Why do I get, variable is not initialized error?

Question

#include <stdio.h>
#include <conio.h>
void Calculator(); 

void main() 
{
    Calculator(); 
    getch();
} 

void Calculator() 
{
 int n,j;
 char f1;
 double t;
 printf("please enter two numbers");
 scanf("%d%d",&n,&j);
 printf("please enter the syboml you want ( * / + or -)");
 scanf("%c",&f1);
 if( f1 == '+')
    t = n + j;
 if (f1 == '-')
    t = n-j;
 if (f1 == '*')
    t = n*j;
 if (f1 == '/')
    t = n/j;
 printf("%f" ,t);
}

You don't always initialize `t`. Also, `void main()` should be `int main(void)`, and `` is not portable. — Keith Thompson, May 29 '21 at 20:26
You should also use `scanf(" %c", &f1)`. Otherwise you'll set `f1` to the newline character after the two numbers. See https://stackoverflow.com/questions/20306659/the-program-doesnt-stop-on-scanfc-ch-line-why — Barmar, May 29 '21 at 20:27
@TomerW after I entered n and j I got the erorr without even getting the chance to enter f1 — user16071647, May 29 '21 at 20:43
@KeithThompson conio never caused me an eror and why must it be int main(void) — user16071647, May 29 '21 at 20:44
@BenM I get the error even before I enter the symbol. right after I enter n and j — user16071647, May 29 '21 at 20:45
I will trying adding else tho I get the eror before entering the symbol for some reason — user16071647, May 29 '21 at 20:46
When you do division, you should convert one of the input variable to `float`. — Barmar, May 29 '21 at 20:51
@Barmar in an 11/3 division o get 3.000000 instead of 3.666667 — user16071647, May 29 '21 at 21:00
You get that error/warning because you do not init any of your variables. If you disagree please explain for each of your variables where you do init it. — Yunnosch, May 29 '21 at 21:03
@Yunnosch the main erorr was I needed a space before %c so that \n does not be taken as input. — user16071647, May 29 '21 at 21:11
@user16071647 It works for me. Did you do `(float)(n/j)` instead of `(float)n/j`? — Barmar, May 29 '21 at 21:11
But the point is that you have to convert one of the variables to `float` *before* you do the division. Otherwise you do integer division, which returns an integer, and then convert that result to float. — Barmar, May 29 '21 at 21:25
`` is specific to MS-DOS compilers, if I recall correctly. `int main(void)` is the form specified by the C standards, Compilers are permitted to support `void main()`, but they don't have to, and there's no good reason to use it. — Keith Thompson, May 29 '21 at 22:45
@Barmar Kinda late but can one input affect the other? What i mean suppose i am inputing a sentence with spaces and all. Then asking for 2nd input. can my first input like the new line get in the first scanf — user16071647, Jul 12 '21 at 00:35
@user16071647 It depends on the specific formats you're using. Formats that are delimited by whitespace such as `%d` and `%s` will not consume the newline. — Barmar, Jul 12 '21 at 16:07

ozerodb · Answer 1 · 2021-05-29T21:19:15.843

In your final printf, you are using t that has never been initialized, and might hold a garbage value if no-one of those if conditions is met.

Consider initializing t (a simple = 0 does the job) or add an else clause somewhere

Edit:

While I was at it, I also made some changes to make sure the second scanf ignores the trailing /n without using fflush.

Edit 2:

As suggested by HAL9000, assuming that an initialization to 0 would be enough is wrong. I modified the second part of the program to make use of a switch-case and eventually reject an invalid operator.

The final code looks like this

#include <conio.h>
#include <stdio.h>
void Calculator();

int main() {
  Calculator();
  getch();
  return 0;
}

void Calculator() {
  int n, j;
  char f1;
  double t;
  printf("please enter two numbers: ");
  scanf("%d%d", &n, &j);
  printf("please enter the symbol you want ( * / + or -): ");
  scanf(" %c", &f1);
  switch (f1) {
    case '+':
      t = n + j;
      break;
    case '-':
      t = n - j;
      break;
    case '*':
      t = n * j;
      break;
    case '/':
      t = (float)n / j;
      break;
    default:
      printf("Invalid symbol, please use ( * / + or -)\n");
      return;
  }
  printf("%f\n", t);
}

What makes you think that `t=0` is less of a garbage value than the the pseudo-random number that is stored in `t` as default? By initializing `t` with a meaningless value, just to get rid of a warning, you are only hiding a bug, not fixing it. — HAL9000, May 29 '21 at 21:05
@HAL9000 I completely agree with you, I just wanted to point out to the OP the reason of the warning. If this was my code, I would add a check on the char passed, and eventually handle an incorrect char — ozerodb, May 29 '21 at 21:08
@HAL9000 I know I know my bad, I added a little switch-case to handle the operator, this way you either input a valid operator or `t` is never used and the initialization doesn't matter — ozerodb, May 29 '21 at 21:21

score 0 · Answer 2 · edited May 29 '21 at 21:11

In some compilers, you might be getting this t is not initialized error as your code will never ask please enter the symbol you want ( * / + or -) because scanf("%c",&f1); will take input as trailing newline char, so t never gets initialized. I ran your code in GCC compiler on mac but got output as 0.0000 as t will never be initialized in your case.

You can just eat up the trailing char for that you can use getchar(); or you can also put a space in the format string, e.g. scanf(" %c",&f1); to consume the newline character.

    void Calculator() 
{
 int n,j;
 char f1;
 double t;
 printf("please enter two numbers");
 scanf("%d%d",&n,&j);
 printf("please enter the syboml you want ( * / + or -)");
 scanf(" %c",&f1);
 if( f1 == '+')
    t = n + j;
 if (f1 == '-')
    t = n-j;
 if (f1 == '*')
    t = n*j;
 if (f1 == '/')
    t = n/j;
 printf("%f" ,t);
}

`fflush(stdin)` is undefined behavior for standard C, see https://stackoverflow.com/questions/2979209/using-fflushstdin. The portable way to achieve this is to actually read the newline character first. — Nate Eldredge, May 29 '21 at 20:57
Thanks for pointing it out.@NateEldredge @user3386109 this way is mentioned in the question comment itself. But yeah this the better way for sure. — Ganesh Jadhav, May 29 '21 at 21:02

score 0 · Answer 3 · answered May 29 '21 at 21:20

Not every program path leads to the assignment of the t variable. So it can be used in the printf not initialized.

switch(f1)
{
 case '+':
    t = n + j;
    break;
 case '-':
    break;
    t = n-j;
 case '*':
    t = n*j;
    break;
 case '/':
    t = n/j;
    break;
 default:
    t = 0;
    break;
}

Now t will always will be assigned with the value.

Some additional remarks:

Always check the return value of the scanf
Your main definition is invalid. If main does not take any parameters is has to be int main(void)

Why do I get, variable is not initialized error?

3 Answers3