CODEWITHSUNDEEP
cprintf
Jesús Ros
Jesús Ros

Reputation: 480

Weird output in printf()

When I run the following code

    #include <stdio.h>
    #include <stdlib.h>

    int main(){

     printf("2/10=%lf 2./10=%lf\n", 2/10, 2./10);
     printf("2/10=%lf\n",2/10);
     printf("2/10=%d\n",2/10);
     printf("2./10=%lf\n",2./10);
     return 0;
    }

I expected it to return

    2/10=0.000000 2./10=0.200000
    2/10=0.000000
    2/10=0
    2./10=0.200000

and instead I get

    2/10=0.200000 2./10=0.000000
    2/10=0.200000
    2/10=0
    2./10=0.200000

after thinking a bit I can understand that in the first line the 2/10 could get interpreted as a float division instead of an integer one because I have put the %lf specifier in the printf() (is that really what happens?) but what I cannot explain is why the following 2./10 prints 0.000000 instead of 0.200000 as it does in the 4th line.

Anyone can explain it to me?

EDIT:

If I change the code slightly I get

    #include <stdio.h>
    #include <stdlib.h>

    int main(){
      printf("2/10=%lf\n",2/10);
      printf("2/10=%lf 2./10=%lf\n", 2/10, 2./10);
      printf("2/10=%d\n",2/10);
      printf("2./10=%lf\n",2./10);
      return 0;
    }

    2/10=0.000000
    2/10=0.200000 2./10=0.000000
    2/10=0
    2./10=0.200000

which seems to support the idea that the 0.2 in the second printf() statement actually corresponds to the calculation of 2./10 and the 0.0 in the same printf() statement to some undefined behaviour...

Upvotes: 2

Views: 168

Answers (1)

haccks
haccks

Reputation: 106102

2/10 (integer division) is producing 0 which is an int type. printing a data with wrong conversion specificationr invokes undefined behavior.

C11: 7.21.6 Formatted input/output functions:

If a conversion specification is invalid, the behavior is undefined.282) If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.

Upvotes: 4

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