Why is (int64_t)-1 + (uint32_t)0 signed?
c, integer-promotion
Solution
Why is (int64_t)-1 + (uint32_t)0 signed?
Because `int64_t` conversion rank is greater than `uin32_t` conversion rank. `(uint32_t)0` is converted to `int64_t` in the `+` expression and `int64_t` is the type of the resulting expression.
Problem
Why is `(int64_t)-1 + (uint32_t)0` signed in C? It looks like it's `int64_t`, but my intuition would say `uint64_t`. FYI When I run ``` #include <stdint.h> #include <stdio.h> #define BIT_SIZE(x) (sizeof(x) * 8) #define IS_UNSIGNED(x) ((unsigned)(((x) * 0 - 1) >> (BIT_SIZE(x) - 1)) < 2) #define DUMP(x) dump(#x, IS_UNSIGNED(x), BIT_SIZE(x)) static void dump(const char *x_str, int is_unsigned, int bit_size) { printf("%s is %sint%d_t\n", x_str, "u" + !is_unsigned, bit_size); } int main(int argc, char **argv) { (void)argc; (void)argv; DUMP(42); DUMP(42U); DUMP(42L); DUMP(42UL); DUMP(42LL); DUMP(42ULL); DUMP('x'); DUMP((char)'x'); DUMP(1 + 2U); DUMP(1 << 2U); DUMP((int32_t)-1 + (uint64_t)0); DUMP((int64_t)-1 + (uint32_t)0); return 0; } ``` I get the following output: ``` 42 is int32_t 42U is uint32_t 42L is int32_t 42UL is uint32_t 42LL is int64_t 42ULL is uint64_t 'x' is int32_t (char)'x' is int8_t 1 + 2U is uint32_t 1 << 2U is int32_t (int32_t)-1 + (uint64_t)0 is uint64_t (int64_t)-1 + (uint32_t)0 is int64_t ```