can u elaborate

I mean when I have: int a=4; a++; Then cpu will add two 4 bytes numbers (I'm assuming int to be 4 bytes) i.e 4 and 1... And when I have: int *b =&a; b++; Then cpu will still be adding two 4 bytes numbers(I'm assuming 32 bit architecture): i.e. address of a and 4. For the cpu it doesn't matter what those numbers represent...it just has to perform addition... That's why I'm not able to understand why it's UB if I don't access those invalid addresses.

the whole calculation is done once

Ohk.. if it's so for such systems where pointer arithmetic is handled differently then normal arithmetic then it makes sense

can you show me where they have mentioned intermediate values should also be within range

Can you tell me one thing?

When you get an undefined behavior?

A) when you do calculation

B) when you store address

C) or when you access the location

Not until you access

Going by your point —— int *p = 10000; The above statement will result in an error

But no it will not

Btw is this also UB: int *a =0xab14aa66 //random address

Btw is this also UB: int *a =0xab14aa66 //random address

I m telling this guy the same thing he is not getting convinced

I'm asking

He's giving the standard reference

Get to know the correct answer by some guy who knows and then tell who is being idiotic

Standard has the final say in what is valid and what is UB

Although I agree that it's weird

My point is simple until and unless you calculate the full expression you can't say whether it is undefined or not

I m gonna show it wait for 1 min

#include <stdio.h> int main() { char s[] = "hello"; printf("%s", s + 1000 - 997); return 0; }

Take this program

.LC0: .string "%s" main: push rbp mov rbp, rsp sub rsp, 16 mov DWORD PTR [rbp-6], 1819043176 mov WORD PTR [rbp-2], 111 lea rax, [rbp-6] add rax, 3 mov rsi, rax mov edi, OFFSET FLAT:.LC0 mov eax, 0 call printf mov eax, 0 leave ret

this is the assembly code for it

Go through and tell me

Compiler : x86-64 gcc 10.2

according to you gcc is not according to standards?

and this is not my machine It is compiler explorer

Quiz me

the answer is there in question itself

it is undefined

it is undefined behavior

https://godbolt.org/

atleast in some machine it is should be a undefined behavior to be classified as undefined behavior right?

Can you tell me how printf prints a string?

I have read and i know

Can you tell me how printf prints a string?

Bro you can argue endlessly on this..it might not make sense to you (to me as well)...but if the standard mandates it to be UB..then it is *UB*

no point in arguing more if you find any machine in which that statement is defined behavior then ping me

I am new in C programming Currently on Pointers.. Can anybody please explain how below program output is geeks main( ) { char g[ ] = "geeksforgeeks"; printf("%s", g + g[6] - g[8]); }

Here, in questions four options were there from which i have to mark correct one a) geeks b)orgeeks c)rgeeks d)forgeeks Correct answer was geeks. Undefined behavior was not mentioned..

yes'

we can argue with talented people not with extremely talented

show me in standard where it says intermediate values should in the bounds

again 'intermediate'

"intermediate" or "temporary" values

where it says intermediate?

god knows how you get UB without accessing the memory location and just by adding some number to address !

where it says intermediate?

So do you agree that the following is UB(even if I don't access it): int main(){ int* a = new int[100]; a+=1000; }

again it is not undefined behavior

again it is not undefined behavior

But it's not a case of temporary value

if you say that as an undefined behavior end() iterator will never exist in C++

By Dennis ritchie

thats is for traversing the array

you need and end limiter to know where to stop

fine I will stop now You can report that question in geekforgeeks

and let them answer your query

then you can prepare using the standard without gfg :)

the standard clearly states when you access

the standard clearly states when you access

Bro...please read point 4.3...and let us know what you think it means

fine char s[] = "hello"; char *p = s + 10;

so according to you this program itself is undefined?

the prrogram itself?

so according to you this program itself is undefined?

What do you think after reading point 4.3

4.3 states that if the pointer is far away from the bound it is undefined but when you subtract it if you come back to your bounds which the program has access It is not undefined, you can use it

fine char s[] = "hello"; char *p = s + 10;

Let's come to the subtraction part later... Here you didn't perform subtraction

So do you agree that it's UB

Let us do one thing : post this on stackoverflow and let some great minds answer

Agree?

fine char s[] = "hello"; char *p = s + 10;

I mean do you agree that this is IB

UB

I mean do you agree that this is IB

no not if you don't access

Here you didn't perform subtraction

But you did read 4.3 didn't you

No let us post new one which the same example of geeksforgeeks

Agree?

Madhu has already posted the question... It's same I think

no no other people will answer

no worry I will post