#define 5 6

Looks like #define true false :D

I think the warn limit would be 3

For such people as you meh

Looks like #define true false :D

false isnt a thing is it #define true 0 maybe #define calloc malloc

false isnt a thing is it #define true 0 maybe #define calloc malloc

Smarty sneaky one

Smarty sneaky one

Argreed.

Thank you. I did and I understand the rules

Hi

Hello!!

nohello.com

nohello.com

I will remember that 😂

well i think its fine to introduce but.. do it rather in Offtopic group

#ot

Hi

Hello

I have a beginner

Could anyone please help me with something

With c programming

#howtoask

Could anyone please help me with something

Yes

I was trying to calculate two integer numbers but when I scan the first variable as a float it suddenly gives me an unwanted number but if I scan the first number as an integer and the second number as a float it gives me the right answer (I notified the two variables as an integers )

I was trying to calculate two integer numbers but when I scan the first variable as a float it suddenly gives me an unwanted number but if I scan the first number as an integer and the second number as a float it gives me the right answer (I notified the two variables as an integers )

Paste a code.

int x,y; scanf(“%d”,&x); scanf(“%d”,&y); printf(“%d”,x+y);

int x,y; scanf(“%d”,&x); scanf(“%d”,&y); printf(“%d”,x+y);

For float use %f.

int x,y; scanf(“%d”,&x); scanf(“%d”,&y); printf(“%d”,&x+y);

I know

But the problem is I am scanning not printing

So i can write whatever I want after running

I was trying to calculate two integer numbers but when I scan the first variable as a float it suddenly gives me an unwanted number but if I scan the first number as an integer and the second number as a float it gives me the right answer (I notified the two variables as an integers )

int x,y; scanf(“%d”,&x); scanf(“%d”,&y); printf(“%d”,&x+y);

Wouldn't it be &x+&y

Yes i know it was a typing mistake

Can anyone tells me what’s wrong

Okay so when you taking float input how's you taking it

Can anyone tells me what’s wrong

https://stackoverflow.com/questions/37858125/input-double-precision-float-in-c/37859213

int x,y; scanf(“%d”,&x); scanf(“%d”,&y); printf(“%d”,&x+y);

Wrong printf.

If I type 3.5 it suddenly gives me 12

Wouldn't it be &x+&y

Why would it be? What's the purpose of adding the address of local variables?

If I type 3.5 it suddenly gives me 12

Watch the link.

But if I right at first 3 and then 4.5 it gives me a right answer

If I type 3.5 it suddenly gives me 12

Can you please provide a full example including your source code as well as in- and output and expected behavior?

Okay

I though that %d is for digit

What happens when you input 3.3, 1 into this program?

Use floats

Non sense

Use floats

I gave the link for him -_-

I though that %d is for digit

Decimal (d) or integer (i) are for ints

Non sense

https://stackoverflow.com/questions/37858125/input-double-precision-float-in-c/37859213

Are you silly or what

Use floats

Decimal (d) or integer (i) are for ints

Oh thanks

When i type a decimal no at first it gives me a weird number but if i type a decimal no. It gives me the right answer

Decimal (d) or integer (i) are for ints

This is for decimal and decimal also called numeric

When i type a decimal no at first it gives me a weird number but if i type a decimal no. It gives me the right answer

Use floats

Okay i will but what’s wrong with that

Okay i will but what’s wrong with that

Console input is line buffered; when you enter 3.5 into a %d format specifier, only the 3 is consumed, the remaining data remains buffered, so the second scanf() call attempts to convert it according to its specifier, it finds a '.' and aborts the conversion leaving y undefined.

When i type a decimal no at first it gives me a weird number but if i type a decimal no. It gives me the right answer

For every conversion specifier other than n, the longest sequence of input characters which does not exceed any specified field width and which either is exactly what the conversion specifier expects or is a prefix of a sequence it would expect, is what's consumed from the stream. The first character, if any, after this consumed sequence remains unread.

When i type a decimal no at first it gives me a weird number but if i type a decimal no. It gives me the right answer

This happens because the y's . is not problematic since it's never used from the buffer

I am not understanding

Sorry but I am a beginner

I am not understanding

It is not a weird number, if you enter 3 and 2.5 it reads the first integer (3) and tries to read the second one and stops and the . since an integral value shall not contain this character, hence the second integer is 2. So the rsult is 5

Please simplify more, sorry for bothering you guys

But I really want to understand

But I really want to understand

Basically when you give 5.6 to an integer variable via scanf it only takes the first part 5

Got it?

Yes

Yes

But here it didnt give me the ability to scan Y

Because it thought the . was Y

And then to print both of the numbers

Ahaaa okayyy

But i didnt press enter

To scan Y after I typed 5.3

Idk about that

Please guys who knows?

It depends on your ide, you're using online that's y

If you want to undestand what's going on, analyze return value of scanf