@programminginc - страница 2664

Patrick

I've had this problem before

Patrick

without getting extremely complicated you will have to use a switch/case or if statements with each operator written out

Patrick

I find the cleanest way is to use a macro

Patrick

something like this: #DEFINE DoMath(OP) \ a OP b; use as: DoMath(<)

Ishikawa

if there no way to do it without if/else or switch/case?

Patrick

if there no way to do it without if/else or switch/case?

Unfortunately it will always be needed in some way because think: translating those ASCII characters < > <= >= into machine code will need some sort of translation no matter what you do.

Ishikawa

okay ty very much pal

Patrick

if there no way to do it without if/else or switch/case?

Patrick

if there no way to do it without if/else or switch/case?

Patrick

not a great illustration, but that's how I've done it in the past :)

Anonymous

double n1,n2; int cnt = 2; while(cnt == scanf("%lf%lf",&n1,&n2)){ // expression }

Yes but when i entered char or smth different ,there is still infinite loop again ,i am controlling this because i dont wanna infinite loop

Shaurya

Yes but when i entered char or smth different ,there is still infinite loop again ,i am controlling this because i dont wanna infinite loop

can you clear your input format ?

Anonymous

can you clear your input format ?

double n1,n2; int cnt = 2; while(cnt == scanf("%lf%lf",&n1,&n2)){ if(cnt==2){ ... } else break: }

Shaurya

double n1,n2; int cnt = 2; while(cnt == scanf("%lf%lf",&n1,&n2)){ if(cnt==2){ ... } else break: }

no it's not clear

Anonymous

no it's not clear

If the user does not enter 2 double, I want it to exit the loop. If he does, I want it to continue. I cannot share the other lines. There are too many lines.

Shaurya

If the user does not enter 2 double, I want it to exit the loop. If he does, I want it to continue. I cannot share the other lines. There are too many lines.

this means there are unknown number of inputs

Shaurya

?

Anonymous

yes the end will happen if anything but double enters

Shaurya

yes the end will happen if anything but double enters

Form the unknown number of inputs you are considering 1. size of input will be only two or, 2. form the inputs , double will be only two ?

Anonymous

Size of input will be only two and ,form the inputs has to be double and will be only two

Ishikawa

not a great illustration, but that's how I've done it in the past :)

helps a lot. TY

ברני

Is there a good site of quiz for practice c?... I'm always getting massive question and when I finish, new tesk of another thing.. and didn't get a chance to practice the first tesk...

Shaurya

Size of input will be only two and ,form the inputs has to be double and will be only two

then my code is working well

Shaurya

In you want to handle integer and double separately then #include<bits/stdc++.h> using namespace std; int main(){ string num1, num2; double n1, n2; int cnt,f; do{ cnt = 2; cin>>num1 >> num2; f = 0; for(auto n : num1){ if(n == '.') { f++; continue; } else if(n <'0' && n >'9'){ cnt--; break; } } if(f != 1) break; for(auto n : num2){ if(n == '.') { f++; continue; } else if(n <'0' && n >'9'){ cnt--; break; } } if(f != 2) break; if(cnt == 2){ n1 = stod(num1); n2 = stod(num2); } }while(cnt == 2); return 0; }

@Samaritan_7 You may try this one also

Anonymous

Thanks i just want to handled char with double

Shaurya

Thanks i just want to handled char with double

Welcome 😊

Shaurya

thank you, the way of calculating is right, but I still dont understand why the code is not working

I reviewed your code 1. In count_perimeter() function check the condition of for loop and bracket 2. check the area calculation approach

Gulshan

Do anyone uses linux ? Who can help me, installing a c++ library

Anonymous

Do anyone uses linux ? Who can help me, installing a c++ library

nometa.xyz

Anonymous

want some books about C++ STL?

Alex

why can`t you just access bit field member by its name?

Alex

you define order by writing fields one by one

Alex

this is called endian https://stackoverflow.com/questions/2100331/c-macro-definition-to-determine-big-endian-or-little-endian-machine

Alex

could you provide code and write two options? can`t understand

Alex

its about endian

Mr.

Please someone help me

Roxifλsz 🇱🇹

/warn that is definitely not cpp

Mr.

Please I don't get you

Linking

Please I don't get you

This is html, not cpp.

Dima

lol

Dima

bruh

V01D

Dude that HTML is really, REALLY malformed

Ammar

If you think that is "Coding C++", then yes, you are crazy. Or just terminally stupid.

Dima

Ammar

You can use htonX/ntohX to uniform the byte order. For example: uint32_t x = htonl(0x01234567); x will always have the same value on any machine "if we see from its bytes and their order". htonl = host to network (long). But the value representation maybe different, depending on the machine endianess. That is the standard function, usually used to communicate over the network, because participant on the network maybe big endian machine or little endian machine, they uniform the byte order when it is transferred over the network (the data byte order always be big endian when the data is on the way). When the data arrives on target machine, they use ntohX to convert the byte order to corresponding machine endianess by using ntohX. uint32_t fix_x = ntohl(x); Where x is the data received from the network. ntohl = network to host (long) fix_x is now valid on the target machine, and its representation will be the same with the sender.

Ammar

htonX and ntohX will be no operation on big endian machine.

Ammar

There's also the (at least theoretical) possibility that your program is running on some unusual hardware that uses a numeric representation that is neither big endian nor little endian. In that case, the htonl() and ntohl() functions for that environment would have been programmed to do the right thing (whatever that might be, as it needs to fullfil the standard network) to convert that machine's internal representation to the standard network representation (which is big endian) and back.

Ammar

https://man7.org/linux/man-pages/man3/byteorder.3.html

Ammar

Although it is related to network. It can be used for non networking activity too.

Anonymous

How to compare two double(a>b or a=b or a<b) number in C,i just did google but did not find any satisfied solution

Ammar

I didn't say that.

Ammar

Yes.

Ammar

The case was inside of a union. He gave those bitfields value from uint32_t foo, and he complained the values of two bitfields can be different (this was caused by endianess, because byte order will affect 8 groupped bits).

Ammar

Correct, %int8_t or char won't be affected by endianess.

Ammar

How to compare two double(a>b or a=b or a<b) number in C,i just did google but did not find any satisfied solution

Can't you just use those comparison operators in C/C++?

Anonymous

At double and float they don't work

Ammar

At double and float they don't work

Could you show the case?

Anonymous

line is y=x+6 point is double a,b; I want to compare this point is right or left side of this line ,i have a code at my pc but yeah if u try ;if double a==b ; u will see it is not working

Ammar

a == b for floating point numbers may be wrong. You should consider the epsilon tolerance in that case.

Anonymous

What about a>b and a<b

Ammar

I think this answer is something you are looking for https://stackoverflow.com/a/32334103/7275114

Anonymous

How to compare this

Anonymous

I think this answer is something you are looking for https://stackoverflow.com/a/32334103/7275114

Thanks but i saw that link before what i want is how to use this epsilon stuff with < smaller > bigger situations

Ammar

Thanks but i saw that link before what i want is how to use this epsilon stuff with < smaller > bigger situations

I think after you verify that your numbers are not equal by the given tolerance. The comparison operator < and > are valid to be used.

Ammar

// is a greater than b? bool is_greater(double a, double b) { if (is_equal(a, b)) { return false; } return a > b; }

Ammar

That looks valid to me.

Ammar

Maybe something wrong with your code.

Ammar

Sorry, I am not accepting private message for technical thing like that. Unless there is a private business between us. Just tell your problem in group. Thanks for your understanding.