Thread: new to C. need basic help

i dont really understand.

gah. ive never been able to do programming.

Your variable aperturef is a char. That can hold ONE character. %s when used in scanf will absolutely read in AT LEAST two characters, one being what you typed, the second being a zero to indicate the end of the string. So the next location in memory after aperturef will be set to zero. If there's some meaningfull data in that location, it will be overwritten by this zero.

The point about printing it is related to the same - printf() will see %s as an indication that the argument is a pointer to char, where there is zero or more characters followed by a zero-character. E.g. "Hello" would be (( 'H', 'e', 'l', 'l', 'o', 0 )). If you try to do this with a char only, printf will continue along until it finds a zero - whcih if you are not lucky, could be severa dozen or hundreds of bytes of memory! If you are "really unlucky", it runs to a point where there is no valid memory any longer, and the application crashes!

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Mats

ok.

thankyou. that actually makes sense.
so in this case. do i need to change my input method of %s.
or should i change my char to something that can hold more characters.

Originally Posted by k4k45hi

ok.

thankyou. that actually makes sense.
so in this case. do i need to change my input method of %s.
or should i change my char to something that can hold more characters.

Yes, one of those two would work. You can choose either.

By the way, it looks like your code expects 1.4f rather than f1.4 - not sure that's what you meant to do.

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Mats

i think i will go with the changing %s.

now what can i change this to to make it work.
and ive changed the inputs around so that aperturef is first.

Originally Posted by k4k45hi

i think i will go with the changing %s.

now what can i change this to to make it work.
and ive changed the inputs around so that aperturef is first.

Try googling something like "format specifiers scanf" perhaps?

One of the important parts of learning to program is "how to find information about what you are doing" - there won't always be someone else to ask.

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Mats

ok well,
i had done that while waiting for a reply and had replaced the %s with a %c.

so the program no longer comes up with a segmentation fault but instead reads back 0.000000. i will give it a shot and see if i can get it on my own.

and let you know how i go.

thanks for all your help mate.

why is this code returning 0.000000.
ive tried everything i could see, but nothing fixed it.

Code:

  float aperture;
  char aperturef;

  fprintf(output,"please enter aperture value: ");
  fscanf(input,"%c%f",&aperturef,&aperture);

  fprintf(output,"the aperture value is %c%f\n ",aperturef,aperture);

must mean the rest of the code must be interfering with it.

heres the whole thing.

Code:

#include <stdio.h>
#include <string.h>

int main(int argc, char ** argv)


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

  float  exposuretime;
  float aperture;
  char  aperturef;

  fprintf(output,"please enter exposure time: ");
  fscanf(input,"%f",&exposuretime);


  fprintf(output,"please enter aperture value: ");
  fscanf(input,"%c%f",&aperturef,&aperture);

  fprintf(output,"the exposure time is: %f\n ",exposuretime);
  fprintf(output,"the aperture value is %c%f\n ",aperture,aperturef);

}

You may want to check the return result from fscanf(), e.g.

Code:

  int res;
  .... 
  res = fscanf(...);
  if (res != number_of_things_wanted) {
      printf("error...");
  }

For example, fscanf(stdin, "%d %d", &i1, &i2) woould return 2 on successfull read of two integers.

--
Mats

hmm.

ive never used return result

well ive changed aperture with aperturef. but didnt make a difference