1

Pretty much what the title says. When I declare a function

void foo(int *x[])

x is considered as a parameter of type int**, what about the second case?

EDIT

The part that I didn't understand was why couldn't I pass a 2D massive using a function with parameter type of int * [], but managed to do it using int ( * )[]. I thought that if the name of an array was converted to a pointer to its first element, then the name of a 2D array would be converted to a pointer of the pointer of its first element, that being said a 2D array is a massive of pointers. And int (*)[] means I am passing a pointer to an integer array. So I'm confused.

Improve this question
4
Related questions
25
Difference between the int * i and int** i
44
Difference Between *(Pointer + Index) and Pointer[]
11
Difference between int* and int[] in C++
Load 7 more related questions Show fewer related questions

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.