Sizeof(SIMD vector) issue in macros [duplicate] - c

Question

This question already has answers here:
C macros and use of arguments in parentheses
(2 answers)
Closed 2 years ago.
I want to define several macros to calculate the size of a type. The following is an example running normally.
#include <stdio.h>
#define A sizeof(long long) / sizeof(int)
#define B 36 / A
int main(){
printf("%zu %zu\n", A, B); // print out: 2 1
}
While it becomes strange when using SIMD vectors, for example (the definition of A)
#include <x86intrin.h>
#include <stdio.h>
#define A sizeof(__m128i) / sizeof(int)
#define B 36 / A
int main(){
printf("%zu %zu\n", A, B); // print out 4 0
}
What's the issue?

Answer 1

Remember that macros are not variables: they're merely substituted tokens. So B expands to
36 / sizeof(__m128i) / sizeof(int)
Division in C and C++ associates left-to-right so this is equivalent to
(36 / sizeof(__m128i)) / sizeof(int)
If sizeof(__m128i) is 16 and sizeof(int) is 4, then this expression is (36 / 16) / 4. Now integer division in C and C++ truncates, so 36/16 is 2, and 2/4 is 0.
For this reason, you should always parenthesize macros that expand to expressions:
#define A (sizeof(__m128i) / sizeof(int))
#define B (36 / A)
If using C++, a better solution is not to use macros at all; constants in C++ are quite fully featured, and avoid this issue.
constexpr std::size_t A = sizeof(__m128i) / sizeof(int);
constexpr std::size_t B = 36 / A;

Answer 2

With
#define A sizeof(__m128i) / sizeof(int)
#define B 36 / A
then the macro B will expand as 36 / sizeof(__m128i) / sizeof(int) which is evaluated as (36 / sizeof(__m128i)) / sizeof(int).
The result of the first division will be zero, and zero divided with anything is zero. Which is the result you get.
The solution, as almost always when having trouble with macros, is to add explicit parentheses in your macro definitions:
#define A (sizeof(__m128i) / sizeof(int))
#define B (36 / A)