const and typedef of arrays in C

First, you are mistaken, the function prototypes

int doSomething(table_t t);int doSomething(int* t);

are exactly equivalent. For function parameters, the first array dimension is always rewritten as a pointer. So there is no guarantee for the size of the array that is received.

const-qualification on arrays always applies to the base type of the array, so the two declarations

const table_t a;int const a[N];

are equivalent, and for functions parameters we have

int doSomething(const table_t t);int doSomething(int const* t);

The content of the table will be constant. Easily checked with this code.

#include<stdio.h>typedef int table_t[3];void doSomething(const table_t t){    t++;    //No error, it's a non-const pointer.    t[1]=3; //Error, it's a pointer to const.}int main(){    table_t t={1,2,3};    printf("%d %d %d %ld",t[0],t[1],t[2],sizeof(t));    t[1]=5;    doSomething(t);    return 0;}

Array types and pointer types are not 100% equivalent, even in this context where you do ultimately get a pointer type for the function parameter. Your mistake is in assuming that const would have acted the same way if it were a pointer type.

To expand on ARBY's example:

typedef int table_t[3];typedef int *pointer_t;void doSomething(const table_t t){    t++;    //No error, it's a non-const pointer.    t[1]=3; //Error, it's a pointer to const.}void doSomethingElse(const pointer_t t){    t++;    //Error, it's a const pointer.    t[1]=3; //No error, it's pointer to plain int}

It does act similarly to const int *, but const pointer_t is instead equivalent to int * const.

(Also, disclaimer, user-defined names ending with _t are not allowed by POSIX, they're reserved for future expansion)