CODEWITHSUNDEEP
c++stringcharacter
guitar_geek
guitar_geek

Reputation: 498

storage of character pointer in memory

1a) There is this code 

char *p;
p[0]='a';
p[1]='b';
printf("%s",p);

When i run this program on ideone.com compiler: c++ 4.3.2, it displays "RUNTIME ERROR" every single time i run it.

1b). however when i edit this code to 

char *p;
//allocate memory using malloc
p[0]='a';
p[1]='b';
printf("%s",p);

It correctly runs and prints "ab" . shouldn't it require the p[2]='\0' at the end?

2) 

char *p;
p="abc"
printf("%s",p);

this correctly runs and prints "abc" . why does this work without allocation.

Can anyone please explain the rules regarding string storage ?

Upvotes: 0

Views: 56

Answers (3)

quantdev
quantdev

Reputation: 23793

1a) undefined behavior because you dereference a non initialized pointer

1b) undefined behavior since you call printf with %s for a non null terminated string

2) works ok: there is an allocation, it just is a string literal (you can't modify it, it is stored in the read-only portion of the program : you should declare it const char* for that reason)

Note:

In C++, use std::string and std::cout .

Upvotes: 4

glezmen
glezmen

Reputation: 554

1a) here you don't allocate memory, so the p pointer points to a random place, therefore causing segfault when you write that random location

1b) if you allocate memory manually with malloc, it will work correctly. If the allocated memory contains 0s, you don't have to add it manually (but you should, because you can't count on the zero filling)

2) here you assign the p pointer to the string literal "abs", so it will point to it, and the allocation is done by the compiler

Upvotes: 1

Marius Bancila
Marius Bancila

Reputation: 16328

In the first example you declare a pointer to char and then you assign values to undefined locations in memory. Undefined because its what the uninitialize p pointer points to. You need to allocate memory for the sequence (with new[] in C++, not malloc). If you do not put a '\0' at the end, printing will stop at the first 0 encountered in memory.

In the third example you are declaring a pointer to char and initialize its value with the address of the literal string "abc". That is stored in the (read-only) data section in the executable and that gets map to the process address space. So that's a valid pointer and your printing works.

Upvotes: 2

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