c++ pointing to array of char

Question

#include<iostream>

using namespace std;



int main(){

char s[10] = "abcde"; 

char* first = s - 1;

cout << first << endl;

return 0;
}

when I run this, I get a blank in my console, but when I say *first = s; I get the whole char array printed to my console. My question is, what exactly is first pointing to when I set it to s - 1?

This is UB. The reason you're seeing the whole array is because you forgot to add \0 at the end of the string. — Borgleader
– Borgleader, Commented Jul 2, 2013 at 1:31

Sergey Kalinichenko · Accepted Answer · 2013-07-02 01:31:20Z

4

When you set a pointer to s-1, the pointer points to a character at a position in memory that is one char before your allocated string. Dereferencing such pointer is undefined behavior - the program may print anything, or even crash.

answered Jul 2, 2013 at 1:31

Sergey Kalinichenko

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1 Comment

Sergey Kalinichenko Over a year ago

@user2510809 Correct - undefined behavior is a wrong thing to do. Programs that invoke undefined behavior are considered invalid.

Daniel · Accepted Answer · 2013-07-02 01:31:06Z

0

first would point to a piece of unallocated memory - the behavior is undefined

answered Jul 2, 2013 at 1:31

Daniel

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