If we declare:
int i;
int *ptr1 = &i;
*ptr1=10;
cout << ptr1;
Here ptr1 will give the address. But:
char *ptr2;
ptr2="Priyesh";
cout << ptr2;
Here it will give the content of the character pointer. Why is there such a difference?
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USE gdb mate, for quick learningashish– ashish2015-12-30 06:14:08 +00:00Commented Dec 30, 2015 at 6:14
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1Possible duplicate of cout << with char* argument prints string, not pointer valueBarmar– Barmar2015-12-30 06:15:46 +00:00Commented Dec 30, 2015 at 6:15
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2 Answers
operator << is overloaded specially for char pointers - the assumption is that if you try to print a char pointer, you actually want to print the string it points to.
If you want to print it the same way as any other pointer, cast it to void* first:
char *ptr2;
ptr2="Priyesh";
cout << static_cast<void*>(ptr2);
(or cout << (void*)ptr2;)
answered Dec 30, 2015 at 6:14
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4 Comments
user3514850
What's your mean by operator << is overloaded specially for char pointers?
PaulMcKenzie
Do you know what operator overloading is? If so, then the explanation is clear.
user3514850
I don't know operator overloading. Please explain a bit and give some link from where i can gain more knowledge.
Stack Exchange Broke The Law
@user3514850 Imagine it was a
print function instead of <<... then there would be separate print(const char*) and print(const void*) functions. And if you did print(p) it would call the first one if p was a pointer to char, and the second one if it was any other kind of pointer.Because there's an operator<< overload that specifically takes a const char* argument to print it as a string.
answered Dec 30, 2015 at 6:14
Some programmer dude
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