Why do I have an error in the pointer in C?

Question

I am starting to learn pointers in C.

Why do I have an error in line 8 at &i?

This is the source:

char * func(char *d, char *str) 
{
    return d;
}

int main(int argc, char *argv[])
{
    char *i = NULL;
    func(&i, "aaa"); // line 8. here I have the error (in "&i")
}

score 3 · Answer 1 · answered Mar 06 '14 at 06:55

3

You are passing char** and the function expects a char*. You need to pass i without the address of & operator, because this way you are taking the address of the pointer.

Just pass func(i, "aaa");

answered Mar 06 '14 at 06:55

concept3d

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score 1 · Accepted Answer · edited May 23 '17 at 10:26

1

The type of &i is not char * it is char * *.

You should go through this How do pointer to pointers work in C?

edited May 23 '17 at 10:26

Community

1
1

answered Mar 06 '14 at 06:55

Sadique

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What is the difference between them? – user2986392 Mar 06 '14 at 06:57
One is a pointer to char and another is a pointer to pointer to char. – Sadique Mar 06 '14 at 06:58

score 1 · Answer 3 · answered Mar 06 '14 at 06:56

When you write &i, you are passing the memory address of i, but since i is already a char*, the type of &i is char** (pointer to pointer to char). You therefore have an extraneous & that is causing a type mismatch. Simply remove the & and pass i to the function with taking its address:

func(i, aaa); //no need to use & on a variable that is already a pointer

score 0 · Answer 4 · answered Mar 06 '14 at 07:04

As per your function prototype func(char *d, char *str) first parameter accepts pointer of char type and in char *i; i is a pointer of char type so pass it directly to function.

If you want to pass &i to the function you need to have pointer to pointer variable that is char **d.

Why do I have an error in the pointer in C?

4 Answers4