Question

This is a newbie C++ pointer question. I have no clue on why this happen…

I found that if i write this code. It is totally valid.

Code1

int *j;    //create an pointer j
*j = 50;   //assign 50 to the *j, meaning value pointed by address 50 is xxx

However, when I wanna try to make it more simple. The compiler give me this error message.

Code2

int *j = 50; //i guess this should be the same with Code1...

Compile error

error: invalid conversion from ‘int*’ to ‘int’

So why would be like that?

Answer 1

There’s a bit of ambiguity in the C syntax there.

int *j = 50;

is equivalent to

int *j;
j = 50;

Actually, what you’re doing in the first snippet of code is dangerous because you haven’t allocated any memory for j before assigning a value to it yet. You’ll need to make memory for it like so:

in C:

int *j = (int*) malloc(sizeof(int));
*j = 50;

in C++

int *j = new int;
*j = 50;

Or point the pointer at some other block of memory that’s already valid:

int k;
int *j = &k;
*j = 50;
printf("%d", k); // 50

Edit: It’s worth pointing out that the ambiguity has to do with the ‘*’ symbol. In the declaration, int *j; the *j means “a pointer named j”. But when you’re using it in the second line “*j = 50”, * becomes the dereference symbol that means “the value at address j”.