Anomaly in printf() function

Question

I am seeing anomalous behavior in print function while dealing with int and float numbers.

float y = 9/5;
printf("%f", y);

printf("%f", 9/5);

First print statement outputs 1.000000 which is acceptable while other outputs 0.000000. Why outputs are different in both cases?

Since both operands of `9/5` are integers, the result of the division is also an integer (1). The `%f` conversion specifier expects a `float` or `double` operand. — John Bode, Jul 24 '19 at 06:07
The duplicate is the first google search result for `printf("%f", 9/5);` — Antti Haapala -- Слава Україні, Jul 24 '19 at 07:41

Jabberwocky · Answer 1 · 2019-07-24T06:17:46.993

8

9/5 is an integer and it's value is 1.

printf("%f", 9/5); is undefined behaviour because %f expects a double but you provide an int.

Try printf("%f", 9.0/5); and the output will be what you expect.

More generally spoken: if the format specifiers of printf don't match the arguments, the behaviour is undefined; in most cases you get unexpected output.

edited Jul 24 '19 at 06:17

answered Jul 24 '19 at 06:05

Jabberwocky

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lol, the question was a value-exact duplicate :D – Antti Haapala -- Слава Україні Jul 24 '19 at 07:36

Anomaly in printf() function

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