Unlogical output of C code

Question

I wrote some lines of following code.

#include "stdio.h"

int main(){
    float t,res;
    char c;
    scanf("%f",&t); 
    getchar();
    scanf("%s",&c);
    if (c=='R') res = 4/5 * t;
    else if (c=='F') res = (9/5 * t) + 32;
    else if (c=='K') res = t + 273;
    printf("%.2f",&res);
    return 0;
}

I have no idea why the output shown 0.00 when I give t = 25 and c = 'R'. The console looks like this.

25
R
0.00

Would any body give me an explanation?

Answer 1

• scanf("%s",&c); has a big chance to cause out-of-range access, which invokes undefined behavior. Use scanf("%c",&c); (%c instead of %s) to read only one character.
• printf("%.2f",&res); will invoke undefined behavior by passing data having wrong type to printf(). Use printf("%.2f",res); (without &) to print data having type double (or float, which will automatically converted to double in variable-length arguments)
• The expressions res = 4/5 * t and res = (9/5 * t) + 32 may not do what you want because 4/5 and 9/5 are integer divisions, in which remainders are truncated. Try using res = 4.f/5 * t and res = (9.f/5 * t) + 32
• You should check whether scanf()s are successful and the data read is valid in order not to invoke undefined behavior by using values of uninitialized variables having automatic-storage duration, which are indeterminate.

Answer 2

• First of all, you use printf in the following way :

  printf("%.2f",&res);
  

  but according to what you want to print, you should change it to :

  printf("%.2f", res);
  

• Secondly, as you are reading a simple character, modify this line :

  scanf("%s",&c);
  

  to :

  scanf("%c", &c);
  

• Finally and most importantly, the statement :

  res = 4/5 * t;
  

  will compute the integer quotient (which will give 0 in case of 4/5), but you want to do the float division. So you should change it to :

  res = (float)4/5 * t;
  

  or :

  res = 4/(float)5 * t;
  

  as one of the two members of the division needs to be float to get the result you want.

  Of course instead of casting you could also do :

  res = 4.0/5 * t;
  

  or :

  res = 4/5.0 * t;
  

  The same should also be applied to 9/5*t, which should change to 9.0/5*t or 9/5.0*t.

Answer 3

Of course,

printf("%.2f",res);  // without &

and

4.0/5   9.0/5  // .0 makes literal of double type

or better

4.0F/5   9.0F/5  // .0F makes literal of float type

in the expressions.

But my second advise is for the following code

scanf("%f",&t); 
getchar(); // cleaning the input stream from one `'\n'`
scanf("%s",&c);

It is better to write this part as

scanf("%f",&t);
while (getchar() != '\n'); // cleaning the input stream from everithing
c = getchar(); // reading one character

this approach allows to read correct character after input something like

1.34abdfsk

R

In that case abdfsk as well as \n will be skipped by loop while (getchar() != '\n');

Answer 4

Change

scanf("%s",&c);

to

scanf("%c",&c); // %c format specifier for character

Change

printf("%.2f",&res);

to

printf("%.2f",res); // No address of operator required

In

res = 4/5 * t;

the / & * have the same precedence, hence the expression is evaluated left to right. So operator occurrence matters.

Since both 4 & 5 are evaluated as integers the division yields zero and that is the reason you've got zero as the result. To force compiler to treat 4/5 as a float, you could explicitly mention at least one value as float:

res = 4/5.0F * t;