problems with printf

This is a discussion on problems with printf within the C Programming forums, part of the General Programming Boards category; i have the following code: Code: #include <stdio.h> float z(float x,float y) { double g; g = -x*x + -y*y ...

  1. #1
    Registered User
    Join Date
    Mar 2012

    problems with printf

    i have the following code:

    #include <stdio.h>
    float z(float x,float y)
    	double g;
    	g = -x*x + -y*y + 2;
    	printf("%f", &g);
    	return g;
    int main()
    	float x;
    	float y;
    	double s;
    	x = 0;
    	y = 0;
    	s = z(x,y);
    	return 0;
    When I try compiling with gcc on a linux mint distribution, I get the following warning:
    Lab5Exercise1.c: In function ‘z’:
    Lab5Exercise1.c:11: warning: format ‘%f’ expects type ‘double’, but argument 2 has type ‘double *’

    Since it is just a warning, an output file was created which returns 0.000000 when it should be 2 if my math is correct.

    Before this, i had a loop to test several coordinates pair into the function, but the returned values where really off from where they should be, so i made this basic test code.

    Thx in advance, Donaldgx

  2. #2
    C++ Witch laserlight's Avatar
    Join Date
    Oct 2003
    You want to print a double, so print it, not a pointer to a double:
    printf("%f", g);
    You probably confused this with scanf, where you want to read a double, so you need to pass a pointer to a double so that your double variable is modified.
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  3. #3
    Programming Wraith GReaper's Avatar
    Join Date
    Apr 2009
    "%f" in printf expect the value of a double, not the pointer of a double. That's what your compiler is saying.
    Devoted my life to programming...

  4. #4
    Registered User
    Join Date
    Mar 2012
    Thanks for the help, it works now.

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