[SOLVED] Casting pointer to pointer **p (malloc(sizeof(int*))) to a pointer *p – Stack Overflow


This Content is from Stack Overflow. Question asked by Vaibhav

int main() {
    int *n = malloc(sizeof(int*));//line 1

I am not sure what is happening here.
Does line 1 (in the snippet above) mean n now simply is a pointer to an int?
And we have started using the memory allocated for a pointer as variable itself?
Is the pointer to pointer cast to pointer to int?

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In this line

int *n = malloc(sizeof(int*));//line 1

you allocated memory of the size equal to sizeof( int * ) and its address is assigned to the pointer n having the type int *.

So using this pointer to access the memory it is interpreted as a memory storing an object of the type int


Actually the kind of the argument expression used to specify the size of the allocated memory is not very important. For example you could just write

int *n = malloc( 8 );//line 1

This Question was asked in StackOverflow by Vaibhav and Answered by Vlad from Moscow It is licensed under the terms of CC BY-SA 2.5. - CC BY-SA 3.0. - CC BY-SA 4.0.

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