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最近博主在后台收到一位朋友的咨询，说他最近参加了一场技术面试，有这么一道笔试题：
请使用C语言的宏定义实现一个功能，求得某个整型数M在N字节对齐的时，它的值大小。
说明：
1.M是一个非负整数；
2.N是2的整数倍，值可能是1，2，4，8，16等等。
要求：
1.不得使用除法(/);
2.不能使用函数实现，只能用宏实现；
3.自行设计测试用例，明确得出测试用例执行成功与否。
刚好，今天比较清闲，茶余饭后，顺手撸了一把代码：
#include <stdio.h>
#include <stdlib.h>
/* max test number for aligned */
#define MAX_TEST_NUM 1000
/* default for 8 bytes */
#define DEF_ALIGNED_BYTES 8
/* n = 2/4/6/8/16/... */
#define GET_ALIGNED_2_POWER_NUM(num, n) (((num) + (n) - 1) & (~((n) - 1)))
int main(int argc, const char *argv[])
{
int i = 0;
int max_cnt = MAX_TEST_NUM;
int aligned_bytes = DEF_ALIGNED_BYTES;
int aligned_num;
if (argc > 1) {
aligned_bytes = atoi(argv[1]);
if (aligned_bytes < 0) {
printf("error aligned_bytes input !\r\n");
return -1;
}
}
/* test cases start */
for (i = 0; i < max_cnt; i++) {
aligned_num = GET_ALIGNED_2_POWER_NUM(i, aligned_bytes);
//printf("%-4d ALIGNED %d => %-4d\r\n", i, aligned_bytes, aligned_num);
if (aligned_num % aligned_bytes != 0) {
printf("error aligned_num get: %-4d ALIGNED %d => %-4d\r\n", i, aligned_bytes, aligned_num);
printf("test cases (0 ~ %d) ALIGNED %d [ FAIL ] !\r\n", max_cnt, aligned_bytes);
return -1;
}
}
printf("test cases (0 ~ %d) ALIGNED %d [ OK ] !\r\n", max_cnt, aligned_bytes);
return 0;
}
ubuntu下面使用gcc编译，得到可执行文件：
gcc -o test main.c
跑了下测试用例：
recan@ubuntu:~/llc/aligned_macro_test$ ./test 2
test cases (0 ~ 1000) ALIGNED 2 [ OK ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 4
test cases (0 ~ 1000) ALIGNED 4 [ OK ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 8
test cases (0 ~ 1000) ALIGNED 8 [ OK ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 16
test cases (0 ~ 1000) ALIGNED 16 [ OK ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 32
test cases (0 ~ 1000) ALIGNED 32 [ OK ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 64
test cases (0 ~ 1000) ALIGNED 64 [ OK ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 128
test cases (0 ~ 1000) ALIGNED 128 [ OK ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 256
test cases (0 ~ 1000) ALIGNED 256 [ OK ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 3
error aligned_num get: 1 ALIGNED 3 => 1
test cases (0 ~ 1000) ALIGNED 3 [ FAIL ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 5
error aligned_num get: 1 ALIGNED 5 => 1
test cases (0 ~ 1000) ALIGNED 5 [ FAIL ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 7
error aligned_num get: 1 ALIGNED 7 => 1
test cases (0 ~ 1000) ALIGNED 7 [ FAIL ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 167
error aligned_num get: 1 ALIGNED 167 => 1
test cases (0 ~ 1000) ALIGNED 167 [ FAIL ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$ ./test 79
error aligned_num get: 1 ALIGNED 79 => 1
test cases (0 ~ 1000) ALIGNED 79 [ FAIL ] !
recan@ubuntu:~/llc/aligned_macro_test$
recan@ubuntu:~/llc/aligned_macro_test$
从log上看，基本实现了功能，对于非对齐的数值，都能做出准确的判断。
大家记住这个宏定义吧！
/* n = 2/4/6/8/16/... */
#define GET_ALIGNED_2_POWER_NUM(num, n) (((num) + (n) - 1) & (~((n) - 1)))
题外话，如果把题目中的N改为【任意非负整数】呢？
又该如何改造这个宏定义呢？
下次有时间，我们再试试看！
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