还有另一种使用数组和指针的方法：
#include <stdio.h>
#include <inttypes.h>
int main(void) {
// Two uint32_t to one uint64_t
uint32_t val1[2] = {1000, 90000};
uint64_t *val1_u64_ptr = (uint64_t*)val1; //intermediate pointer cast to avoid Wstrict-aliasing warnings
uint64_t val2 = *val1_u64_ptr;
printf("val2: %" PRIu64 "\n", val2);
// val2: 386547056641000
// back to uint32_t array from uint64_t
uint64_t val3 = 386547056641000ull;
uint32_t *val4 = (uint32_t*)&val3;
printf("val4: %" PRIu32 ", %" PRIu32 "\n", val4[0], val4[1]);
// val4: 1000, 90000
return 0;
}
对我来说，这段代码更容易理解和阅读。您只是在内存中创建一个具有两个32-bit unsigned int 的连续空间，然后将相同的内存空间读取为单个64-bit unsigned int 值，反之亦然。不涉及仅将内存作为不同类型读取的操作。
编辑
忘了提到，如果您已经从某个地方读取了 64-bit array，那么您可以轻松地将所有内容读取为 32-bit array 对：
#include <stdio.h>
#include <inttypes.h>
int main() {
uint64_t array64[] = {
386547056641000ull,
93929935171414ull,
186655006591110ull,
73141496240875ull,
161460097995400ull,
351282298325439ull,
97310615654411ull,
104561732955680ull,
383587691986172ull,
386547056641000ull
};
int n_items = sizeof(array64) / sizeof(array64[0]);
uint32_t* array32 = (uint32_t*)&array64;
for (int ii = 0; ii < n_items * 2; ii += 2) {
printf("[%" PRIu32 ", %" PRIu32 "]\n", array32[ii], array32[ii + 1]);
}
return 0;
}
输出：
[1000, 90000]
[3295375190, 21869]
[22874246, 43459]
[2498157291, 17029]
[3687404168, 37592]
[1218152895, 81789]
[3836596235, 22656]
[754134560, 24345]
[4162780412, 89310]
[1000, 90000]
使用联合结构
如果使用 https://stackoverflow.com/a/2810339/2548351 中的结构联合，则更好、更易读：
#include <stdio.h>
#include <inttypes.h>
typedef union {
int64_t big;
struct {
int32_t x;
int32_t y;
};
} xy_t;
int main() {
// initialize from 64-bit
xy_t value = {386547056641000ull};
printf("[%" PRIu32 ",%" PRIu32 "]\n", value.x, value.y);
// [1000, 90000]
// initialize as two 32-bit
xy_t value2 = {.x = 1000, .y = 90000};
printf("%" PRIu64, value.big);
// 386547056641000
return 0;
}