Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
menu search
person
Welcome To Ask or Share your Answers For Others

Categories

When I run the following code:

int i[] = {1,2,3};
int* pointer = i;
cout << i << endl;

char c[] = {'a','b','c',''};
char* ptr = c;
cout << ptr << endl;

I get this output:

0x28ff1c
abc

Why does the int pointer return the address while the char pointer returns the actual content of the array?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
1.1k views
Welcome To Ask or Share your Answers For Others

1 Answer

This is due to overload of << operator. For char * it interprets it as null terminated C string. For int pointer, you just get the address.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
...