I am really confused with understanding array pointer in C

Question

#include <stdio.h> int main (void) { char *str = "Hello"; //defining and initializing the str pointer, which is directing to 'H'. printf("%s\n", str); printf("%p\n", str); return 0; }

The Result is:

Hello 0000000000404000

My question is where did 0000000000404000 come from?

no newline in the output? You should do printf(%p\n", (void*)str); instead, %p expects a void pointer. — yano
– yano, Commented Dec 14, 2016 at 23:46
That's just where the string happened to end up in the address space. It's completely arbitrary and not something you should rely on. — user2357112
– user2357112, Commented Dec 14, 2016 at 23:46
You have two printfs. What did you expect the second printf to do? — user3386109
– user3386109, Commented Dec 14, 2016 at 23:47
Please explain in the question what output you expected instead — M.M
– M.M, Commented Dec 14, 2016 at 23:48
It happens to be the address where "Hello" is stored in memory. It must be stored somewhere – so why not there? — Jongware
– Jongware, Commented Dec 14, 2016 at 23:49

Brandon83 · Accepted Answer · 2016-12-15 00:07:26Z

2

The format specifier %p will print the address contained within the char* str variable whereas the %s specifier will print the actual string literal Hello. The address in memory 0x00000000 00404000 is where Hello resides.

answered Dec 15, 2016 at 0:07

Brandon83

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2 Comments

Chris Lee Over a year ago

Thank you!! I finally understood what it is.

Ahmed Masud Over a year ago

@ChrisLee if you understand from an answer you should accept it and vote for it

Saurav Sahu · Accepted Answer · 2016-12-15 03:42:09Z

In Bjarne Stroustrup words:

%s The argument is taken to be a string (character pointer), and characters from the string are printed until a null character or until the number of characters indicated by the precision specification is reached; however, if the precision is 0 or missing, all characters up to a null are printed;

%p The argument is taken to be a pointer. The representation printed is implementation dependent;

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