the C language for a couple days and am trying to just ...

  1. #1
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    Question Help: Really strange output for simple calculation.

    Hey everyone,

    I have been reading up on the C language for a couple days and am trying to just play around with it a little bit and get the feel for using functions and whatnot. Well I tried to create a simple program that greets you and then asks you to enter two numbers (which another function in the program would be called on to multiply it and then output it. However, when i build and run the program, it outputs "-1115707584" for 2*2.

    If someone could point out maybe what I'm doing wrong that would be super helpful.

    Thanks,

    gpix13

    Code:
    #include <stdio.h>
    
    int multiply();
    int pause();
    
    int main () {
    	int x;
    	int y;
    	printf("Hello there!\n");
    	pause();
    	printf("Please enter a number:");
    	scanf("%i", &x);
    	printf("Please enter a second number:");
    	scanf("%i", &y);
    	multiply();
    }
    
    int multiply (int x, int y) {
    	int a = x * y;
    	printf("The product of your numbers is: %i", a);
    	return 0;
    }
    
    int pause() {
    	int move_on;
    	printf("Press 'Enter' to continue");
    	move_on=getchar();
    	return(0);
    }

  2. #2
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    You are protyping your multiply function without any arguments.
    Code:
    int multiply();
    When you call multiply you don't provide any arguments.
    Code:
    multiply();
    But your multiply function is expecting two arguments, so your mutilply is working with two undefined variables.
    Code:
    int multiply (int x, int y) {
    So fix your prototype, and pass your function x and y.

    While we are at it, you want to add a return 0 at the end of your main function.

    Dylan

  3. #3
    Algorithm Dissector iMalc's Avatar
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    C++ would not allow you to compile such a mistake, but C just ignores the extra parameters unless you declare the function as multiply(void).

    Make sure you're compiling at a very high warning level.
    My homepage
    Advice: Take only as directed - If symptoms persist, please see your debugger

    Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"

  4. #4
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    Welcome to the forum Gpix13!

    First post, and you already are using *code tags* and *good indentation*.

    Well done! What a pleasure to see.
    Last edited by Adak; 01-16-2010 at 08:03 AM.

  5. #5
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    @Dylan

    Thanks so much. I think I understand what I did wrong now. Here is the new code that actually multiplies 'correctly'.

    Code:
    #include <stdio.h>
    
    int multiply(int x, int y);
    int pause();
    
    int main () {
    	int x;
    	int y;
    	printf("Hello there!\n");
    	pause();
    	printf("Please enter a number:");
    	scanf("%i", &x);
    	printf("Please enter a second number:");
    	scanf("%i", &y);
    	multiply (x, y);
    	return 0;
    }
    
    int multiply (int x, int y) {
    	int a = x * y;
    	printf("The product of your numbers is: %i", a);
    	return 0;
    }
    
    int pause() {
    	int move_on;
    	printf("Press 'Enter' to continue");
    	move_on=getchar();
    	return(0);
    }
    @iMalc
    Completely understandable, I'll be sure to check my warning settings. Thank you

    @Adak
    Thank you very much, I really appreciate it. Hopefully I keep with it to where I can actually do something with a program I make.

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