fscanf() not asking for input ?

Instead of asking me for the aperture value , the fprintf gets seemingly skipped , and the next thing that prints is that parture = 0.0f , when it hasnt even prompted me for an input :

Code:

#include <stdio.h>
int main(int argc, char * argv[])

{


// Mainline Variable Declarations
FILE * output = stdout;
FILE * input = stdin;



	int exposure=0;		
	float aperture=0;	//Initiate the aperture variable - floating point number


	fprintf(output,"\nEnter Exposure Time: ");	// Prompt the user for exposure time
	fscanf(input,"&#37;5sd",&exposure);			// Store exposure time in the exposure variable
							

		fprintf(output,"Please enter the aperture size - like so : f1.2\n");	// Prompt the user for the aperture size.         			             
                fscanf(input,"f%f",&aperture);    // Store the aperture value in the aperture variable. 
		

		
		   if ((aperture==1.2) || (aperture==1.4) || (aperture==1.8) || (aperture==2.0)  || (aperture==2.8) || (aperture==4.0)  || (aperture==5.4) || (aperture==8.0) || (aperture==11.0) || (aperture==16.0)|| (aperture==22.0) || (aperture==32.0))
	
		{

				fprintf(output,"You have entered a defined aperture.\n");
				fprintf(output,"::::::Values::::::Entered::::::\n\n");
					

				fprintf(output,"The selected Exposure time is %d\n", exposure);         // Print the exposure time entered on screen
				fprintf(output,"The selected Aperture is  : f%.1f\n",aperture);		// Print the aperture on screen
		
                }


			
				
                   else

	        {
					fprintf(output,"You have entered aperture value of f%.1f \n", aperture);
					fprintf(output,"This is an invalid aperture, the program will now exit\n"); 								                    
			                return 0;
				
                }


}

Well the first step is to learn how to indent code.
Set your IDE to only use spaces for indenting, and set the tab stops to 4 spaces. If you use spaces and TABS, the result when posted on any forum is invariably a mess.

> fscanf(input,"f%f",&aperture);
Did you know this is part of the comment line?
Your IDE has perhaps line wrapped this into another line, making you think it's OK. But it isn't.

Consider this, your code reformatted

Code:

#include <stdio.h>
int main(int argc, char *argv[])
{
    // Mainline Variable Declarations
    FILE *output = stdout;
    FILE *input = stdin;

    int exposure = 0;
    float aperture = 0;         //Initiate the aperture variable - floating point number

    fprintf(output, "\nEnter Exposure Time: "); // Prompt the user for exposure time
    fscanf(input, "%5sd", &exposure); // Store exposure time in the exposure variable

    fprintf(output, "Please enter the aperture size - like so : f1.2\n"); // Prompt the user for the aperture size.
    fscanf(input,"f%f",&aperture);                                        // Store the aperture value in the aperture variable. 

    if ((aperture == 1.2) || (aperture == 1.4) || (aperture == 1.8)
        || (aperture == 2.0) || (aperture == 2.8) || (aperture == 4.0)
        || (aperture == 5.4) || (aperture == 8.0) || (aperture == 11.0)
        || (aperture == 16.0) || (aperture == 22.0) || (aperture == 32.0))
    {
        fprintf(output, "You have entered a defined aperture.\n");
        fprintf(output, "::::::Values::::::Entered::::::\n\n");

        fprintf(output, "The selected Exposure time is %d\n", exposure);  // Print the exposure time entered on screen
        fprintf(output, "The selected Aperture is  : f%.1f\n", aperture); // Print the aperture on screen
    }
    else
    {
        fprintf(output, "You have entered aperture value of f%.1f \n",
                aperture);
        fprintf(output, "This is an invalid aperture, the program will now exit\n");  // Alert the user that incorrect value was entered.
        return 0;
    }
}

Perhaps move the return 0 to the actual end of the function.

> fscanf(input, "%5sd", &exposure);
What kind of format is that supposed to be?

Code:

$ gcc -Wall foo.c
foo.c: In function `main':
foo.c:12: warning: char format, different type arg (arg 3)
foo.c:35: warning: control reaches end of non-void function

And to make your floating point comparisons work OK, you REALLY NEED 1.2f, 2.8f, etc to make your compiler avoid converting your aperture to double, which causes it to have greater precision. Certain numbers (such as 1.4, 2.8, 5.6) can not be described EXACTLY in floating point (because they are 2x multiples of 0.7, which becomes 0.69999999...). When the float is converted to double precision it's just filled out with zero's. As you can perhaps understand, the value 0.69999990000 isn't exactly the same as 0.699999999999 - so the compiler says it's not the same).

--
Mats