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Comment by kbp | Hacker News Reader

> I always thought [] was just syntactic sugar over a regular pointer, but seems like I was wrong.

For another example showing their differences, if a.c has:

    int n[] = { 1 };

And b.c has:

    extern int *n;
    printf("%d\n", *n);

Then running b.c will segfault, whereas if it said extern int n[], it would work as expected (hint: extern int n and printf("%d\n", n) would also work). Arrays degrade to pointers at the drop of a hat, but they're distinct from them.

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> Then running b.c will segfault

why would it not ? On one side you declare an int array of size 1, on the other side you declare a pointer. Think about what's going on with the bytes :

    int n[] = { 1 };

will look exactly the same in memory than

    int n = 1;

which is of course not compatible with

    int* n = 1;

Arrays by themselves have no indirection in C.

=> https://godbolt.org/g/xwT7uW

Right, that's what I was explaining. A lot of people get tripped up by this because arr is equivalent to &arr, int arr[] is equivalent to int *arr as a formal parameter, etc, so many people go around with the idea that arrays are just pointers in C, like in the comment I quoted; it's an easy thing to get confused by. I was giving an example to demonstrate how they differ.

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