Why is it giving 'hexa' and not 'he'? (the same output for cpp too)

Question

#include <stdio.h>

int main() {
//code
char str1[]="hexa";
char str2[2]=" ";
char *t,*s;
s=str1;
t=str2;
while(*t=*s)
*t++=*s++;
printf("%s\n",str2);
return 0;
}

Answer 1

No.. Aayu

char str1[]="Hexa";
char str2[5]="ui";
char *t,*s;
s=str1;
t=str2;
cout<<*(t+1); //output str2+1 => i

*t=*s; //*t =*s = *str1 => H, srr2=Hi
cout<<*t; //output H

*(t+1)=*(s+1); // s+1 =>str+1 =>'e' to *(t+1), str2=He
cout<<*(t+1) ; //output e

*(t+2)=*(s+2); //*(s+2) =>str+2 =>x into *(t+2), str2=Hex
cout<<(*t+2); //*t = H => *t+2 => 72(ascii value) +2 =74, ( you are not referring to *(t+2))

*(t+3)=*(s+2); //*(s+2) is x. Assigned into *(t+3) = x so str2= Hexx

cout<<"\n"<<str2; //this output Hexx

Answer 2

In c, you can access beyond array length, it won't stop you but warning you only.. And you get garbage values if you not assigned but here your assigning charecter with also \0 at end so it returning assigned value....

Answer 3

Yes. In c/c++.
It only stop you at the time of initialization like
char str1[2] = " hexa" ; is error...
char str1[3] ="he"; is correct..

but after it won't stop you access beyond initial length but you may not get desired accurate values if you do so..

Answer 4

~ swim ~ yes. Wrong calculation.. Thanks.. Edited..
But we get correct result because null will added automatically...

Answer 5

Its returning the same for cpp too.

Answer 6

#include <iostream>
using namespace std;

int main() {
//code
char str1[]="Hexa";
char str2[5]="ui";
char *t,*s;
s=str1;
t=str2;
cout<<*(t+1);
*t=*s;
cout<<*t;
*(t+1)=*(s+1);
cout<<*(t+1);
*(t+2)=*(s+2);
cout<<(*t+2);
*(t+3)=*(s+2);

cout<<"\n"<<str2;

return 0;
}

output
iHe74
Hexx

is 74 taken as a garbage value?

Answer 7

Jayakrishna🇮🇳

Okay, got it. Thanks a lot 😊.