[英]printf() format specifier for int64_t and uint64_t in 32-bit ANSI C
[英]Why is (int64_t)-1 + (uint32_t)0 signed?
为什么(int64_t)-1 + (uint32_t)0
在C中签名? 看起来它是int64_t
,但我的直觉会说uint64_t
。
仅供参考我跑的时候
#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;
}
我得到以下输出:
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
为什么(int64_t)-1 +(uint32_t)0签名?
因为int64_t
转换等级大于uin32_t
转换等级。 (uint32_t)0
在+
表达式中转换为int64_t
, int64_t
是结果表达式的类型。
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