## 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.

why result of (double + int) is 0 (C language), 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  Double vision, or diplopia, can result from a range of underlying conditions. Diplopia can affect just one eye or both. A childhood squint, or eye turn, can sometimes recur and cause double vision.

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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• 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.