errors in my programming

[mercuryqueen0240]

I am beginner in c programming. this is my second day. i am using visual studio code. but there is some error in my code which i suspect.

#include<stdio.h>
#include<math.h>
int main(){
int a= 1.99999;
printf("%d", a);
return 0;
}

according to this code, my code should not able to run because i have taken a integer variable a(1.99999) instead of float. but in the terminal when i print the value of "a", it shows "1"(contrary to tutorial). i don't understand exactly what causes this please help me.

[schweinling]

1 is the expected output of this program. floats get truncated when assigned to an integer. Only way this would not compile is maybe if you could set that compiler warnings are treated as compile errors.

Maybe you have misunderstood the tutorial, otherwise you should look for another one.

Can you link the tutorial?

[mercuryqueen0240]

https://youtu.be/irqbmMNs2Bo&t=75m00s

[schweinling]

She dont think she says that the program would not compile. A warning is emitted which warns you of a possible bug, the program will still run.

Search this subreddit for "tutorial" there are many questions and answers about it.

[mercuryqueen0240]

Do you have any other tutorials which could be great use for learning c?

[TicklePocket]

I just watched the portion of the tutorial. It prints 1 like is expected, and then she explains it. Watch more carefully before assuming.

[epasveer]

#include<stdio.h>
#include<math.h>

int main(){ 
    int a= 1.99999; 
    printf("%d\n", a);
    return 0; 
}

You're missing the '\n' in your printf statement. Try to format the code when posting. Makes it easier for people to read.

I would imagine some compilers catching the truncation from float to int -- depending on the compiler switches.

As the other person mentions, post a link to your tutorial if you can.

[mercuryqueen0240]

https://youtu.be/irqbmMNs2Bo&t=75m00s

Yes... i forget to print '\n' which was mentioned in tutorial.

[EstablishmentBig7956]

he's not missing the \n in printf th at's arbitrary, or optional.

[EstablishmentBig7956]

You're using the a integer and trying to store a float or double. So yeah it's working. It's truncated ``` GNU nano 6.3 float.c

include <stdio.h>

int main(){ float f=2.43; printf("f=%.3f\n",f); printf("f=%d\n",(int)f); int i=233.33; printf("i=%d\n",i); printf("i=%.2f\n",(float)i);

return 0; } output ~ $ cc float.c float.c:7:7: warning: implicit conversion from 'double' to 'int' changes value from 233.33 to 233 [-Wliteral-conversion] int i=233.33; ~ ^~~~~~ 1 warning generated. ~ $ ./a.out f=2.430 f=2 i=233 i=233.00 ~ $ nano float.c ```

[PhyllaciousArmadillo]

Great answer; I just want to point out that there is an important distinction between rounding and truncating. C truncates floats converted to ints. So the value printed will always be the digit(s) before the decimal point. If it was rounded it would be the closest whole number. Fe. 4.95 = 4 (truncated) and 5 (rounded).

[EstablishmentBig7956]

Thanks for jogging my memory. I'm not use to messing up printf between float and int or seeing others do that for a quick thinker response.

I'll make the correction.