Why the output of code below is `Gate2019`? String created in function call are stored where? closed

Question

The Following code has output `Gate2019`

#include <stdio.h>

char* fun(){
    char* temp = "Gate2019";
    return temp;
}

int main(void) {
	// your code goes here
	puts(fun());
	return 0;
}

Comments

This is a dangling pointer right?? It will give a sagmentation fault I think!

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Why it is dangling?

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I'm so sorry!! Its a string constant so it will remain in the data section. accessing that memory location won't cause any problem.

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1)will it not problem main is calling fun() and not address of fun()

but while declaring it is pointer to a function..

2) Secondly  also, though temp is a pointer variable, we r returning temp, and not address of temp

Answer 1

Here when function is called then temp pointer array of char is been created and is stored the value 'Gate2019'

and return statement contains the starting address of temp

so for string from obtained address till end until it finds null value is printed

Therefore from temp starting address till end value will be printed  so Gate2019 will be printed

Comments

what is the case of getting dangling pointer error??