@programminginc - страница 1071

Mat

I can reply with Rust, that i like :P

Anonymous

C is still the most used language in OS dev, and that will not change that fast. Yelling about it won't change it.

Anonymous

Sometimes all the C hate and C++ hype + C yelling is just annoying. Both are programming languages with different aims and use cases. There is no reason to yell at one or the other. Programming is more than just forcing one language in every use case.

Anonymous

I also can't understand why anyone want to use C++ for web related stuff. Just a dumb decision which language to choose for solving this problem.

Mat

Sometimes all the C hate and C++ hype + C yelling is just annoying. Both are programming languages with different aims and use cases. There is no reason to yell at one or the other. Programming is more than just forcing one language in every use case.

I totally agree. It's like using scissors for everything

Gabriel

For apps

Gabriel

C++ and swift?

Mat

C++ and swift?

?

Mat

For an app you can use plenty of languages. It depends on what you want to do

Gabriel

For developing apps

VJ

int main() { char arr[] = "geeksforgeeks"; char *ptr = arr; while(*ptr != '\0') ++*ptr++; printf("%s %s", arr, ptr); getchar(); return 0; }

Mat

An app is what you use to do something with the computer or phone

Mat

It's not something defined

VJ

int main() { char arr[] = "geeksforgeeks"; char *ptr = arr; while(*ptr != '\0') ++*ptr++; printf("%s %s", arr, ptr); getchar(); return 0; }

anyon explain me??

Mat

int main() { char arr[] = "geeksforgeeks"; char *ptr = arr; while(*ptr != '\0') ++*ptr++; printf("%s %s", arr, ptr); getchar(); return 0; }

Oh my..

VJ

++*ptr++ how this works in above program

Mat

++*ptr++... probably it is read as ++(*ptr++) = ++((*ptr)++) but I'm not sure

Mat

What a mess, i have to read the output to understand it🤔

Ariana

++*ptr++... probably it is read as ++(*ptr++) = ++((*ptr)++) but I'm not sure

i approve XD

VJ

Output: hffltgpshfflt

Mat

So it increase every letter one time

Ariana

++*ptr++ increments i tby 2 doesnt it?

VJ

So it increase every letter one time

how??

Mat

g -> h, e -> f

Mat

One ++ increases the pointer value, the other one the char the pointer is pointing to

Ariana

++*ptr++ becomes mov rax, QWORD PTR [rbp-8] lea rdx, [rax+1] mov QWORD PTR [rbp-8], rdx movzx edx, BYTE PTR [rax] add edx, 1 mov BYTE PTR [rax], dl in godbolt

Ariana

Ah

Ariana

So it increase every letter one time

yeah

VJ

tq

Mat

One ++ increases the pointer value, the other one the char the pointer is pointing to

So you can move the ptr through the word

Mat

Probably if first increment the letter, after that the pointer

Mat

So *ptr += 1; ptr +=1;

Mat

(++*ptr)++ where the input of the ++ to the right is ptr

VJ

thanks @Occupato for explanation

Mat

++*ptr++ becomes mov rax, QWORD PTR [rbp-8] lea rdx, [rax+1] mov QWORD PTR [rbp-8], rdx movzx edx, BYTE PTR [rax] add edx, 1 mov BYTE PTR [rax], dl in godbolt

I need a bigger screen :/

Sujith

Is any Android app for coding C#?

Anonymous

Is any Android app for coding C#?

This is a c/c++ group, not C#

Mat

Is any Android app for coding C#?

Termux. You can do a lot of things with it

Sujith

Termux. You can do a lot of things with it

Thanks

VJ

#include<stdio.h> int fun(char *str1) { char *str2 = str1; while(*++str1); return (str1-str2); } int main() { char *str = "geeksforgeeks"; printf("%d", fun(str)); //getchar(); return 0; }

VJ

output is 13

VJ

anyone explain while loop in that program

Ariana

?

Ariana

oh

Ariana

++ changes str1

Ariana

while terminates iff *++str1 == '\0'

VJ

so the value of str1 is size of characters in the str2 ??

Mat

#include<stdio.h> int fun(char *str1) { char *str2 = str1; while(*++str1); return (str1-str2); } int main() { char *str = "geeksforgeeks"; printf("%d", fun(str)); //getchar(); return 0; }

Like the programme before, *(++str) You increment the pointer, than you read what's pointing. If it is '\0' it stops

Mat

so the value of str1 is size of characters in the str2 ??

It's pointing to the end of the string. The diference between the start and the end of the string gives you its length

VJ

will /0 ,will be counted as length???

Mat

will /0 ,will be counted as length???

Nope, as false

Mat

You operate with the pointers, not with what they're pointing

VJ

#include<stdio.h> int fun(char *str1) { char *str2 = str1; while(*++str1); return (str1-str2); } int main() { char *str = "geeksforgeeks"; printf("%d", fun(str)); //getchar(); return 0; }

in return statement wat is the value of str2???

Mat

in return statement wat is the value of str2???

The address of the first char of the string

VJ

then how th o/p is 13??

Mat

Length("geeksforgeeks") == 13

Mat

The address pointing to the first g and the address of the last char ('\0') have a difference of 13 positions

VJ

so while loop goest until it reaches 13('\0') in str1 ,str2 points out address of fst char in str1,in return statement str1-str2 (13-address) how is possible??

Mat

Nope

Mat

It's like str1 == (address + 13)

Mat

And str2 == address

Ariana

XD YES

Mat

address == address of the first char of the string

VJ

It's like str1 == (address + 13)

got it tq once again

Mat

got it tq once again

👍

Mat

To hate java you need to care about it

Mat

got it tq once again

Remember that when when you're dealing w/ pointers they works with sizes. Let's take int* p; You'll have p++; That will be evaluated as p = p + sizeof(int);

VJ

in case *p++??

VJ

in case *p++??

will it be same as above??

Mat

Wrong answer😂

VJ

lol

Mat

will it be same as above??

It depends on what is done first (*p)++ or *(p++)