why result of (double + int) is 0 (C language)

Question

result of

printf("%d\n", 5.0 + 2);

is 0

but

int num = 5.0 + 2;
printf("%d\n", num);

is 7

What's the difference between the two?

Answer 1

The result of 5.0 + 2 is 7.0 and is of type double.

The "%d" format is to print int.

Mismatching format specification and argument type leads to undefined behavior.

To print a float or double value with printf you should use the format "%f".

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With

int num = 5.0 + 2;

you convert the result of 5.0 + 2 into the int value 7. Then you print the int value using the correct format specifier.

Answer 2

In all expressions, every operand has a type. 5.0 has type double. 2 has type int.

Whenever a double and an integer are used as operands of the same operator, the integer is silently converted to a double before calculation. The result is of type double.

And so you pass double to printf, but you have told it to expect an int, since you used %d. The result is a bug, the outcome is not defined.

But in case of int num = 5.0 + 2;, you first get a result as double, 7.0. Then force a conversion back to int. That code is equivalent to:

int num = (int)((double)5.0 + (double)2);

More details here: Implicit type promotion rules