In char **function() , dynamic allocation of char array gives undesired result

Question

This is my C++ code .
According to me , it should give output:
abc
Garbage
abc

But it is giving output:
abc
Garbage
Garbage

#include<bits/stdc++.h>
using namespace std;


char **func()
{
char* PA = new char[10];
PA[0]='a';
PA[1]='b';
PA[2]='c';
PA[3]='\0';

printf("%s\n",PA);
printf("Garbage\n");

char **PPA = &PA;
return PPA;

}

int main()
{
 printf("%s\n",*func());
 return 0;
}

Where am I doing wrong?

After your suggestions, This is working char **result = new char*(); *result = (char *)malloc(sizeof(char)*10000); *(*result+0) = 'a'; *(*result+1) = 'b'; *(*result+2) = 'c'; *(*result+3) = '\0'; — Badman, Oct 03 '16 at 20:45

score 3 · Answer 1 · answered Oct 03 '16 at 14:03

3

char **PPA = &PA;

Retrieves the address of the variable PA itsself, which is an automatic variable and goes out of scope as soon as the function terminates. That means you have undefined behavior here. The C standard doesn't guarantee any consistent behavior, so anything may happen, including what you experienced.

To fix that, you could change the function prototype to char* func() and return PA directly and remove PPA altogether.

answered Oct 03 '16 at 14:03

cadaniluk

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Or since this is supposed to be C++, use std::string – Peter Oct 03 '16 at 14:41
Agree with Peter - very much self-inflicted. C++ is object-oriented, C is pointer-oriented. – MSalters Oct 03 '16 at 15:06

In char **function() , dynamic allocation of char array gives undesired result

1 Answers1