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  • returning a heap block by reference in c++

    - by basicR
    I was trying to brush up my c++ skills. I got 2 functions: concat_HeapVal() returns the output heap variable by value concat_HeapRef() returns the output heap variable by reference When main() runs it will be on stack,s1 and s2 will be on stack, I pass the value by ref only and in each of the below functions, I create a variable on heap and concat them. When concat_HeapVal() is called it returns me the correct output. When concat_HeapRef() is called it returns me some memory address (wrong output). Why? I use new operator in both the functions. Hence it allocates on heap. So when I return by reference, heap will still be VALID even when my main() stack memory goes out of scope. So it's left to OS to cleanup the memory. Right? string& concat_HeapRef(const string& s1, const string& s2) { string *temp = new string(); temp->append(s1); temp->append(s2); return *temp; } string* concat_HeapVal(const string& s1, const string& s2) { string *temp = new string(); temp->append(s1); temp->append(s2); return temp; } int main() { string s1,s2; string heapOPRef; string *heapOPVal; cout<<"String Conact Experimentations\n"; cout<<"Enter s-1 : "; cin>>s1; cout<<"Enter s-2 : "; cin>>s2; heapOPRef = concat_HeapRef(s1,s2); heapOPVal = concat_HeapVal(s1,s2); cout<<heapOPRef<<" "<<heapOPVal<<" "<<endl; return -9; }

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