我可以将 unsigned long int 转换为 uintptr_t 吗？

Question

uintptr_t uint_to_uintptr(unsigned int n) {
    return (uintptr_t)n;
}

uintptr_t ulint_to_uintptr(unsigned long int n) {
    return (uintptr_t)n;
}

uintptr_t ullint_to_uintptr(unsigned long long int n) {
    return (uintptr_t)n;
}

上面 3 种类型转换中哪一种不会导致信息（位）丢失？

c99标准说：

以下类型指定无符号整数类型，其属性是任何指向

void

的有效指针都可以转换为该类型，然后转换回指向

void

的指针，结果将与原始指针进行比较：

uintptr_t

但它没有指定这个

unsigned integer type

的大小是多少。

它是指最大宽度无符号整数类型（

UINTMAX_MAX

）的最大值还是简单的

unsigned int

？

Answer 1

您可以添加断言来检查您的代码是否不会在特定实现中调用 UB。

uintptr_t uint_to_uintptr(unsigned int n) {
    _Static_assert(sizeof(uintptr_t) >= sizeof(unsigned int), "uintptr_t is smaller than unsigned int");
    return (uintptr_t)n;
}

uintptr_t ulint_to_uintptr(unsigned long int n) {
    _Static_assert(sizeof(uintptr_t) >= sizeof(unsigned long int), "uintptr_t is smaller than unsigned long int");
    return (uintptr_t)n;
}

uintptr_t ullint_to_uintptr(unsigned long long int n) {
    _Static_assert(sizeof(uintptr_t) >= sizeof(unsigned long long int), "uintptr_t is smaller than unsigned long long int");
    return (uintptr_t)n;
}