[rbranson]

This works, sure, but if you do:

        char test[1024][32];
        printf("%i\n", sizeof(test));

You'll get 1024 * 32 (32,768). It didn't know that it's a 2D array. When you pass a stack-allocated array, it passes by pointer. It works this way because C is portable assembler.

[barrkel]

There is no logical connection between the use of the `sizeof` operator and differentiating between 1D and 2D arrays, so "it didn't know that it's a 2D array" makes no sense.

Secondly, it's not really meaningful to talk about "pass by pointer". Parameters are generally either passed by value, by reference, or by name. C only does pass by value; to emulate pass by reference, you need to pass the address by value, as a pointer - or as an "array". But the only array type you can declare is missing the crucial dimension length aspect.

Finally, saying that C works the way it does because it's portable assembler does not have strong explaining power. Arrays are not usually machine-level concepts; pointers with appropriate declaration syntaxes for static and automatic block allocations could take their place in C. Since C does have a higher-level construct called an array, it's not a big leap to imagine it being better designed.

[nitrogen]

sizeof() is used to determine the actual in-memory size of something for use by memcpy/malloc/etc., not the number of elements. There are commonly-used array size macros that can help with this, such as

  #define ARRAY_SIZE(x) sizeof(x)/sizeof((x)[0])