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

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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?

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".


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.

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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

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The result of expression 5.0 + 2 is of type double, since at least one of the two operands of operator + here is a floating point / double value (so the other one will be converted to double before adding).

If you write printf("%d\n", 5.0 + 2), you will pass a floating point value where the format specifier actually expects an int. This mismatch is undefined behaviour, and the 0 you receive could be something else (another number, a crash, a .... what ever), too.

int num = 5.0 + 2, in contrast, will convert the double-value resulting from 5.0 + 2 back to an integral value (discarding any fractional part). So the value of num will be 7 and will be - since num is an integral type - valid in conjunction with format specifier %d then.

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5.0+2 is typed double.

The compiler warning for

int main() { return _Generic(5.0 + 2, struct foo: 0); }

should tell you as much if

int main() { return _Generic(5.0 + 2, double: 5.0+2); }

compiling without error doesn't.

Matching "%d" with a double in printf results in undefined behavior.

Any result is legal, including your harddrive getting erased (unlikely to happen unless your program already has such functionality somewhere in it; if it does, UB can well result it it being inadvertently invoked).

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The usual arithmetic conversions are implicitly performed to cast their values to a common type. The compiler first performs integer promotion; if the operands still have different types, then they are converted to the type that appears highest in the following hierarchy -

In int num = 5.0 + 2; this code snippet you are adding a float with integer and storing back to integer again. So, c automatically casts the result into integer to store in an integer type variable. So, while printing using %d, it prints fine.

But, in printf("%d\n", 5.0 + 2); this code snippet, the addition is casted into floating point number as float has higher priority over integer, but you are printing it using %d. Here mismatch of format specifier causing the unexpected result.

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Comments
  • Refer to this answer it will help more stackoverflow.com/a/9027033/7865621
  • result of printf("%d\n", 1.1 ); is garbage value result of printf("%d\n", 5.0 ); is 0 What's the difference between the two? Thank you very much for your reply.
  • @PWS - 0 is not less garbage. Undefined is undefined, period.
  • @PWS Not "garbage", just not what you expect it to be. As mentioned, mismatching format and argument type leads to undefined behavior, which theoretically could summon nasal demons or reformat your hard-drive.