lvalue-to-rvalue conversion of an array in ISO C

Answer 1

ISO C is the C standard. The current one is C99 but C1x is right around the corner. If by rapid, you mean a new standard every decade or so, then yes, it is rapidly evolving :-)

Section 6.5.3.4/3 of ISO C99 states:

When applied to an operand that has type char, unsigned char, or signed char, (or a qualified version thereof) the result is 1.

When applied to an operand that has array type, the result is the total number of bytes in the array.

Answer 2

For arrays, the sizeof returns the total size. Be careful about arrays passed as pointers.

C99 standard:

When applied to an operand that has array type, the result is the total number of bytes in the array

Answer 3

Or just microsoft's C is not ISO C but some other standard C (if there exists any).

Microsoft Visual C still supports C89 [only] whereas other compilers like gcc/clang etc support C99 too which is the current Standard.

C99 [Section 6.5.17/2] says

The left operand of a comma operator is evaluated as a void expression; there is a sequence point after its evaluation. Then the right operand is evaluated; the result has its type and value.⁹⁵

Thus the result of sizeof (0,arr) would be sizeof(char*)[due to the implicit lvalue to rvalue conversion /automatic decay to pointer type] not 100*sizeof(char)

sizeof(arr) would have given 100*sizeof(char) from 6.5.3.4/3

_{95) A comma operator does not yield an lvalue.}

---

decided that this would be another solution to the above problem, which I tried on Microsoft Visual Studio 2008, but regardless of whether it is compiled as C or C++ code sizeof(0, arr) always yields 4.

C++03 [5.18/1] Comma Operator

The type and value of the result are the type and value of the right operand; the result is an lvalue if its right operand is.

So sizeof(0, arr) = sizeof (arr) and which would be equal to 100* sizeof(char) and not = sizeof(char*).

So MSVC++ is giving incorrect result (in case of C++ code).