What's the difference between int as a parameter and int in main function?

Asked Oct 27 '17 at 13:24

Active Oct 27 '17 at 17:10

Viewed 64 times

A question in following codes. The result is :

9232312
2

The type of y in fun is 'a pointer to a pointer to int'.Then *y must be a pointer to int. So it should printed an address number(like 9232312 etc.), but why is the result 2?

void fun(int **y){
    printf("%d\n", *(y+1));
}
int main(){
    int a[2][2] = {1,2,3,4};
    printf("%d\n", *(a+1));
    fun((int **)a);
    return 0;
}

It is not the question of pointer or array that I am confused of. I dont understand why the result is different？why the converting from array to pointer occurs error? Apparently your comments and marking my question duplicate are not helping ..

edited Oct 27 '17 at 17:10

Jerry Stratton

3,287
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asked Oct 27 '17 at 13:24

张浥东

7

There is no `y` variable in `main` to print. I don't see how this can compile. – François Andrieux Oct 27 '17 at 13:24
3

arrays != pointers!! – Sourav Ghosh Oct 27 '17 at 13:25
1

There is no `int**` in your `main`, it is as simple as that. – Jens Gustedt Oct 27 '17 at 13:26
2

The reason why you needed an explicit cast to convert `a` to an `int**` is because `a` is not an `int**`. Compiler warnings exist for a reason. – Kevin Oct 27 '17 at 13:27

What's the difference between int** as a parameter and int** in main function?

0 Answers0

What's the difference between int as a parameter and int in main function?