Why this code is printing string rather than printing address?

Question

Here in this code, cout<<q<<endl; is returning string "mani"?. q contain the address of first character 'm' so it should print address not string.please explain.

int main(){
    char *q;
    char b[5]={'m','a','n','i'};
    q=&b[0];
    cout<<b<<endl;
    cout<<q<<endl;

Bathsheba · Answer 1 · 2018-01-04T11:37:51.320

3

std::cout has a special overload for const char*, which outputs the memory as an array of chars starting at the pointer passed up to the next NUL terminator (it's your job to make sure that the appropriate memory is available for that).

If you want to switch off this behaviour and output the pointer address then use a cast:

std::cout << (const void*)b << endl;

edited Jan 04 '18 at 11:37

answered Jan 04 '18 at 11:36

Bathsheba

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I suppose a static cast will be more C++ like. – Stargateur Jan 04 '18 at 11:37
@Stargateur: I prefer my way as it sorts out any `const_cast` too. Every C++ programmer ought to know how the C-style cast behaves in C++. – Bathsheba Jan 04 '18 at 11:38
2

IIRC a static cast may be used to add a const qualifier. That's deemed "safe", supposedly. But I wouldn't wager a beer on it. – StoryTeller - Unslander Monica Jan 04 '18 at 11:41

Why this code is printing string rather than printing address?

1 Answers1