CODEWITHSUNDEEP
cpointersmalloc
Mumfordwiz
Mumfordwiz

Reputation: 1545

changing the pointer in C

While writing my code I came upon a problem

char* a = malloc (sizeof(char));
a = "blabla";
free(a);

gives me an error, because I changed the value of the pointer a to something else.

So, why does this work

char* a = "blabla";
printf("%s", a);

again I'm changing the value of the pointer to blabla, I'm ruining the pointers address. It should be error, shouldn't it?

Upvotes: 4

Views: 164

Answers (4)

maccartm
maccartm

Reputation: 2115

In the second one you are creating a pointer to the string literal "blabla", it doesn't affect the pointer in any way, a new pointer is created. In the first example you change the value of the pointer itself.

Upvotes: 2

Aniket
Aniket

Reputation: 21

The first snippet where your address gets lost where 'blaba' is stored It will give error.In the second one you assinged directly the value to blabla

Upvotes: -1

haccks
haccks

Reputation: 106012

The first snippet causes memory leak. Once the statement a = "blabla"; is executed, the pointer a will no longer point to the memory allocated by malloc and its address gets lost.

Upvotes: 1

David Schwartz
David Schwartz

Reputation: 182753

The first case has a bug because you are passing an address to free other than one you got from malloc. That is an error.

The second case has no such problem, so why shouldn't it work? You aren't "ruining" the pointer's address, you are setting it to a valid address and then using that valid address for a valid purpose.

The second example is no different from:

int i = 1;
printf ("%d", i);

You set its value and used the new legal value in a proper way, so no problem.

Upvotes: 4

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